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Unformatted text preview: 5E17(pp 10901099) 1/19/06 2:27 PM Page 1090 CHAPTER 17 The collection of wind velocity vectors
at any given time during a tornado is an
example of a vector ﬁeld. V ector Calculus 5E17(pp 10901099) 1/19/06 2:27 PM Page 1091 In this chapter we study the calculus of vector ﬁelds. (These
are functions that assign vectors to points in space.) In particular we deﬁne line integrals (which can be used to ﬁnd
the work done by a force ﬁeld in moving an object along
a curve). Then we deﬁne surface integrals (which can be
used to ﬁnd the rate of ﬂuid ﬂow across a surface). The connections between these
new types of integrals and the single, double, and triple integrals that we have
already met are given by the higherdimensional versions of the Fundamental Theorem of Calculus: Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem.  17.1 Vector Fields
The vectors in Figure 1 are air velocity vectors that indicate the wind speed and direction
at points 10 m above the surface elevation in the San Francisco Bay area. We see at a
glance from the largest arrows in part (a) that the greatest wind speeds at that time occurred
as the winds entered the bay across the Golden Gate Bridge. Part (b) shows the very different wind pattern at a later date. Associated with every point in the air we can imagine a
wind velocity vector. This is an example of a velocity vector ﬁeld. (a) 12:00 P.M., June 11, 2002 (b) 4:00 P.M., June 30, 2002 FIGURE 1 Velocity vector ﬁelds showing San Francisco Bay wind patterns 1091 5E17(pp 10901099) 1092 ❙❙❙❙ 1/19/06 2:27 PM Page 1092 CHAPTER 17 VECTOR CALCULUS Other examples of velocity vector ﬁelds are illustrated in Figure 2: ocean currents and
ﬂow past an airfoil. Nova Scotia (a) Ocean currents off the coast of Nova Scotia (b) Airflow past an inclined airfoil FIGURE 2 Velocity vector fields Another type of vector ﬁeld, called a force ﬁeld, associates a force vector with each
point in a region. An example is the gravitational force ﬁeld that we will look at in
Example 4.
In general, a vector ﬁeld is a function whose domain is a set of points in 2 (or 3 ) and
whose range is a set of vectors in V2 (or V3 ).
(a plane region). A vector ﬁeld on 2 is a function F that assigns to each point x, y in D a twodimensional vector F x, y .
1 Definition Let D be a set in y
F(x, y)
(x, y)
x 0 The best way to picture a vector ﬁeld is to draw the arrow representing the vector F x, y
starting at the point x, y . Of course, it’s impossible to do this for all points x, y , but we
can gain a reasonable impression of F by doing it for a few representative points in D as
in Figure 3. Since F x, y is a twodimensional vector, we can write it in terms of its component functions P and Q as follows:
F x, y
or, for short, FIGURE 3 2 P x, y i Q x, y j F Pi P x, y , Q x, y Qj Notice that P and Q are scalar functions of two variables and are sometimes called scalar
ﬁelds to distinguish them from vector ﬁelds. Vector ﬁeld on R@ . A vector ﬁeld on 3 is a function F that
assigns to each point x, y, z in E a threedimensional vector F x, y, z .
2 Definition Let E be a subset of z 3 F (x, y, z) 0 A vector ﬁeld F on 3 is pictured in Figure 4. We can express it in terms of its component functions P, Q, and R as (x, y, z)
y F x, y, z P x, y, z i Q x, y, z j R x, y, z k x FIGURE 4 Vector ﬁeld on R# As with the vector functions in Section 14.1, we can deﬁne continuity of vector ﬁelds
and show that F is continuous if and only if its component functions P, Q, and R are
continuous. 5E17(pp 10901099) 1/19/06 2:28 PM Page 1093 S ECTION 17.1 VECTOR FIELDS ❙❙❙❙ 1093 We sometimes identify a point x, y, z with its position vector x
x, y, z and write
F x instead of F x, y, z . Then F becomes a function that assigns a vector F x to a
vector x.
EXAMPLE 1 A vector ﬁeld on 2 is deﬁned by
F x, y yi xj Describe F by sketching some of the vectors F x, y as in Figure 3.
SOLUTION Since F 1, 0
j, we draw the vector j
0, 1 starting at the point 1, 0 in
Figure 5. Since F 0, 1
i, we draw the vector
1, 0 with starting point 0, 1 .
Continuing in this way, we calculate several other representative values of F x, y in the
table and draw the corresponding vectors to represent the vector ﬁeld in Figure 5. y F (2, 2) F(0, 3) F (1, 0) x, y
x 0 F x, y 1, 0
2, 2
3, 0
0, 1
2, 2
0, 3 x, y
1, 0
2, 2
3, 0
0, 1
2, 2
0, 3 0, 1
2, 2
0, 3
1, 0
2, 2
3, 0 F x, y
0, 1
2, 2
0, 3
1, 0
2, 2
3, 0 F IGURE 5 It appears from Figure 5 that each arrow is tangent to a circle with center the origin.
To conﬁrm this, we take the dot product of the position vector x x i y j with the
vector F x
F x, y :
x Fx
xi yj
yi xj F(x, y)=_y i+x j xy yx 0 This shows that F x, y is perpendicular to the position vector x, y and is therefore
tangent to a circle with center the origin and radius x
s x 2 y 2. Notice also that
F x, y y s 2 x2 sx 2 y2 x so the magnitude of the vector F x, y is equal to the radius of the circle.
Some computer algebra systems are capable of plotting vector ﬁelds in two or three
dimensions. They give a better impression of the vector ﬁeld than is possible by hand
because the computer can plot a large number of representative vectors. Figure 6 shows a
computer plot of the vector ﬁeld in Example 1; Figures 7 and 8 show two other vector
ﬁelds. Notice that the computer scales the lengths of the vectors so they are not too long
and yet are proportional to their true lengths.
5 _5 6 5 _6 _5 5 6 _5 _6 5 _5 FIGURE 6 FIGURE 7 FIGURE 8 F(x, y)=k _y, x l F(x, y)=k y, sin x l F(x, y)=k ln(1+¥), ln(1+≈)l 5E17(pp 10901099) ❙❙❙❙ 1094 1/19/06 2:28 PM Page 1094 CHAPTER 17 VECTOR CALCULUS z EXAMPLE 2 Sketch the vector ﬁeld on 3 given by F x, y, z z k. SOLUTION The sketch is shown in Figure 9. Notice that all vectors are vertical and point
upward above the xyplane or downward below it. The magnitude increases with the
distance from the xyplane.
0
y
x FIGURE 9 F(x, y, z)=z k We were able to draw the vector ﬁeld in Example 2 by hand because of its particularly
simple formula. Most threedimensional vector ﬁelds, however, are virtually impossible to
sketch by hand and so we need to resort to a computer algebra system. Examples are
shown in Figures 10, 11, and 12. Notice that the vector ﬁelds in Figures 10 and 11 have similar formulas, but all the vectors in Figure 11 point in the general direction of the negative
yaxis because their ycomponents are all 2. If the vector ﬁeld in Figure 12 represents a
velocity ﬁeld, then a particle would be swept upward and would spiral around the zaxis in
the clockwise direction as viewed from above. 1 1
z 0 z _1 5 0 z3 1
1
_1 0
y 1 1 0 _1
x FIGURE 10
F(x, y, z)=y i+z j+x k
In Visual 17.1 you can rotate the vector
ﬁelds in Figures 10–12 as well as
additional ﬁelds. z 0
x 1 0
y 1 1 0 1
x _1
_1 FIGURE 11
F(x, y, z)=y i2 j+x k y0 0
1 1 x FIGURE 12
y
x
z
F(x, y, z)= i j+ k
z
z
4 EXAMPLE 3 Imagine a ﬂuid ﬂowing steadily along a pipe and let V x, y, z be the velocity vector at a point x, y, z . Then V assigns a vector to each point x, y, z in a
certain domain E (the interior of the pipe) and so V is a vector ﬁeld on 3 called a
velocity ﬁeld. A possible velocity ﬁeld is illustrated in Figure 13. The speed at any given
point is indicated by the length of the arrow.
Velocity ﬁelds also occur in other areas of physics. For instance, the vector ﬁeld in
Example 1 could be used as the velocity ﬁeld describing the counterclockwise rotation
of a wheel. We have seen other examples of velocity ﬁelds in Figures 1 and 2. y FIGURE 13 EXAMPLE 4 Newton’s Law of Gravitation states that the magnitude of the gravitational force between two objects with masses m and M is Velocity ﬁeld in ﬂuid ﬂow F m MG
r2 where r is the distance between the objects and G is the gravitational constant. (This
is an example of an inverse square law.) Let’s assume that the object with mass M is
located at the origin in 3. (For instance, M could be the mass of the Earth and the origin
would be at its center.) Let the position vector of the object with mass m be x
x, y, z .
Then r
x , so r 2
x 2. The gravitational force exerted on this second object acts 5E17(pp 10901099) 1/19/06 2:28 PM Page 1095 S ECTION 17.1 VECTOR FIELDS ❙❙❙❙ 1095 toward the origin, and the unit vector in this direction is
x
x
x, y, z is Therefore, the gravitational force acting on the object at x z x y m MG
x
x3 Fx 3 [Physicists often use the notation r instead of x for the position vector, so you may see
Formula 3 written in the form F
m MG r 3 r.] The function given by Equation 3 is
an example of a vector ﬁeld, called the gravitational ﬁeld, because it associates a vector
[the force F x ] with every point x in space.
Formula 3 is a compact way of writing the gravitational ﬁeld, but we can also write it
in terms of its component functions by using the facts that x x i y j z k and
x
sx 2 y 2 z 2 :
F x, y, z x 2 m MGx
y 2 z2 32 i x 2 m MGy
y 2 z2 32 j x 2 m MGz
y 2 z2 32 k F IGURE 14 Gravitational force ﬁeld The gravitational ﬁeld F is pictured in Figure 14.
EXAMPLE 5 Suppose an electric charge Q is located at the origin. According to
Coulomb’s Law, the electric force F x exerted by this charge on a charge q located at a
point x, y, z with position vector x
x, y, z is Fx 4 qQ
x
x3 where is a constant (that depends on the units used). For like charges, we have qQ 0
and the force is repulsive; for unlike charges, we have qQ 0 and the force is attractive. Notice the similarity between Formulas 3 and 4. Both vector ﬁelds are examples of
force ﬁelds.
Instead of considering the electric force F, physicists often consider the force per unit
charge:
Ex
Then E is a vector ﬁeld on 3 1
Fx
q Q
x
x3 called the electric ﬁeld of Q. Gradient Fields
If f is a scalar function of two variables, recall from Section 15.6 that its gradient ∇ f (or
grad f ) is deﬁned by
f x, y fx x, y i fy x, y j Therefore, ∇ f is really a vector ﬁeld on 2 and is called a gradient vector ﬁeld. Likewise,
if f is a scalar function of three variables, its gradient is a vector ﬁeld on 3 given by
f x, y, z fx x, y, z i fy x, y, z j fz x, y, z k 5E17(pp 10901099) 1096 ❙❙❙❙ 1/19/06 2:29 PM Page 1096 CHAPTER 17 VECTOR CALCULUS x 2 y y 3. Plot the gradient vector
ﬁeld together with a contour map of f. How are they related?
EXAMPLE 6 Find the gradient vector ﬁeld of f x, y 4 SOLUTION The gradient vector ﬁeld is given by f
i
x f x, y
_4 f
j
y x2 2 xy i 3y 2 j 4 Figure 15 shows a contour map of f with the gradient vector ﬁeld. Notice that the gradient vectors are perpendicular to the level curves, as we would expect from Section 15.6.
Notice also that the gradient vectors are long where the level curves are close to each
other and short where they are farther apart. That’s because the length of the gradient
vector is the value of the directional derivative of f and close level curves indicate a
steep graph. _4 FIGURE 15 A vector ﬁeld F is called a conservative vector ﬁeld if it is the gradient of some scalar
function, that is, if there exists a function f such that F ∇ f . In this situation f is called
a potential function for F.
Not all vector ﬁelds are conservative, but such ﬁelds do arise frequently in physics. For
example, the gravitational ﬁeld F in Example 4 is conservative because if we deﬁne
f x, y, z sx 2 m MG
y2 z2 then
f
i
x f x, y, z x 2 f
j
y
m MGx
y2 z2 f
k
z
32 i x m MGy
y2 z2 2 32 j x 2 m MG z
y2 z2 k 32 F x, y, z
In Sections 17.3 and 17.5 we will learn how to tell whether or not a given vector ﬁeld is
conservative.  17.1 Exercises
9. F x, y, z 1–10  Sketch the vector ﬁeld F by drawing a diagram like
Figure 5 or Figure 9. 1. F x, y 1
2 i 3. F x, y yi 1
2 5. F x, y yi
sx 2 2. F x, y j i ■ xj j 4. F x, y x xj
y2 6. F x, y yi
sx 2 ■ 10. F x, y, z yj
■ ■ ■ ■ 11–14 yi
xj
y2 xj ■ ■ j
■ i
■ ■  Match the vector ﬁelds F with the plots labeled I–IV.
Give reasons for your choices. 11. F x, y y, x 12. F x, y 1, sin y 7. F x, y, z j 13. F x, y x 8. F x, y, z zj 14. F x, y y, 1 x 2, x 1 ■ 5E17(pp 10901099) 1/19/06 2:29 PM Page 1097 ❙❙❙❙ S ECTION 17.1 VECTOR FIELDS 3 I Mathematica), use it to plot 5 II y2 F x, y _3 3 _5 CAS 5 IV 21–24  Find the gradient vector ﬁeld of f .
ln x 23. f x, y, z
3 _5 5 ■ ■ 25–26 ■  _3
■ ■ _5
■ ■ ■ ■ ■ ■ ■ ■ ■ Match the vector ﬁelds F on
I–IV. Give reasons for your choices.
 3 z2 y2 ■ ■ ■ 24. f x, y, z
■ ■ x xe x cos y z ■ ■ ■ ■ Find the gradient vector ﬁeld ∇ f of f and sketch it. ■ xy
■ ■ 1
4 26. f x, y 2x
■ ■ ■ ■ x ■ 2 y
■ ■ ■ ■ CAS 15–18 22. f x, y 2y sx 2 25. f x, y
■ 6x2 j r 2 2r x, where x
x, y and r
x . Use a
CAS to plot this vector ﬁeld in various domains until you can
see what is happening. Describe the appearance of the plot and
explain it by ﬁnding the points where F x
0. 21. f x, y _3 3x y 20. Let F x _5 3 III 2xy i Explain the appearance by ﬁnding the set of points x, y such
that F x, y
0. 5 _3 1097 with the plots labeled 27–28  Plot the gradient vector ﬁeld of f together with a contour
map of f . Explain how they are related to each other. 27. f x, y sin x 28. f x, y sin y sin x y 15. F x, y, z i 2j 3k 16. F x, y, z i 2j zk 17. F x, y, z xi yj 3k  Match the functions f with the plots of their gradient vector ﬁelds (labeled I–IV). Give reasons for your choices. 18. F x, y, z xi yj zk 29. f x, y xy 31. f x, y x2 I 4 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 29–32 I II 1 32. f x, y sx 2 II y2 x2 y2 4 y2 z0 _1 30. f x, y 1 z0 _1 _4
_1 y 0 _1 0
1
y _1
1 0x 1 III 4 _4 4 _1
x _4
1 z0 _4 4 III 4 IV z0 _1 _1 _1 0
1
y
■ 1 0 IV 1 C AS ■ ■ ■ 1
■ 0 ■ _1
x _1 ■ ■ y
■ 0 ■ _4 _1
1 0x 1
■ ■ 4 4 ■ _4 19. If you have a CAS that plots vector ﬁelds (the command is ﬁeldplot in Maple and PlotVectorField in _4 ■ ■ ■ _4
■ ■ ■ ■ ■ ■ ■ ■ ■ 5E17(pp 10901099) 1098 ❙❙❙❙ 1/19/06 2:30 PM Page 1098 CHAPTER 17 VECTOR CALCULUS equations to ﬁnd an equation of the ﬂow line that passes
through the point (1, 1).
i x j and then sketch
34. (a) Sketch the vector ﬁeld F x, y
some ﬂow lines. What shape do these ﬂow lines appear to
have?
(b) If parametric equations of the ﬂow lines are x x t ,
y y t , what differential equations do these functions
satisfy? Deduce that dy d x x.
(c) If a particle starts at the origin in the velocity ﬁeld given
by F, ﬁnd an equation of the path it follows. 33. The ﬂow lines (or streamlines) of a vector ﬁeld are the paths followed by a particle whose velocity ﬁeld is the given vector
ﬁeld. Thus, the vectors in a vector ﬁeld are tangent to the ﬂow
lines.
x i y j to draw
(a) Use a sketch of the vector ﬁeld F x, y
some ﬂow lines. From your sketches, can you guess the
equations of the ﬂow lines?
(b) If parametric equations of a ﬂow line are x x t , y y t ,
explain why these functions satisfy the differential equations
dx dt x and d y dt
y. Then solve the differential  17.2 Line Integrals
In this section we deﬁne an integral that is similar to a single integral except that instead
of integrating over an interval a, b , we integrate over a curve C. Such integrals are called
line integrals, although “curve integrals” would be better terminology. They were invented
in the early 19th century to solve problems involving ﬂuid ﬂow, forces, electricity, and
magnetism.
We start with a plane curve C given by the parametric equations
1 y Pi1 Pi
Pn P™ xt y P¡
P¸
x 0 FIGURE 1 t i1 a t b n f x i*, yi* t*
i
a yt x t i y t j, and we assume that C is a
or, equivalently, by the vector equation r t
0. See Section 14.2.] If we
smooth curve. [This means that r is continuous and r t
divide the parameter interval a, b into n subintervals ti 1, ti of equal width and we let
x i x ti and yi y ti , then the corresponding points Pi x i , yi divide C into n subarcs
with lengths s1, s2 , . . . , sn . (See Figure 1.) We choose any point Pi* x i*, yi* in the i th
subarc. (This corresponds to a point t* in ti 1, ti .) Now if f is any function of two varii
ables whose domain includes the curve C, we evaluate f at the point x i*, yi* , multiply by
the length si of the subarc, and form the sum P i*(x *, y *)
i
i C x si i1 ti bt which is similar to a Riemann sum. Then we take the limit of these sums and make the following deﬁnition by analogy with a single integral.
2 Definition If f is deﬁned on a smooth curve C given by Equations 1, then the
line integral of f along C is
n y C f x, y ds f x i*, yi* lim nl si i1 if this limit exists.
In Section 11.2 we found that the length of C is
L y b a dx
dt 2 dy
dt 2 dt 5E17(pp 10901099) 1/19/06 2:30 PM Page 1099 SECTION 17.2 LINE INTEGRALS ❙❙❙❙ 1099 A similar type of argument can be used to show that if f is a continuous function, then the
limit in Deﬁnition 2 always exists and the following formula can be used to evaluate the
line integral: y 3 C y f x, y ds b a 2 dx
dt f x t ,y t dy
dt 2 dt The value of the line integral does not depend on the parametrization of the curve, provided that the curve is traversed exactly once as t increases from a to b.
If s t is the length of C between r a and r t , then
 The arc length function s is
discussed in Section 14.3. ds
dt 2 dx
dt 2 dy
dt So the way to remember Formula 3 is to express everything in terms of the parameter t :
Use the parametric equations to express x and y in terms of t and write ds as
dx
dt ds
z 2 dy
dt 2 dt In the special case where C is the line segment that joins a, 0 to b, 0 , using x as the
parameter, we can write the parametric equations of C as follows: x x, y 0,
a x b. Formula 3 then becomes 0 C f(x, y)
(x, y) y y f x, y ds C y b a f x, 0 dx and so the line integral reduces to an ordinary single integral in this case.
Just as for an ordinary single integral, we can interpret the line integral of a positive
0, xC f x, y ds represents the area of one side of
function as an area. In fact, if f x, y
the “fence” or “curtain” in Figure 2, whose base is C and whose height above the point
x, y is f x, y . x FIGURE 2 EXAMPLE 1 Evaluate x 2 y 2 xC 2 x 2 y ds, where C is the upper half of the unit circle 1. SOLUTION In order to use Formula 3 we ﬁrst need parametric equations to represent C. Recall that the unit circle can be parametrized by means of the equations
x
y y C FIGURE 3 0 1 y sin t and the upper half of the circle is described by the parameter interval 0
Figure 3.) Therefore, Formula 3 gives ≈+¥=1
(y˘0) _1 cos t x 2 x 2 y ds dx
dt y 2 cos 2 t sin t y 2 cos 2 t sin t ssin 2 t y 2 cos 2 t sin t dt 2 0 0 0 2 2
3 dy
dt t
2 dt cos 2 t dt
2t cos 3t
3 0 . (See 5E17(pp 11001109) ❙❙❙❙ 1100 1/19/06 2:32 PM Page 1100 CHAPTER 17 VECTOR CALCULUS y Suppose now that C is a piecewisesmooth curve; that is, C is a union of a ﬁnite number of smooth curves C1, C2 , . . . , Cn , where, as illustrated in Figure 4, the initial point of
Ci 1 is the terminal point of Ci. Then we deﬁne the integral of f along C as the sum of the
integrals of f along each of the smooth pieces of C : C¢ C∞
C£ C™ y C¡ C y f x, y ds y f x, y ds C1 y f x, y ds C2 Cn f x, y ds x 0 xC 2 x ds, where C consists of the arc C1 of the parabola y EXAMPLE 2 Evaluate
FIGURE 4 x 2 from 0, 0 to 1, 1 followed by the vertical line segment C2 from 1, 1 to 1, 2 . A piecewisesmooth curve
y SOLUTION The curve C is shown in Figure 5. C1 is the graph of a function of x, so we can
choose x as the parameter and the equations for C1 become (1, 2) x y x x2 0 x 1 4x 2 321
0 C™ Therefore (1, 1) C¡ y C1 x (0, 0) y 2 x ds 0 y FIGURE 5 C=C¡ 1 1 0 dx
dx 2x 2 1
4 4x 2 dx 2 x s1 2 dy
dx dx
2
3 1 5 s5
6 1 C™ On C2 we choose y as the parameter, so the equations of C2 are
x y and C2 Thus y 2 x ds y C 2 1 y 1 y 2 x ds C1 1 y dx
dy 21 y
2 dy
dy 2 2 x ds y C2 2 y dy
5 s5
6 2 x ds 2 1 2 dy 1 2 2 Any physical interpretation of a line integral xC f x, y ds depends on the physical interpretation of the function f . Suppose that x, y represents the linear density at a point
x, y of a thin wire shaped like a curve C. Then the mass of the part of the wire from Pi 1
to Pi in Figure 1 is approximately x i*, yi* si and so the total mass of the wire is approxx i*, yi* si . By taking more and more points on the curve, we obtain the mass
imately
m of the wire as the limiting value of these approximations:
n m x i*, yi* lim nl si i1 y x, y ds C 2 x 2 y represents the density of a semicircular wire, then the
[For example, if f x, y
integral in Example 1 would represent the mass of the wire.] The center of mass of the
wire with density function is located at the point x, y , where
4 x 1
m y C x x, y ds y 1
m y C y x, y ds Other physical interpretations of line integrals will be discussed later in this chapter. 5E17(pp 11001109) 1/19/06 2:33 PM Page 1101 SECTION 17.2 LINE INTEGRALS ❙❙❙❙ 1101 E XAMPLE 3 A wire takes the shape of the semicircle x 2 y 2 1, y 0, and is thicker
near its base than near the top. Find the center of mass of the wire if the linear density at
any point is proportional to its distance from the line y 1.
SOLUTION As in Example 1 we use the parametrization x
and ﬁnd that ds dt. The linear density is x, y k1 cos t, y sin t, 0 t , y where k is a constant, and so the mass of the wire is y m C k1 [ kt y y ds cos t k1 0 k 0 sin t dt 2 From Equations 4 we have
y 1
m y C 1
2 y
1 y 0 FIGURE 6 0 y 2 C yk 1
1 sin 2 t dt 2 y ds [ 1
2 cos t t 1
4 sin 2 t 0 4 center of
mass 1 k sin t 2 2 By symmetry we see that x
_1 1 y x, y ds 0, so the center of mass is x 0, 4
2 0, 0.38 2 See Figure 6.
Two other line integrals are obtained by replacing si by either x i x i x i 1 or
yi yi yi 1 in Deﬁnition 2. They are called the line integrals of f along C with
respect to x and y:
n 5 y C f x, y dx f x i*, yi* nl xi f x i*, yi* lim yi i1
n 6 y C f x, y dy lim nl i1 When we want to distinguish the original line integral xC f x, y ds from those in Equations 5 and 6, we call it the line integral with respect to arc length.
The following formulas say that line integrals with respect to x and y can also be
evaluated by expressing everything in terms of t : x x t , y y t , dx x t dt,
dy y t dt.
7 y f x, y dx y y f x, y dy y C C b a b a f x t , y t x t dt f x t , y t y t dt 5E17(pp 11001109) 1102 ❙❙❙❙ 1/19/06 2:33 PM Page 1102 CHAPTER 17 VECTOR CALCULUS It frequently happens that line integrals with respect to x and y occur together. When
this happens, it’s customary to abbreviate by writing y y P x, y d x C C y Q x, y d y C P x, y d x Q x, y dy When we are setting up a line integral, sometimes the most difﬁcult thing is to think of
a parametric representation for a curve whose geometric description is given. In particular,
we often need to parametrize a line segment, so it’s useful to remember that a vector representation of the line segment that starts at r0 and ends at r1 is given by
rt 8 1 t r0 t r1 0 t 1 (See Equation 13.5.4.)
y EXAMPLE 4 Evaluate x d y, where (a) C C1 is the line segment from 5, 3
C2 is the arc of the parabola x 4 y 2 from 5, 3 to 0, 2 . to 0, 2 and (b) C
(See Figure 7.) (0, 2) C™ C¡ xC y 2 d x SOLUTION
0 4 x (a) A parametric representation for the line segment is x=4¥ x 5t 5 y 5t 3 0 t 1 (_5, _3) FIGURE 7 (Use Equation 8 with r0
Formula 7 gives y C1 y 2 dx 5, x dy 3 and r1 y 1 0, 2 .) Then dx
3 2 5 dt 5t 0 1 5 y 25t 2 25t 0 25t 3
5
3 25t 2
2 5t 5 dt, d y 5 dt, and 5 5 dt 4 dt
1 5
6 4t
0 (b) Since the parabola is given as a function of y, let’s take y as the parameter and write
C2 as
x 4 y2
yy
3y2
Then dx 2y d y and by Formula 7 we have y C2 y 2 dx x dy y
y 2
3
2
3 y2 2y d y
2y 3 y4
2 y2
y3
3 y 2 dy 4 4 dy
2 40 5
6 4y
3 Notice that we got different answers in parts (a) and (b) of Example 4 even though the
two curves had the same endpoints. Thus, in general, the value of a line integral depends
not just on the endpoints of the curve but also on the path. (But see Section 17.3 for conditions under which the integral is independent of the path.) 5E17(pp 11001109) 1/19/06 2:34 PM Page 1103 SECTION 17.2 LINE INTEGRALS ❙❙❙❙ 1103 Notice also that the answers in Example 4 depend on the direction, or orientation, of the
curve. If C1 denotes the line segment from 0, 2 to 5, 3 , you can verify, using the
parametrization
x 5t that
B A b _C y 2 dx C1 5t 0 x dy t 1 5
6 t y B A y 2 In general, a given parametrization x x t , y y t , a t b, determines an orientation of a curve C, with the positive direction corresponding to increasing values of the
parameter t. (See Figure 8, where the initial point A corresponds to the parameter value a
and the terminal point B corresponds to t b.)
If C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point B to terminal point A in Figure 8), then we have C a y C y f x, y dx C y f x, y dx C y f x, y dy C f x, y dy But if we integrate with respect to arc length, the value of the line integral does not change
when we reverse the orientation of the curve: FIGURE 8 y C y f x, y ds C This is because si is always positive, whereas
the orientation of C. f x, y ds
x i and yi change sign when we reverse Line Integrals in Space
We now suppose that C is a smooth space curve given by the parametric equations
x xt y z yt zt a t b or by a vector equation r t
x t i y t j z t k. If f is a function of three variables
that is continuous on some region containing C, then we deﬁne the line integral of f
along C (with respect to arc length) in a manner similar to that for plane curves:
n y C f x, y, z ds f x i*, yi*, zi* lim nl si i1 We evaluate it using a formula similar to Formula 3:
9 y C f x, y, z ds y b a f x t ,y t ,z t dx
dt 2 dy
dt 2 dz
dt 2 dt Observe that the integrals in both Formulas 3 and 9 can be written in the more compact
vector notation y b a f rt r t dt 5E17(pp 11001109) 1104 ❙❙❙❙ 1/19/06 2:35 PM Page 1104 CHAPTER 17 VECTOR CALCULUS For the special case f x, y, z 1, we get y C y ds b r t dt a L where L is the length of the curve C (see Formula 14.3.3).
Line integrals along C with respect to x, y, and z can also be deﬁned. For example,
n y C f x, y, z dz f x i*, yi*, zi* lim nl y b a zi i1 f x t , y t , z t z t dt Therefore, as with line integrals in the plane, we evaluate integrals of the form y 10 C P x, y, z d x Q x, y, z d y R x, y, z dz by expressing everything x, y, z, dx, d y, dz in terms of the parameter t.
6 E XAMPLE 5 Evaluate x 4
z xC y sin z ds, where C is the circular helix given by the equations sin t, z cos t, y t, 0 2 . (See Figure 9.) t SOLUTION Formula 9 gives
2 C y C 0
_1 y sin z ds y 0 _1
0 y 0 y 2 2 0 x dx
dt sin 2 t ssin 2 t cos 2 t 11 s2 y FIGURE 9 2 sin t sin t 2 0 1
2 1 cos 2 t dt 2 dy
dt 2 dz
dt dt 1 dt
s2
t
2 [ 1
2 sin 2 t 2
0 s2 xC y d x z d y x dz, where C consists of the line segment C1 from
2, 0, 0 to 3, 4, 5 followed by the vertical line segment C2 from 3, 4, 5 to 3, 4, 0 . EXAMPLE 6 Evaluate SOLUTION The curve C is shown in Figure 10. Using Equation 8, we write C1 as z rt (3, 4, 5) C¡
(2, 0, 0)
x FIGURE 10 x y
(3, 4, 0) t 2, 0, 0 2 t t 3, 4, 5 2 t, 4 t, 5t z 0 t or, in parametric form, as C™
0 1 y 4t 5t 1 Thus y C1 y dx z dy x dz y 1 0 y 1 0 4 t dt
10 5t 4 dt
29t dt 10 t 2 t 5 dt
29 t2
2 1 24.5
0 5E17(pp 11001109) 1/19/06 2:36 PM Page 1105 SECTION 17.2 LINE INTEGRALS ❙❙❙❙ 1105 Likewise, C2 can be written in the form
rt
or 1
x Then dx 0 t 3, 4, 5 3 y t 3, 4, 0
z 4 5 3, 4, 5
0 5t 5t
1 t d y, so y z dy y dx C2 y x dz 1 0 3 5 dt 15 Adding the values of these integrals, we obtain y C y dx z dy x dz 24.5 15 9.5 Line Integrals of Vector Fields
Recall from Section 6.4 that the work done by a variable force f x in moving a particle
from a to b along the xaxis is W xab f x d x. Then in Section 13.3 we found that the
work done by a constant force F in moving an object from a point P to another point Q in
l
space is W F D, where D PQ is the displacement vector.
Now suppose that F P i Q j R k is a continuous force ﬁeld on 3, such as the
gravitational ﬁeld of Example 4 in Section 17.1 or the electric force ﬁeld of Example 5 in
Section 17.1. (A force ﬁeld on 2 could be regarded as a special case where R 0 and P
and Q depend only on x and y.) We wish to compute the work done by this force in moving a particle along a smooth curve C.
We divide C into subarcs Pi 1Pi with lengths si by dividing the parameter interval
a, b into subintervals of equal width. (See Figure 1 for the twodimensional case or
Figure 11 for the threedimensional case.) Choose a point Pi* x i*, yi*, zi* on the i th subarc
corresponding to the parameter value t*. If si is small, then as the particle moves from
i
Pi 1 to Pi along the curve, it proceeds approximately in the direction of T t* , the unit tani
gent vector at Pi*. Thus, the work done by the force F in moving the particle from Pi 1 to
Pi is approximately z F(x *, y*, z *)
i
i
i
T(t * )
i
Pi1
0 Pi
P *(x *, y*, z *)
iiii Pn
y F x i*, yi*, zi* x si T t*
i F x i*, yi*, zi* T t*
i si P¸ and the total work done in moving the particle along C is approximately
FIGURE 11 n F x i*, yi*, zi* 11 T x i*, yi*, zi* si i1 where T x, y, z is the unit tangent vector at the point x, y, z on C. Intuitively, we see that
these approximations ought to become better as n becomes larger. Therefore, we deﬁne the
work W done by the force ﬁeld F as the limit of the Riemann sums in (11), namely,
12 W y C F x, y, z T x, y, z ds y C F T ds Equation 12 says that work is the line integral with respect to arc length of the tangential
component of the force. 5E17(pp 11001109) 1106 ❙❙❙❙ 1/19/06 2:36 PM Page 1106 CHAPTER 17 VECTOR CALCULUS If the curve C is given by the vector equation r t
x t i y t j z t k, then
Tt
r t r t , so using Equation 9 we can rewrite Equation 12 in the form
W y rt
rt b Frt a y r t dt b a Frt r t dt This integral is often abbreviated as xC F d r and occurs in other areas of physics as well.
Therefore, we make the following deﬁnition for the line integral of any continuous vector
ﬁeld.
13 Definition Let F be a continuous vector ﬁeld deﬁned on a smooth curve C
given by a vector function r t , a t b. Then the line integral of F along C is y C y F dr b a Frt y r t dt C F T ds When using Definition 13, remember that F r t is just an abbreviation for
F x t , y t , z t , so we evaluate F r t simply by putting x x t , y y t , and z z t
in the expression for F x, y, z . Notice also that we can formally write d r r t dt.
 Figure 12 shows the force ﬁeld and the curve
in Example 7. The work done is negative because
the ﬁeld impedes movement along the curve. ticle along the quartercircle r t
SOLUTION Since x y x2 i EXAMPLE 7 Find the work done by the force ﬁeld F x, y cos t i cos t and y 1 sin t j, 0 x y j in moving a par2. t sin t, we have
cos 2t i Frt
and rt sin t i Frt cos t sin t j
cos t j r t dt Therefore, the work done is y C 0 1 F dr x y 0 2 cos 3t
2
3 FIGURE 12 2 0 y 2 2 cos 2t sin t dt 0 2
3 NOTE
Even though xC F d r xC F T ds and integrals with respect to arc length are
unchanged when orientation is reversed, it is still true that
● y C y F dr C F dr because the unit tangent vector T is replaced by its negative when C is replaced by
E XAMPLE 8 Evaluate xC F d r, where F x, y, z
cubic given by x t y t2 z yz j xy i
t3 0 t C. z x k and C is the twisted 1 5E17(pp 11001109) 1/19/06 2:38 PM Page 1107 S ECTION 17.2 LINE INTEGRALS  Figure 13 shows the twisted cubic C in
Example 8 and some typical vectors acting at
three points on C. (1, 1, 1)
F { r(3/4)} 0.5 0
0
y1 2
2 i y Thus C y F dr t3 k
3t 2 k ti t3 i Frt F { r (1)} t2 j
2t j rt
rt z1 1107 SOLUTION We have 2
1.5 ❙❙❙❙ 1 0 Frt t5 j t4 k r t dt C y F { r(1/ 2)}
1
x FIGURE 13 1 0 0 t 3 6 5t dt t4
4 5t 7
7 1 27
28 0 Finally, we note the connection between line integrals of vector ﬁelds and line integrals
of scalar ﬁelds. Suppose the vector ﬁeld F on 3 is given in component form by the equation F P i Q j R k. We use Deﬁnition 13 to compute its line integral along C : y C y F dr b a y b a y b a Frt
Pi r t dt
Qj Rk xti P x t ,y t ,z t x t ytj z t k dt Q x t ,y t ,z t y t R x t , y t , z t z t dt But this last integral is precisely the line integral in (10). Therefore, we have y C F dr y C P dx For example, the integral
d r where xC F R dz Q dy xC y d x z dy F x, y, z  17.2
1–16 1.
2.
3.
4.
5.  yi where F Pi Qj Rk x dz in Example 6 could be expressed as
zj xk Exercises Evaluate the line integral, where C is the given curve. xC y ds, C : x t , y t, 0 t 2
xC y x ds, C : x t 4, y t 3, 1 t 1
2
xC x y 4 ds, C is the right half of the circle x 2 y 2 16
xC y e x ds, C is the line segment joining (1, 2) to (4, 7)
xC x y ln x dy, 7. xC x y d x 8. xC sin x d x 2 C is the arc of the parabola y x 2 from (1, 1) to (3, 9)
6. xC x e y d x,
C is the arc of the curve x e y from (1, 0) to e, 1 x y d y, C consists of line segments from
0, 0 to 2, 0 and from 2, 0 to 3, 2 x2 9.
10.
11.
12. cos y d y, C consists of the top half of the circle
y 2 1 from 1, 0 to 1, 0 and the line segment from
1, 0 to 2, 3 xC x y 3 ds, C : x 4 sin t, y 4 cos t, z 3t, 0 t
2
2
xC x z ds, C is the line segment from (0, 6, 1) to (4, 1, 5)
xC x e y z ds, C is the line segment from (0, 0, 0) to (1, 2, 3)
xC 2 x 9z ds, C: x t, y t 2, z t 3, 0 t 1 5E17(pp 11001109) 1108 13. ❙❙❙❙ 1/19/06 2:39 PM Page 1108 CHAPTER 17 VECTOR CALCULUS xC x 2 y sz d z, 14. xC z d x 15. t 3, y C: x t 2, 0 t, z t 21. F x, y, z 1 rt xC 16. ■ y d z, x dy t 2, y C: x t 3, z t 2, 0 t 1 22. F x, y, z rt x yz d x 2 x d y x y z d z, C consists of line
segments from 1, 0, 1 to 2, 3, 1 and from 2, 3, 1 to
2, 5, 2 ■ CAS xC x 2 d x y 2 d y z 2 d z, C consists of line segments from
0, 0, 0 to 1, 2, 1 and from 1, 2, 1 to 3, 2, 0
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ t3 i ■ ti
■ sin x i cos y j x z k,
t 2 j t k, 0 t 1
z i y j x k,
sin t j cos t k,
■ ■ ■ 0
■ t
■ ■ ■ ■ ■ 23–24  Use a graph of the vector ﬁeld F and the curve C to guess
whether the line integral of F over C is positive, negative, or zero.
Then evaluate the line integral.
23. F x, y ■ x y i x y j,
C is the arc of the circle x 2
clockwise from (2, 0) to 0, 17. Let F be the vector ﬁeld shown in the ﬁgure. (a) If C1 is the vertical line segment from 3, 3 to 3, 3 ,
determine whether xC F dr is positive, negative, or zero.
(b) If C2 is the counterclockwiseoriented circle with radius 3
and center the origin, determine whether xC F dr is positive, negative, or zero. y2
2 4 traversed counter y
x
i
j,
sx 2 y 2
sx 2 y 2
C is the parabola y 1 x 2 from 1, 2 to (1, 2) 24. F x, y 1 2 ■ y
3 ; 1
_2 _1 0
_1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 25. (a) Evaluate the line integral xC F d r, where 2 _3 ■ 1 2 3x 26. (a) Evaluate the line integral xC F d r, where _2 ; _3 18. The ﬁgure shows a vector ﬁeld F and two curves C1 and C2. Are the line integrals of F over C1 and C2 positive, negative,
or zero? Explain. CAS F x, y, z
x i z j y k and C is given by
rt
2 t i 3 t j t 2 k, 1 t 1.
(b) Illustrate part (a) by using a computer to graph C and
the vectors from the vector ﬁeld corresponding to t
and 1 (as in Figure 13).
2
cos 3t, y sin 3t in the ﬁrst quadrant. 28. Find the exact value of xC F d r, where F x, y, z
x 4e y i ln z j sy 2
segment from 1, 2, 1 to 6, 4, 5 . C¡ 1 27. Find the exact value of xC x 3 y 5 ds, where C is the part of the astroid x
CAS y F x, y
e x 1 i x y j and C is given by
2
rt
t i t 3 j, 0 t 1.
(b) Illustrate part (a) by using a graphing calculator or computer to graph C and the vectors from the vector ﬁeld
corresponding to t 0, 1 s2, and 1 (as in Figure 13). z 2 k and C is the line ln t, y e t,
t 2, use a calculator or CAS to evaluate the line integral
xC x sin y ds correct to three decimal places. 29. If C is the curve with parametric equations x C™ 1 x x2 i xy j
on a particle that moves once around the circle x 2 y 2 4
oriented in the counterclockwise direction.
(b) Use a computer algebra system to graph the force ﬁeld and
circle on the same screen. Use the graph to explain your
answer to part (a). 30. (a) Find the work done by the force ﬁeld F x, y CAS 19–22  Evaluate the line integral xC F d r, where C is given by
the vector function r t . 19. F x, y rt x 2 y 3 i y sx j,
t i t 3 j, 0 t
2 20. F x, y, z rt ti yz i
t2 j xz j
t 3 k, 1 x y k,
0t 31. A thin wire is bent into the shape of a semicircle x 2 y 2 4,
x 0. If the linear density is a constant k, ﬁnd the mass and
center of mass of the wire. 32. Find the mass and center of mass of a thin wire in the shape 2 of a quartercircle x 2 y 2 r 2, x
function is x, y
x y. 0, y 0, if the density 5E17(pp 11001109) 1/19/06 2:40 PM Page 1109 S ECTION 17.2 LINE INTEGRALS 33. (a) Write the formulas similar to Equations 4 for the center of mass x, y, z of a thin wire with density function x, y, z
in the shape of a space curve C.
(b) Find the center of mass of a wire in the shape of the helix
x 2 sin t, y 2 cos t, z 3 t, 0 t 2 , if the density
is a constant k.
34. Find the mass and center of mass of a wire in the shape of the helix x t, y cos t, z sin t, 0 t 2 , if the density at
any point is equal to the square of the distance from the origin.
35. If a wire with linear density x, y lies along a plane curve C,
its moments of inertia about the x and yaxes are deﬁned as
Ix y C y 2 x, y ds Iy y C x 2 x, y ds (b) Is this also true for a force ﬁeld F x
constant and x
x, y ? x 10 cos t, y 10 sin t. The height of the fence at position
x, y is given by the function h x, y
4 0.01 x 2 y 2 , so
the height varies from 3 m to 5 m. Suppose that 1 L of paint
covers 100 m2. Sketch the fence and determine how much paint
you will need if you paint both sides of the fence.
45. An object moves along the curve C shown in the ﬁgure from (1, 2) to (9, 8). The lengths of the vectors in the force ﬁeld F
are measured in newtons by the scales on the axes. Estimate
the work done by F on the object. C x, y, z lies along a space curve
C, its moments of inertia about the x, y, and z axes are
deﬁned as
y2 z2 x, y, z ds Iy y x2 z2 x, y, z ds Iz y k x, where k is a 44. The base of a circular fence with radius 10 m is given by 36. If a wire with linear density y 1109 y
(meters) Find the moments of inertia for the wire in Example 3. Ix ❙❙❙❙ C C C C 1 x2 y2 0 x, y, z ds x
(meters) 1 Find the moments of inertia for the wire in Exercise 33.
37. Find the work done by the force ﬁeld F x, y y 38. Find the work done by the force ﬁeld F x, y x sin y i
x 2 from xi
in moving an object along an arch of the cycloid
rt
t sin t i
1 cos t j, 0 t 2 .
on a particle that moves along the parabola y
1, 1 to 2, 4 . 2j yj 46. Experiments show that a steady current I in a long wire pro duces a magnetic ﬁeld B that is tangent to any circle that lies in
the plane perpendicular to the wire and whose center is the axis
of the wire (as in the ﬁgure). Ampère’s Law relates the electric
current to its magnetic effects and states that y C 39. Find the work done by the force ﬁeld F x, y, z
y z, x z, x y on a particle that moves
along the line segment from 1, 0, 0 to 3, 4, 2 .
40. The force exerted by an electric charge at the origin on a charged particle at a point x, y, z with position vector
r
x, y, z is F r
Kr r 3 where K is a constant. (See
Example 5 in Section 17.1.) Find the work done as the particle
moves along a straight line from 2, 0, 0 to 2, 1, 5 . B that encircles a silo with a radius of 20 ft. If the silo is 90 ft
high and the man makes exactly three complete revolutions,
how much work is done by the man against gravity in climbing
to the top? 0I
2r I 42. Suppose there is a hole in the can of paint in Exercise 41 and 9 lb of paint leaks steadily out of the can during the man’s
ascent. How much work is done?
particle that moves once uniformly around the circle
x 2 y 2 1. 0 I where I is the net current that passes through any surface
bounded by a closed curve C and 0 is a constant called the
permeability of free space. By taking C to be a circle with
radius r , show that the magnitude B
B of the magnetic
ﬁeld at a distance r from the center of the wire is 41. A 160lb man carries a 25lb can of paint up a helical staircase 43. (a) Show that a constant force ﬁeld does zero work on a B dr B 5E17(pp 11101119) 1110 ❙❙❙❙ 1/19/06 2:41 PM Page 1110 CHAPTER 17 VECTOR CALCULUS  17.3 The Fundamental Theorem for Line Integrals
Recall from Section 5.3 that Part 2 of the Fundamental Theorem of Calculus can be written as y 1 b a F x dx Fb Fa where F is continuous on a, b . We also called Equation 1 the Net Change Theorem: The
integral of a rate of change is the net change.
If we think of the gradient vector ∇ f of a function f of two or three variables as a sort
of derivative of f , then the following theorem can be regarded as a version of the Fundamental Theorem for line integrals. t b.
2 Theorem Let C be a smooth curve given by the vector function r t , a
Let f be a differentiable function of two or three variables whose gradient vector
∇ f is continuous on C. Then y f dr C f rb f ra NOTE
Theorem 2 says that we can evaluate the line integral of a conservative vector
ﬁeld (the gradient vector ﬁeld of the potential function f ) simply by knowing the value of
f at the endpoints of C. In fact, Theorem 2 says that the line integral of ∇ f is the net
change in f. If f is a function of two variables and C is a plane curve with initial point
A x 1, y1 and terminal point B x 2 , y2 , as in Figure 1, then Theorem 2 becomes
■ y C f dr f x 2 , y2 f x 1, y1 If f is a function of three variables and C is a space curve joining the point A x 1, y1, z1
to the point B x 2 , y2 , z2 , then we have y C f dr f x 2 , y2 , z2 f x 1, y1, z1 Let’s prove Theorem 2 for this case.
y z B(x™, y™) A(x¡, y¡) C
A(x¡, y¡, z¡) 0 FIGURE 1 C x B(x™, y™, z™) 0
x y 5E17(pp 11101119) 1/19/06 2:41 PM Page 1111 SECTION 17.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ❙❙❙❙ 1111 Proof of Theorem 2 Using Deﬁnition 17.2.13, we have y y f dr C b f rt a y f dx
x dt b a y r t dt
f dz
z dt f dy
y dt dt d
f r t dt
dt b a f rb (by the Chain Rule) f ra The last step follows from the Fundamental Theorem of Calculus (Equation 1).
Although we have proved Theorem 2 for smooth curves, it is also true for piecewisesmooth curves. This can be seen by subdividing C into a ﬁnite number of smooth curves
and adding the resulting integrals.
EXAMPLE 1 Find the work done by the gravitational ﬁeld Fx m MG
x
x3 in moving a particle with mass m from the point 3, 4, 12 to the point 2, 2, 0 along a
piecewisesmooth curve C. (See Example 4 in Section 17.1.)
SOLUTION From Section 17.1 we know that F is a conservative vector ﬁeld and, in fact, F ∇ f , where
f x, y, z m MG
y2 sx 2 z2 Therefore, by Theorem 2, the work done is
W y C F dr f 2, 2, 0
m MG
s2 2 2 2 y f dr C f 3, 4, 12
m MG
4 2 12 2
s3 m MG 2 1
2 s2 1
13 Independence of Path
Suppose C1 and C2 are two piecewisesmooth curves (which are called paths) that have the
same initial point A and terminal point B. We know from Example 4 in Section 17.2 that,
in general, xC F d r xC F d r. But one implication of Theorem 2 is that
1 2 y C1 f dr y f dr C2 whenever ∇ f is continuous. In other words, the line integral of a conservative vector ﬁeld
depends only on the initial point and terminal point of a curve.
In general, if F is a continuous vector ﬁeld with domain D, we say that the line integral
xC F d r is independent of path if xC F d r xC F d r for any two paths C1 and C2 in
1 2 5E17(pp 11101119) 1112 ❙❙❙❙ 1/19/06 2:42 PM Page 1112 CHAPTER 17 VECTOR CALCULUS D that have the same initial and terminal points. With this terminology we can say that line
integrals of conservative vector ﬁelds are independent of path.
A curve is called closed if its terminal point coincides with its initial point, that is,
rb
r a . (See Figure 2.) If xC F d r is independent of path in D and C is any closed
path in D, we can choose any two points A and B on C and regard C as being composed
of the path C1 from A to B followed by the path C2 from B to A. (See Figure 3.) Then
C y FIGURE 2 C A closed curve
C™
B C1 F dr y C2 F dr y C1 F dr y C2 F dr 0 since C1 and C2 have the same initial and terminal points.
Conversely, if it is true that xC F d r 0 whenever C is a closed path in D, then we
demonstrate independence of path as follows. Take any two paths C1 and C2 from A to B
in D and deﬁne C to be the curve consisting of C1 followed by C2. Then
0 A y C C¡ and so xC F d r FIGURE 3 y F dr 1 y F dr xC 2 C1 y F dr C2 y F dr C1 y F dr C2 F dr F d r. Thus, we have proved the following theorem. 3 Theorem xC F d r is independent of path in D if and only if xC F d r
every closed path C in D. 0 for Since we know that the line integral of any conservative vector ﬁeld F is independent
of path, it follows that xC F d r 0 for any closed path. The physical interpretation is that
the work done by a conservative force ﬁeld (such as the gravitational or electric ﬁeld in
Section 17.1) as it moves an object around a closed path is 0.
The following theorem says that the only vector ﬁelds that are independent of path are
conservative. It is stated and proved for plane curves, but there is a similar version for
space curves. We assume that D is open, which means that for every point P in D there is
a disk with center P that lies entirely in D. (So D doesn’t contain any of its boundary
points.) In addition, we assume that D is connected. This means that any two points in D
can be joined by a path that lies in D.
4 Theorem Suppose F is a vector ﬁeld that is continuous on an open connected
region D. If xC F d r is independent of path in D, then F is a conservative vector
ﬁeld on D ; that is, there exists a function f such that ∇ f F. Proof Let A a, b be a ﬁxed point in D. We construct the desired potential function f by deﬁning
y f x, y
C™
(x¡, y) C¡ (x, y) D
(a, b)
0 FIGURE 4 x y x, y
a, b F dr for any point x, y in D. Since xC F d r is independent of path, it does not matter
which path C from a, b to x, y is used to evaluate f x, y . Since D is open, there
exists a disk contained in D with center x, y . Choose any point x 1, y in the disk with
x 1 x and let C consist of any path C1 from a, b to x 1, y followed by the horizontal
line segment C2 from x 1, y to x, y . (See Figure 4.) Then
f x, y y C1 F dr y C2 F dr y x1, y
a, b F dr y C2 F dr 5E17(pp 11101119) 1/19/06 2:43 PM Page 1113 SECTION 17.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ❙❙❙❙ 1113 Notice that the ﬁrst of these integrals does not depend on x, so
f x, y x
If we write F Pi 0 F dr y C2 On C2 , y is constant, so dy
(x, y) C™
C¡ (x, y¡) x D y C2 F dr Q j, then y
y x P dx Q dy 0. Using t as the parameter, where x 1 y
x f x, y C2 P dx C2 y
x Q dy P t, y dt x1 x, we have P x, y by Part 1 of the Fundamental Theorem of Calculus (see Section 5.3). A similar argument, using a vertical line segment (see Figure 5), shows that (a, b)
x 0 x t y y
y f x, y C2 P dx y
y Q dy y y1 Q x, t dt Q x, y FIGURE 5 Thus F Pi f
i
x Qj f
j
y ∇f which says that F is conservative.
The question remains: How is it possible to determine whether or not a vector ﬁeld
F is conservative? Suppose it is known that F P i Q j is conservative, where P and
Q have continuous ﬁrstorder partial derivatives. Then there is a function f such that
F ∇ f , that is,
f
x P and Q f
y Therefore, by Clairaut’s Theorem,
P
y 2 f yx 2 f xy Q
x P x, y i Q x, y j is a conservative vector ﬁeld, where
5 Theorem If F x, y
P and Q have continuous ﬁrstorder partial derivatives on a domain D, then
throughout D we have
simple,
not closed simple,
closed
FIGURE 6 Types of curves not simple,
not closed not simple,
closed P
y Q
x The converse of Theorem 5 is true only for a special type of region. To explain this, we
ﬁrst need the concept of a simple curve, which is a curve that doesn’t intersect itself anywhere between its endpoints. [See Figure 6; r a
r b for a simple closed curve, but
r t1
r t2 when a t1 t2 b.]
In Theorem 4 we needed an open connected region. For the next theorem we need a
stronger condition. A simplyconnected region in the plane is a connected region D such 5E17(pp 11101119) 1114 ❙❙❙❙ 1/19/06 2:43 PM Page 1114 CHAPTER 17 VECTOR CALCULUS simplyconnected region that every simple closed curve in D encloses only points that are in D. Notice from Figure
7 that, intuitively speaking, a simplyconnected region contains no hole and can’t consist
of two separate pieces.
In terms of simplyconnected regions we can now state a partial converse to Theorem 5
that gives a convenient method for verifying that a vector ﬁeld on 2 is conservative. The
proof will be sketched in the next section as a consequence of Green’s Theorem.
P i Q j be a vector ﬁeld on an open simplyconnected
6 Theorem Let F
region D. Suppose that P and Q have continuous ﬁrstorder derivatives and regions that are not simplyconnected P
y FIGURE 7 Q
x throughout D Then F is conservative. EXAMPLE 2 Determine whether or not the vector ﬁeld 10 F x, y
_10 10 x yi x 2j is conservative.
SOLUTION Let P x, y x y and Q x, y C P
y _10 x 2. Then
Q
x 1 1 FIGURE 8 Since P y
 Figures 8 and 9 show the vector ﬁelds in
Examples 2 and 3, respectively. The vectors in
Figure 8 that start on the closed curve C all
appear to point in roughly the same direction as
C. So it looks as if xC F d r 0 and therefore F is not conservative. The calculation in
Example 2 conﬁrms this impression. Some of the
vectors near the curves C1 and C2 in Figure 9
point in approximately the same direction as the
curves, whereas others point in the opposite
direction. So it appears plausible that line integrals around all closed paths are 0. Example 3
shows that F is indeed conservative. C™ _2 EXAMPLE 3 Determine whether or not the vector ﬁeld F x, y 3 x2 2 xy i 3y 2 j is conservative.
SOLUTION Let P x, y 3 2 xy and Q x, y
P
y 2x x2 3y 2. Then
Q
x 2
Also, the domain of F is the entire plane D
, which is open and simplyconnected. Therefore, we can apply Theorem 6 and conclude that F is conservative. 2 C¡ Q x, F is not conservative by Theorem 5. 2 In Example 3, Theorem 6 told us that F is conservative, but it did not tell us how to ﬁnd
the (potential) function f such that F ∇ f . The proof of Theorem 4 gives us a clue as to
how to ﬁnd f . We use “partial integration” as in the following example.
EXAMPLE 4 _2 FIGURE 9 (a) If F x, y
3 2 xy i
x 2 3y 2 j, ﬁnd a function f such that F
(b) Evaluate the line integral xC F d r, where C is the curve given by
rt
e t sin t i e t cos t j, 0 t
. ∇f . 5E17(pp 11101119) 1/19/06 2:44 PM Page 1115 SECTION 17.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ❙❙❙❙ 1115 S OLUTION (a) From Example 3 we know that F is conservative and so there exists a function f
with ∇ f F, that is,
7 fx x, y 3 2 xy 8 fy x, y x2 3y 2 Integrating (7) with respect to x, we obtain
f x, y 9 x2y 3x ty Notice that the constant of integration is a constant with respect to x, that is, a function
of y, which we have called t y . Next we differentiate both sides of (9) with respect to y :
x2 fy x, y 10 ty Comparing (8) and (10), we see that
3y 2 ty
Integrating with respect to y, we have
y3 ty K where K is a constant. Putting this in (9), we have
f x, y x2y 3x y3 K as the desired potential function.
(b) To use Theorem 2 all we have to know are the initial and terminal points of C,
namely, r 0
0, 1 and r
0, e . In the expression for f x, y in part (a), any
value of the constant K will do, so let’s choose K 0. Then we have y C F dr y C f dr e3 f 0, e e3 1 f 0, 1
1 This method is much shorter than the straightforward method for evaluating line integrals that we learned in Section 17.2.
A criterion for determining whether or not a vector ﬁeld F on 3 is conservative is given
in Section 17.5. Meanwhile, the next example shows that the technique for ﬁnding the
potential function is much the same as for vector ﬁelds on 2.
EXAMPLE 5 If F x, y, z ∇f y2 i 2 xy e 3z j 3ye 3z k, ﬁnd a function f such that F. SOLUTION If there is such a function f , then
11 fx x, y, z y2 12 fy x, y, z 2 xy 13 fz x, y, z 3ye 3z e 3z 5E17(pp 11101119) 1116 ❙❙❙❙ 1/19/06 2:45 PM Page 1116 CHAPTER 17 VECTOR CALCULUS Integrating (11) with respect to x, we get
xy2 f x, y, z 14 t y, z where t y, z is a constant with respect to x. Then differentiating (14) with respect to y,
we have
fy x, y, z t y y, z 2 xy and comparison with (12) gives
t y y, z
Thus, t y, z ye 3z e 3z h z and we rewrite (14) as
xy2 f x, y, z ye 3z hz Finally, differentiating with respect to z and comparing with (13), we obtain h z
and, therefore, h z
K , a constant. The desired function is
xy2 f x, y, z
It is easily veriﬁed that ∇ f ye 3z 0 K F. Conservation of Energy
Let’s apply the ideas of this chapter to a continuous force ﬁeld F that moves an object
along a path C given by r t , a t b, where r a
A is the initial point and r b
B
is the terminal point of C. According to Newton’s Second Law of Motion (see Section 14.4), the force F r t at a point on C is related to the acceleration a t
r t by the
equation
Frt mr t So the work done by the force on the object is
W y C y b a y F dr b a mr t Frt r t dt r t dt m
2 y m
2 y m
2 [ rt 2b
a m
2 ( rb 2 b a
b a d
rt
dt
d
rt
dt r t dt
2 (Theorem 14.2.3, Formula 4) dt (Fundamental Theorem of Calculus) ra 2 ) 5E17(pp 11101119) 1/19/06 2:45 PM Page 1117 S ECTION 17.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ❙❙❙❙ 1117 Therefore
W 15 1
2 1
2 2 m vb m va 2 where v r is the velocity.
The quantity 1 m v t 2, that is, half the mass times the square of the speed, is called the
2
kinetic energy of the object. Therefore, we can rewrite Equation 15 as
W 16 KB KA which says that the work done by the force ﬁeld along C is equal to the change in kinetic
energy at the endpoints of C.
Now let’s further assume that F is a conservative force ﬁeld; that is, we can write
F ∇ f . In physics, the potential energy of an object at the point x, y, z is deﬁned as
P x, y, z
f x, y, z , so we have F
∇P. Then by Theorem 2 we have
W y C y F dr
Prb PA P dr C Pra PB Comparing this equation with Equation 16, we see that
PA KA PB KB which says that if an object moves from one point A to another point B under the inﬂuence
of a conservative force ﬁeld, then the sum of its potential energy and its kinetic energy
remains constant. This is called the Law of Conservation of Energy and it is the reason
the vector ﬁeld is called conservative.  17.3 Exercises 1. The ﬁgure shows a curve C and a contour map of a function f y 0 1 2 0 1
6 4 1 whose gradient is continuous. Find xC 3 5 7 2 8 2 9 x f d r. y 40 C 50 60 30 3–10  Determine whether or not F is a conservative vector ﬁeld.
If it is, ﬁnd a function f such that F
f. 20
10 3. F x, y 6x 4. F x, y 3 x x 2. A table of values of a function f with continuous gradient is given. Find xC
x t 2 1, y f d r, where C has parametric equations
t 3 t, 0 t 1. 5. F x, y xe y i 6. F x, y 0 ey i 5y i 5x 4xy i 4y j 4xy y3 j ye x j
xe y j 7. F x, y 2 x cos y 8. F x, y 1 ln x i 9. F x, y ye x x 2 sin y y cos x i 2xy sin y i x2 j
ex x cos y j sin x j 5E17(pp 11101119) 1118 ❙❙❙❙ 2:47 PM Page 1118 CHAPTER 17 VECTOR CALCULUS 10. F x, y
■ 1/19/06 ■ x y cosh x y
■ ■ x 2 cosh x y j sinh x y i ■ ■ ■ ■ ■ ■ y2 x2 i 22. F x, y
■ ■ ■ 2 x y, x 2 and
three curves that start at (1, 2) and end at (3, 2).
(a) Explain why xC F d r has the same value for all three
curves.
(b) What is this common value? 11. The ﬁgure shows the vector ﬁeld F x, y ■ ■ 2 y x j; P 1, 1 , Q 4, ■ ■ ■ ■ ■ 2 ■ ■ ■ ■ 23. Is the vector ﬁeld shown in the ﬁgure conservative? Explain.
y y
3 x 2 1
CAS 0 1 2 24. F x, y x 3 24–25  From a plot of F guess whether it is conservative. Then
determine whether your guess is correct. (a) Find a function f such that F ∇ f and (b) use
part (a) to evaluate xC F d r along the given curve C. x 25. F x, y 12. F x, y 2xy x2 sin y i x cos y j 2y i
x 2j
s1 x 2 y 2 26. Let F 12–18  ■ yi
x 2 y j,
C is the upper semicircle that starts at (0, 1) and ends at (2, 1)
x 3y 4 i
st i 13. F x, y C: r t x 4 y 3 j,
1 t 3 j, 0 y2 14. F x, y 2 i t ■ (a) 1 y C1 C: x 1 C: r t
C: r t
19–20 x2
0 3z 2 k,
t1 ■ ■ ■ ■ ■ xC y d x ■ 29.
■ ■ ■ ■ Show that the line integral is independent of path and
evaluate the integral. 19.
20.
■ xC tan y d x
xC 1 ye
■ C is any path from 1, 0 to 2, x sec y d y,
x ■ dx
■ x e d y,
■ ■ ■ ■ ■ ■ 21–22  Find the work done by the force ﬁeld F in moving an
object from P to Q. 21. F x, y 2y 32 i 3 x sy j; P 1, 1 , Q 2, 4 31.
32.
■ 4 0 (b) y C2 Q
x P
z ■ F dr R
x ■ ■ ■ 1 R
y Q
z x yz dz is not independent of path. x dy x, y x 0, y x, y 1 2 x, y x
■ x
2 ■ y y
2 ■ 30. 0
2 x, y x ■ 0 4
x2 1 or 4
■ ■ y2
■ 9
■ ■ ■ ■ yi xj
.
x2 y2
Q x.
(a) Show that P y
(b) Show that xC F d r is not independent of path.
[Hint: Compute xC F d r and xC F d r, where C1 and C2
are the upper and lower halves of the circle x 2 y 2 1
from 1, 0 to 1, 0 .] Does this contradict Theorem 6? 33. Let F x, y C is any path from 0, 1 to 1, 2
■ ■ 29–32  Determine whether or not the given set is (a) open,
(b) connected, and (c) simplyconnected.  2 ■ 28. Use Exercise 27 to show that the line integral z 1 e z k,
e y i xe y j
t i t 2 j t 3 k, 0 t 1 18. F x, y, z
■ 2x y j
2t 1, F dr P
y y 2 cos z i 2 x y cos z j x y 2 sin z k,
2
t i sin t j t k, 0 t 17. F x, y, z ■ y2 i
1, z 2 xz
t,y t ■ P i Q j R k is conservative and P, Q, R have continuous ﬁrstorder partial derivatives,
then yz i xz j
x y 2 z k,
C is the line segment from 1, 0, 2 to 4, 6, 3
2 ■ 27. Show that if the vector ﬁeld F 15. F x, y, z
16. F x, y, z ■ f , where f x, y
sin x 2y . Find curves C1 and
C2 that are not closed and satisfy the equation. 2 y arctan x j, 1x
t 2 i 2 t j, 0 C: r t t ■ 1 2 ■ 5E17(pp 11101119) 1/19/06 2:48 PM Page 1119 SECTION 17.4 GREEN’S THEOREM cr
r3 for some constant c, where r x i y j z k. Find the
work done by F in moving an object from a point P1 along
a path to a point P2 in terms of the distances d1 and d2 from
these points to the origin.
(b) An example of an inverse square ﬁeld is the gravitational
ﬁeld F
m MG r r 3 discussed in Example 4 in
Section 17.1. Use part (a) to ﬁnd the work done by the
gravitational ﬁeld when the Earth moves from aphelion  17.4 Green’s Theorem y D
C
0 1119 (at a maximum distance of 1.52 10 8 km from the Sun)
to perihelion (at a minimum distance of 1.47 10 8 km).
(Use the values m 5.97 10 24 kg, M 1.99 10 30 kg,
and G 6.67 10 11 N m 2 kg2.
(c) Another example of an inverse square ﬁeld is the
electric ﬁeld E
qQ r r 3 discussed in Example 5 in
Section 17.1. Suppose that an electron with a charge of
1.6 10 19 C is located at the origin. A positive unit
charge is positioned a distance 10 12 m from the electron and moves to a position half that distance from the
electron. Use part (a) to ﬁnd the work done by the electric
ﬁeld. (Use the value
8.985 10 10.) 34. (a) Suppose that F is an inverse square force ﬁeld, that is, Fr ❙❙❙❙ x Green’s Theorem gives the relationship between a line integral around a simple closed
curve C and a double integral over the plane region D bounded by C. (See Figure 1. We
assume that D consists of all points inside C as well as all points on C.) In stating Green’s
Theorem we use the convention that the positive orientation of a simple closed curve C
refers to a single counterclockwise traversal of C. Thus, if C is given by the vector function r t , a t b, then the region D is always on the left as the point r t traverses C.
(See Figure 2.)
y FIGURE 1 y C
D D
C
0 FIGURE 2 x 0 (a) Positive orientation x (b) Negative orientation Green’s Theorem Let C be a positively oriented, piecewisesmooth, simple closed curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then
 Recall that the left side of this equation is
another way of writing xC F d r, where
F P i Q j. y C P dx Q dy yy Q
x P
y dA D NOTE ■ The notation y C P dx Q dy or gC P d x Q dy is sometimes used to indicate that the line integral is calculated using the positive orientation of the closed curve C. Another notation for the positively oriented boundary curve of 5E17(pp 11201129) 1120 ❙❙❙❙ 1/19/06 2:55 PM Page 1120 CHAPTER 17 VECTOR CALCULUS D is D, so the equation in Green’s Theorem can be written as
Q
x yy 1 P
y y dA D P dx Q dy D Green’s Theorem should be regarded as the counterpart of the Fundamental Theorem of
Calculus for double integrals. Compare Equation 1 with the statement of the Fundamental
Theorem of Calculus, Part 2, in the following equation: y b a F x dx Fb Fa In both cases there is an integral involving derivatives (F , Q x, and P y) on the left
side of the equation. And in both cases the right side involves the values of the original
functions (F , Q, and P) only on the boundary of the domain. (In the onedimensional case,
the domain is an interval a, b whose boundary consists of just two points, a and b.)
Green’s Theorem is not easy to prove in the generality stated in Theorem 1, but we can
give a proof for the special case where the region is both of type I and of type II (see
Section 16.3). Let’s call such regions simple regions.
 Green’s Theorem is named after the
selftaught English scientist George Green
(1793–1841). He worked fulltime in his father’s
bakery from the age of nine and taught himself
mathematics from library books. In 1828 he
published privately An Essay on the Application
of Mathematical Analysis to the Theories of
Electricity and Magnetism, but only 100 copies
were printed and most of those went to his
friends. This pamphlet contained a theorem
that is equivalent to what we know as Green’s
Theorem, but it didn’t become widely known
at that time. Finally, at age 40, Green entered
Cambridge University as an undergraduate
but died four years after graduation. In 1846
William Thomson (Lord Kelvin) located a copy
of Green’s essay, realized its signiﬁcance, and
had it reprinted. Green was the ﬁrst person to
try to formulate a mathematical theory of electricity and magnetism. His work was the basis
for the subsequent electromagnetic theories of
Thomson, Stokes, Rayleigh, and Maxwell. Theorem will be proved if we can show that y 2 C P
dA
y yy P dx D and y 3 C Q dy yy
D Q
dA
x We prove Equation 2 by expressing D as a type I region:
D x, y a x b, t1 x y t2 x where t1 and t 2 are continuous functions. This enables us to compute the double integral
on the right side of Equation 2 as follows:
4 yy
D y D Is a Simple Region Notice that Green’s Proof of Green’s Theorem for the Case in Which P
dA
y b yy
a P
x, y d y d x
y t2 x t1 x y b a P x, t 2 x P x, t1 x dx where the last step follows from the Fundamental Theorem of Calculus.
Now we compute the left side of Equation 2 by breaking up C as the union of the
four curves C1 , C2 , C3 , and C4 shown in Figure 3. On C1 we take x as the parameter and
write the parametric equations as x x, y t1 x , a x b. Thus y=g™(x)
C£
C¢ D y C™ C1 C¡ FIGURE 3 a y b a P x, t1 x d x Observe that C3 goes from right to left but C3 goes from left to right, so we can write
the parametric equations of C3 as x x, y t 2 x , a x b. Therefore y=g¡(x)
0 P x, y d x b x y C3 P x, y d x y C3 P x, y d x y b a P x, t 2 x d x 5E17(pp 11201129) 1/19/06 2:55 PM Page 1121 ❙❙❙❙ SECTION 17.4 GREEN’S THEOREM 1121 On C2 or C4 (either of which might reduce to just a single point), x is constant, so
dx 0 and y C2 P x, y d x y 0 C4 P x, y d x Hence y C P x, y d x y C1 y b a y P x, y d x y P x, y d x C2 y P x, t1 x d x b y P x, y d x C3 C4 P x, y d x P x, t 2 x d x a Comparing this expression with the one in Equation 4, we see that y C P
dA
y yy P x, y d x D Equation 3 can be proved in much the same way by expressing D as a type II region (see
Exercise 28). Then, by adding Equations 2 and 3, we obtain Green’s Theorem.
EXAMPLE 1 Evaluate xC x 4 d x x y d y, where C is the triangular curve consisting of the
line segments from 0, 0 to 1, 0 , from 1, 0 to 0, 1 , and from 0, 1 to 0, 0 .
y SOLUTION Although the given line integral could be evaluated as usual by the methods of
Section 17.2, that would involve setting up three separate integrals along the three sides
of the triangle, so let’s use Green’s Theorem instead. Notice that the region D enclosed
by C is simple and C has positive orientation (see Figure 4). If we let P x, y
x 4 and
Q x, y
x y, then we have y=1x (0, 1) C
D y x (1, 0) C x 4 dx x y dy Q
x P
y dA y[y] (0, 0) dx 1
2 yy 1 yy
0 1x 0 y 0 dy dx D FIGURE 4 1 1
6 EXAMPLE 2 Evaluate x 2 y 2 xC 3y 2y1x
y0 1
2 0 1 31
0 x 1 x 2 dx 1 0 1
6 (7x e sin x d x y 1 ) d y, where C is the circle sy 4 9. SOLUTION The region D bounded by C is the disk x 2 y2 9, so let’s change to polar coordinates after applying Green’s Theorem: y C 3y e sin x d x (7x 1 ) dy sy 4 yy x (7x 1) sy 4 D  Instead of using polar coordinates, we could
simply use the fact that D is a disk of radius 3
and write yy 4 d A
D 4 3 2 36 2 yy
0 3 0 4y 2 0 d 7 3 r dr d y 3 0 r dr 36 y 3y e sin x dA 5E17(pp 11201129) 1122 ❙❙❙❙ 1/19/06 2:55 PM Page 1122 CHAPTER 17 VECTOR CALCULUS In Examples 1 and 2 we found that the double integral was easier to evaluate than the
line integral. (Try setting up the line integral in Example 2 and you’ll soon be convinced!)
But sometimes it’s easier to evaluate the line integral, and Green’s Theorem is used in the
Q x, y
0 on the curve C,
reverse direction. For instance, if it is known that P x, y
then Green’s Theorem gives
Q
x yy P
y dA y P dx C Q dy 0 D no matter what values P and Q assume in the region D.
Another application of the reverse direction of Green’s Theorem is in computing areas.
Since the area of D is xxD 1 d A, we wish to choose P and Q so that
Q
x P
y 1 There are several possibilities:
P x, y 0 P x, y Q x, y x y Q x, y 1
2 P x, y 0 1
2 Q x, y y x Then Green’s Theorem gives the following formulas for the area of D : y A 5 C y x dy C 1
2 y dx EXAMPLE 3 Find the area enclosed by the ellipse y C x dy x2
a2 y2
b2 y dx 1. SOLUTION The ellipse has parametric equations x 0 t a cos t and y
2 . Using the third formula in Equation 5, we have
1
2 y 1
2 A y C
2 D¡ D™
C£ _C£ C™ y dx a cos t b cos t dt 0 ab
2 C¡ x dy y 2 0 dt b sin t, where b sin t a sin t dt ab Although we have proved Green’s Theorem only for the case where D is simple, we can
now extend it to the case where D is a ﬁnite union of simple regions. For example, if D is
the region shown in Figure 5, then we can write D D1 D2, where D1 and D2 are both
C3 so, applysimple. The boundary of D1 is C1 C3 and the boundary of D2 is C2
ing Green’s Theorem to D1 and D2 separately, we get y FIGURE 5 C1 C3 P dx Q dy yy Q
x P
y dA Q
x P
y dA D1 y C2 C3 P dx Q dy yy
D2 5E17(pp 11201129) 1/19/06 2:55 PM Page 1123 SECTION 17.4 GREEN’S THEOREM C If we add these two equations, the line integrals along C3 and y C1 C2 P dx Q
x yy Q dy P
y ❙❙❙❙ 1123 C3 cancel, so we get
dA D which is Green’s Theorem for D D1 D2 , since its boundary is C C1 C2 .
The same sort of argument allows us to establish Green’s Theorem for any ﬁnite union
of nonoverlapping simple regions (see Figure 6).
FIGURE 6 EXAMPLE 4 Evaluate xC y 2 d x 3 xy d y, where C is the boundary of the semiannular
region D in the upper halfplane between the circles x 2 y 2 1 and x 2 y 2 4.
SOLUTION Notice that although D is not simple, the yaxis divides it into two simple
regions (see Figure 7). In polar coordinates we can write y ≈+¥=4 D C D r, 1 r 2, 0 Therefore, Green’s Theorem gives
0 ≈+¥=1 x y C y 2 dx yy 3xy d y x 3xy y2 y dA D FIGURE 7 yy y d A y y
0 2 r sin 1 r dr d D y sin d 0 C™
D
C¡
FIGURE 8 y 2 1 [ r 2 dr cos 14
3
[r]
0 1
3 32
1 Green’s Theorem can be extended to apply to regions with holes, that is, regions that
are not simplyconnected. Observe that the boundary C of the region D in Figure 8 consists of two simple closed curves C1 and C2 . We assume that these boundary curves are
oriented so that the region D is always on the left as the curve C is traversed. Thus, the
positive direction is counterclockwise for the outer curve C1 but clockwise for the inner
curve C2 . If we divide D into two regions D and D by means of the lines shown in
Figure 9 and then apply Green’s Theorem to each of D and D , we get yy
Dª Q
x P
y dA D yy Q
x P
y yy dA D y Q
x P
y dA D D P dx Q dy y D P dx Q dy Dªª FIGURE 9 Since the line integrals along the common boundary lines are in opposite directions, they
cancel and we get yy Q
x P
y dA y C1 P dx Q dy y C2 P dx Q dy y C P dx Q dy D which is Green’s Theorem for the region D.
y i x j x 2 y 2 , show that xC F d r
positively oriented simple closed path that encloses the origin.
EXAMPLE 5 If F x, y 2 for every SOLUTION Since C is an arbitrary closed path that encloses the origin, it’s difﬁcult to compute the given integral directly. So let’s consider a counterclockwiseoriented circle 5E17(pp 11201129) 1124 ❙❙❙❙ 1/19/06 2:55 PM Page 1124 CHAPTER 17 VECTOR CALCULUS y C with center the origin and radius a, where a is chosen to be small enough that C lies
inside C. (See Figure 10.) Let D be the region bounded by C and C . Then its positively
C and so the general version of Green’s Theorem gives
oriented boundary is C C
Cª
D x y C P dx y Q dy P dx C Q dy Q
x P
y y2
x2 yy x2
y2 2 dA D yy F IGURE 10 y2
x2 x2
y2 2 dA D 0 y Therefore C Q dy y P dx F dr y F dr P dx y that is, C C C Q dy We now easily compute this last integral using the parametrization given by
rt
a cos t i a sin t j, 0 t 2 . Thus y C F dr y C y y F dr
a sin t 2 0 y 2 0 dt 2 0 Frt a sin t
a 2 cos 2 t r t dt
a cos t a cos t
dt
a 2 sin 2 t 2 We end this section by using Green’s Theorem to discuss a result that was stated in the
preceding section.
P i Q j is a vector ﬁeld on an
Sketch of Proof of Theorem 17.3.6 We’re assuming that F
open simplyconnected region D, that P and Q have continuous ﬁrstorder partial derivatives, and that
P
y Q
x throughout D If C is any simple closed path in D and R is the region that C encloses, then Green’s
Theorem gives y C F dr y C P dx Q dy yy
R Q
x P
y dA yy 0 d A 0 R A curve that is not simple crosses itself at one or more points and can be broken up into
a number of simple curves. We have shown that the line integrals of F around these
simple curves are all 0 and, adding these integrals, we see that xC F d r 0 for any
closed curve C. Therefore, xC F d r is independent of path in D by Theorem 17.3.3. It
follows that F is a conservative vector ﬁeld. 5E17(pp 11201129) 1/19/06 2:55 PM Page 1125 S ECTION 17.4 GREEN’S THEOREM  17.4
xC x y 2 16. F x, y x
■ 3 d x x d y,
C is the rectangle with vertices (0, 0), (2, 0), (2, 3), and (0, 3) 3. xC x y d x x 2 y 3 d y,
C is the triangle with vertices (0, 0), (1, 0), and (1, 2) 4. xC x d x ■ ■ ■ ■ ■ ■ 3
■ ■ 5–6 ■ ■ ■ Verify Green’s Theorem by using a computer algebra system to evaluate both the line integral and the double integral.
 4 5. P x, y 5 7 x y , Q x, y
C is the circle x 2 y 2 1 ■ ■ ■ xy, ■ ■ ■ ■ ■ 8. xC x 9. 10. 11. xC y 3 d x
xC sin y d x 12.
■ 0, x 1, y 0, and y 1 ■ x cos y d y, C is the ellipse x ■ ■ ■ ■ ■ ■ y 3, x 2 sx 2 y 2 cos x, x 2 ■ C is the triangle from 0, 0 to 2, 6 to 2, 0 to 0, 0
15. F x, y ex C is the circle x x 2 y, e y
2 y 2 xy 2 , 25 oriented clockwise x 2 y1 x 2 y3
x n yn 1 x 3 y2
x n y1 x 1 yn y C x 2 dy 1
2A y y C y 2 dx where A is the area of D.
23. Use Exercise 22 to ﬁnd the centroid of the triangle with vertices 0, 0 , 1, 0 , and 0, 1 .
24. Use Exercise 22 to ﬁnd the centroid of a semicircular region sy , 2 y sin x , x 1 y2 1
2A 1
■ C consists of the arc of the curve y sin x from 0, 0 to
and the line segment from , 0 to 0, 0
14. F x, y x 2 y1 x yplane. Use Green’s Theorem to prove that the coordinates
of the centroid x, y of D are 13 –16  Use Green’s Theorem to evaluate xC F d r. (Check the
orientation of the curve before applying the theorem.) 13. F x, y x 1 y2 22. Let D be a region bounded by a simple closed path C in the y
■ y dx x n 1 yn x
xy cos t. 1 (c) Find the area of the pentagon with vertices 0, 0 , 2, 1 ,
1, 3 , 0, 2 , and 1, 1 . 4
■ x dy A 2x
dx
x 4 2 x 2 y 2 d y,
C is the boundary of the region between the circles
x 2 y 2 1 and x 2 y 2 4 2 1
2 A e sx ) d x
2 x cos y 2 d y,
C is the boundary of the region enclosed by the parabolas
y x 2 and x y 2 y2 sin t, y (b) If the vertices of a polygon, in counterclockwise order, are
x 1, y1 , x 2 , y2 , . . . , x n , yn , show that the area of the
polygon is y d x 4 x y d y,
C is the triangle with vertices (0, 0), (1, 3), and (0, 3) C is the circle x 2 t C 3 x 3 d y, ■ 19. Use one of the formulas in (5) to ﬁnd the area under one arch y xC xe 2 ■ ■ xC ( y 2 ■ point x 2 , y2 , show that ■ xC e y d x 2 xe y d y,
C is the square with sides x ■ 21. (a) If C is the line segment connecting the point x 1, y1 to the 7–12  Use Green’s Theorem to evaluate the line integral along
the given positively oriented curve. 7. ■ x 2 y 2 16, a ﬁxed point P on C traces out a curve called
an epicycloid, with parametric equations x 5 cos t cos 5 t,
y 5 sin t sin 5 t. Graph the epicycloid and use (5) to ﬁnd
the area it encloses. 6 y 2 sin x, Q x, y
x 2 sin y,
C consists of the arc of the parabola y x 2 from (0, 0) to
(1, 1) followed by the line segment from (1, 1) to (0, 0)
■ ■ ; 20. If a circle C with radius 1 rolls along the outside of the circle 6. P x, y ■ ■ 2, 0 , moves along the xaxis to
2, 0 , and then along the semicircle y s4 x 2 to the starting point. Use Green’s Theorem to ﬁnd the work done on this
particle by the force ﬁeld F x, y
x, x 3 3 x y 2 .
of the cycloid x CAS 1
y x , C is the circle
oriented counterclockwise
1 18. A particle starts at the point y d y, C consists of the line segments from 0, 1 to
0, 0 and from 0, 0 to 1, 0 and the parabola y 1 x 2
from 1, 0 to 0, 1
■ y ■ y 2 , 2 tan 2 F x, y
x x y i x y 2 j in moving a particle from the
origin along the xaxis to 1, 0 , then along the line segment
to 0, 1 , and then back to the origin along the yaxis. x d y,
C is the circle with center the origin and radius 1 ■ 2
■ ln x 2 y
2 17. Use Green’s Theorem to ﬁnd the work done by the force xC y d x 2. ■ 1125 Exercises 1–4  Evaluate the line integral by two methods: (a) directly and
(b) using Green’s Theorem. 1. ❙❙❙❙ ,0 of radius a.
25. A plane lamina with constant density x, y
occupies a
region in the x yplane bounded by a simple closed path C.
Show that its moments of inertia about the axes are
Ix 3 y C y 3 dx Iy 3 y C x 3 dy 5E17(pp 11201129) 1126 ❙❙❙❙ 1/19/06 2:56 PM Page 1126 CHAPTER 17 VECTOR CALCULUS 26. Use Exercise 25 to ﬁnd the moment of inertia of a circular disk where f x, y of radius a with constant density about a diameter. (Compare
with Example 4 in Section 16.5.) yy d x d y yy 0
27. If F is the vector ﬁeld of Example 5, show that xC F d r
for every simple closed path that does not pass through or
enclose the origin. R S x, y
u, v du dv Here R is the region in the xyplane that corresponds to the
region S in the uvplane under the transformation given by
x t u, v , y h u, v .
[Hint: Note that the left side is A R and apply the ﬁrst part
of Equation 5. Convert the line integral over R to a line integral over S and apply Green’s Theorem in the uvplane.] 28. Complete the proof of the special case of Green’s Theorem by proving Equation 3.
29. Use Green’s Theorem to prove the change of variables formula for a double integral (Formula 16.9.9) for the case  17.5 1: Curl and Divergence
In this section we deﬁne two operations that can be performed on vector ﬁelds and that
play a basic role in the applications of vector calculus to ﬂuid ﬂow and electricity and magnetism. Each operation resembles differentiation, but one produces a vector ﬁeld whereas
the other produces a scalar ﬁeld. Curl
If F P i Q j R k is a vector ﬁeld on 3 and the partial derivatives of P, Q, and R
all exist, then the curl of F is the vector ﬁeld on 3 deﬁned by 1 R
y curl F Q
z P
z i R
x Q
x j P
y k As an aid to our memory, let’s rewrite Equation 1 using operator notation. We introduce
the vector differential operator ∇ (“del”) as
∇ i j x y k z It has meaning when it operates on a scalar function to produce the gradient of f :
∇f f
x i j f
y k f
z f
i
x f
j
y f
k
z x,
y, and
z, we can also consider
If we think of ∇ as a vector with components
the formal cross product of ∇ with the vector ﬁeld F as follows:
i
∇ F j k x
P y
Q z
R R
y
curl F Q
z i P
z R
x j Q
x P
y k 5E17(pp 11201129) 1/19/06 2:56 PM Page 1127 SECTION 17.5 CURL AND DIVERGENCE ❙❙❙❙ 1127 Thus, the easiest way to remember Deﬁnition 1 is by means of the symbolic expression
∇ curl F 2 EXAMPLE 1 If F x, y, z xz i F y 2 k, ﬁnd curl F. x yz j SOLUTION Using Equation 2, we have i
curl F x
xz F  Most computer algebra systems have commands that compute the curl and divergence of
vector ﬁelds. If you have access to a CAS, use
these commands to check the answers to the
examples and exercises in this section. j
y
xy z y2 y 2y i x yz xy i y2 z
y2 x yz z
x k 0 xi xj y y2 x xz xz j k
yz xj z 0k yz k Recall that the gradient of a function f of three variables is a vector ﬁeld on 3 and so
we can compute its curl. The following theorem says that the curl of a gradient vector ﬁeld
is 0.
3 Theorem If f is a function of three variables that has continuous secondorder
partial derivatives, then curl f 0 Proof We have i
 Notice the similarity to what we know
from Section 13.4: a a 0 for every
threedimensional vector a. curl f 2 f
yz
0i 0j k x
f
x f j
y
f
y z
f
z 2 2 f
zy i 0k f
zx 2 2 f
xz f
xy 2 0 j f
yx k by Clairaut’s Theorem.
Since a conservative vector ﬁeld is one for which F
as follows:
 Compare this with Exercise 27 in
Section 17.3. ∇ f , Theorem 3 can be rephrased If F is conservative, then curl F 0. This gives us a way of verifying that a vector ﬁeld is not conservative. 5E17(pp 11201129) 1128 ❙❙❙❙ 1/19/06 2:56 PM Page 1128 CHAPTER 17 VECTOR CALCULUS EXAMPLE 2 Show that the vector ﬁeld F x, y, z xz i x yz j y 2 k is not conservative.
SOLUTION In Example 1 we showed that curl F
This shows that curl F y2 xi yz k xj 0 and so, by Theorem 3, F is not conservative. The converse of Theorem 3 is not true in general, but the following theorem says the
converse is true if F is deﬁned everywhere. (More generally it is true if the domain is
simplyconnected, that is, “has no hole.”) Theorem 4 is the threedimensional version
of Theorem 17.3.6. Its proof requires Stokes’ Theorem and is sketched at the end of
Section 17.8.
3 4 Theorem If F is a vector ﬁeld deﬁned on all of whose component functions
0, then F is a conservative vector have continuous partial derivatives and curl F
ﬁeld.
EXAMPLE 3 (a) Show that F x, y, z
y 2z3 i
(b) Find a function f such that F 2 xy z 3 j
f. 3 xy 2 z 2 k is a conservative vector ﬁeld. SOLUTION (a) We compute the curl of F:
i
curl F F j x
yz 23 6 xy z 2 k y
z
3
2 xy z 3 xy 2 z 2 6 xy z 2 i 3y 2 z 2 3y 2 z 2 j 2yz3 2yz3 k 0
Since curl F 0 and the domain of F is 3, F is a conservative vector ﬁeld by
Theorem 4.
(b) The technique for ﬁnding f was given in Section 17.3. We have
5 fx x, y, z y 2z3 6 fy x, y, z 2 xy z 3 7 fz x, y, z 3 xy 2 z 2 Integrating (5) with respect to x, we obtain
8 f x, y, z x y 2z3 t y, z Differentiating (8) with respect to y, we get fy x, y, z
with (6) gives ty y, z
0. Thus, t y, z
h z and
fz x, y, z 3 xy 2 z 2 2 xy z 3
hz ty y, z , so comparison 5E17(pp 11201129) 1/19/06 2:56 PM Page 1129 SECTION 17.5 CURL AND DIVERGENCE Then (7) gives h z (x, y, z) FIGURE 1 1129 0. Therefore
x y 2z3 f x, y, z curl F (x, y, z) ❙❙❙❙ K The reason for the name curl is that the curl vector is associated with rotations. One
connection is explained in Exercise 35. Another occurs when F represents the velocity
ﬁeld in ﬂuid ﬂow (see Example 3 in Section 17.1). Particles near (x, y, z) in the ﬂuid tend
to rotate about the axis that points in the direction of curl F x, y, z and the length of this
curl vector is a measure of how quickly the particles move around the axis (see Figure 1).
If curl F 0 at a point P, then the ﬂuid is free from rotations at P and F is called irrotational at P. In other words, there is no whirlpool or eddy at P. If curl F 0, then a
tiny paddle wheel moves with the ﬂuid but doesn’t rotate about its axis. If curl F 0, the
paddle wheel rotates about its axis. We give a more detailed explanation in Section 17.8 as
a consequence of Stokes’ Theorem. Divergence
If F P i Q j R k is a vector ﬁeld on 3 and P x, Q y, and R z exist, then
the divergence of F is the function of three variables deﬁned by P
x div F 9 Q
y R
z Observe that curl F is a vector ﬁeld but div F is a scalar ﬁeld. In terms of the gradient operator
xi
yj
z k, the divergence of F can be written symbolically
as the dot product of and F: div F 10 EXAMPLE 4 If F x, y, z xz i F y 2 k, ﬁnd div F. x yz j SOLUTION By the deﬁnition of divergence (Equation 9 or 10) we have div F F
z x xz y x yz z y2 xz If F is a vector ﬁeld on 3, then curl F is also a vector ﬁeld on
compute its divergence. The next theorem shows that the result is 0.
11 Theorem If F P i Q j R k is a vector ﬁeld on
continuous secondorder partial derivatives, then
div curl F 0 3 3 . As such, we can and P, Q, and R have 5E17(pp 11301139) 1130 ❙❙❙❙ 1/19/06 2:57 PM Page 1130 CHAPTER 17 VECTOR CALCULUS Proof Using the deﬁnitions of divergence and curl, we have
 Note the analogy with the scalar triple
product: a a b
0. div curl F F
R
y x Q
z 2 y 2 R
xy P
z 2 Q
xz R
x
2 P
yz z
2 R
yx Q
x P
y 2 Q
zx P
zy 0
because the terms cancel in pairs by Clairaut’s Theorem.
EXAMPLE 5 Show that the vector ﬁeld F x, y, z xz i
curl G. the curl of another vector ﬁeld, that is, F y 2 k can’t be written as x yz j SOLUTION In Example 4 we showed that z div F
and therefore div F 0. If it were true that F
div F which contradicts div F
 The reason for this interpretation of div F
will be explained at the end of Section 17.9 as a
consequence of the Divergence Theorem. xz
curl G, then Theorem 11 would give div curl G 0 0. Therefore, F is not the curl of another vector ﬁeld. Again, the reason for the name divergence can be understood in the context of ﬂuid
ﬂow. If F x, y, z is the velocity of a ﬂuid (or gas), then div F x, y, z represents the net rate
of change (with respect to time) of the mass of ﬂuid (or gas) ﬂowing from the point x, y, z
per unit volume. In other words, div F x, y, z measures the tendency of the ﬂuid to diverge
from the point x, y, z . If div F 0, then F is said to be incompressible.
Another differential operator occurs when we compute the divergence of a gradient vector ﬁeld f . If f is a function of three variables, we have
2 div f f x f 2 2 f y 2 2 z
2 and this expression occurs so often that we abbreviate it as f
2 f . The operator 2 is called the Laplace operator because of its relation to Laplace’s equation
2
2 f x f 2 We can also apply the Laplace operator
F Pi 2 y
2 f
2 2 f
z2 0 to a vector ﬁeld
Qj Rk in terms of its components:
2 F 2 Pi 2 Qj 2 Rk 5E17(pp 11301139) 1/19/06 2:57 PM Page 1131 SECTION 17.5 CURL AND DIVERGENCE ❙❙❙❙ 1131 V ector Forms of Green ’s Theorem
The curl and divergence operators allow us to rewrite Green’s Theorem in versions that
will be useful in our later work. We suppose that the plane region D, its boundary curve
C, and the functions P and Q satisfy the hypotheses of Green’s Theorem. Then we consider the vector ﬁeld F P i Q j. Its line integral is y C y F dr C P dx Q dy and its curl is
i
curl F x
P x, y curl F Therefore j k
z
0 y
Q x, y P
kk
y Q
x k Q
x P
y Q
x k P
y and we can now rewrite the equation in Green’s Theorem in the vector form y 12 C yy F dr curl F k dA D Equation 12 expresses the line integral of the tangential component of F along C as the
double integral of the vertical component of curl F over the region D enclosed by C. We
now derive a similar formula involving the normal component of F.
If C is given by the vector equation
rt xt i yt j a t b then the unit tangent vector (see Section 14.2) is
Tt
y D r (t) nt n (t) C
x yt
rt j yt
rt i xt
rt j (See Figure 2.) Then, from Equation 17.2.3, we have y
FIGURE 2 i You can verify that the outward unit normal vector to C is given by
T (t) 0 xt
rt C F n ds y b a y b a y b a y C F nt rt dt P x t ,y t y t
rt
P x t , y t y t dt
P dy Q dx yy
D Q x t ,y t x t
rt
Q x t , y t x t dt
P
x Q
y dA rt dt 5E17(pp 11301139) ❙❙❙❙ 1132 1/19/06 2:57 PM Page 1132 CHAPTER 17 VECTOR CALCULUS by Green’s Theorem. But the integrand in this double integral is just the divergence of F.
So we have a second vector form of Green’s Theorem. y 13 C F n ds yy div F x, y dA D This version says that the line integral of the normal component of F along C is equal to
the double integral of the divergence of F over the region D enclosed by C.  17.5
1–8 Exercises Find (a) the curl and (b) the divergence of the vector ﬁeld.  x 2y k 1. F x, y, z x yz i 2. F x, y, z x 2 yz i x y 2z j 3. F x, y, z i yz j 4. F x, y, z cos x z j x2 7. F x, y, z (x y sz ) k e x cos y j x
y2 z2 i zk
y
y2 x2 z2 y x e , x z, z e
■ ■ ■ j x2 z2 k ■ y ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ yz i
2 3z i cos y j 2xy i 2 16. F x, y, z ez i 17. F x, y, z ye 18. F x, y, z
■ ■ xz j x i e y cos x y i
■ ■ 2yz j y2 k xe z k j
x 2 xz k ■ x 2z k j x cos x y j
■ ■ sin z k
■ ■ ■ 3
such that
z x 2 k? Explain. 19. Is there a vector ﬁeld G on curl G xy2 i yz2 j 20. Is there a vector ﬁeld G on curl G yz i x yz j 3 such that
x y k? Explain. 21. Show that any vector ﬁeld of the form
x ■ xy k y 0 ■  Determine whether or not the vector ﬁeld is conservative.
If it is conservative, ﬁnd a function f such that F ∇ f . 15. F x, y, z 10. ■ 13–18 14. F x, y, z x ■ each expression is meaningful. If not, explain why. If so, state
whether it is a scalar ﬁeld or a vector ﬁeld.
(a) curl f
(b) grad f
(c) div F
(d) curl grad f
(e) grad F
(f ) grad div F
(g) div grad f
(h) grad div f
(i) curl curl F
(j) div div F
(k) grad f
(l) div curl grad f
div F ■ 13. F x, y, z 0 ■ 12. Let f be a scalar ﬁeld and F a vector ﬁeld. State whether 9–11  The vector ﬁeld F is shown in the x yplane and looks the
same in all other horizontal planes. (In other words, F is independent of z and its zcomponent is 0.)
(a) Is div F positive, negative, or zero? Explain.
(b) Determine whether curl F 0. If not, in which direction does
curl F point? 9. ■ y
■ x 0
z
y2 ln x, ln x y , ln x y z 8. F x, y, z
■ x yz 2 k sin x y k e sin y i 6. F x, y, z ■ x x 5. F x, y, z y 11. F x, y, z fxi tyj hz k where f , t, h are differentiable functions, is irrotational. ■ ■ 5E17(pp 11301139) 1/19/06 2:58 PM Page 1133 S ECTION 17.5 CURL AND DIVERGENCE 22. Show that any vector ﬁeld of the form F x, y, z f y, z i t x, z j h x, y k 23–29  Prove the identity, assuming that the appropriate partial
derivatives exist and are continuous. If f is a scalar ﬁeld and F, G
are vector ﬁelds, then f F, F G, and F G are deﬁned by f x, y, z F x, y, z F G x, y, z F x, y, z G x, y, z F x, y, z F
23. div F G 24. curl F G 25. div f F curl F G
f ■ 30–32 f
f r P ¨
2 grad div F F 0 ■ ■ ■ xi yj z k and r ■ ■ ■ ■ ■ y r.
x (b) 3
12r 23 r rr 4r
36. Maxwell’s equations relating the electric ﬁeld E and magnetic ﬁeld H as they vary with time in a region containing no charge
and no current can be stated as follows: 31. Verify each identity. (a)
(c) r rr
1r (b)
(d) r r3 r
ln r 0
r r2 div E r r p, ﬁnd div F. Is there a value of p for which
div F 0 ? 0 32. If F
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Green’s ﬁrst identity:
2 y t dA C f t n ds D yy f t dA 34. Use Green’s ﬁrst identity (Exercise 33) to prove Green’s second identity: yy f 2 t curl H H
t 0
1
c (a) E (b) H (c) 2 E (d) 2 H 1
c2
1
c2 1
c2
1
c2 2 E
t2
2
H
t2 2 E
[Hint: Use Exercise 29.]
t2
2
H
t2 37. We have seen that all vector ﬁelds of the form F t 2 f dA y C ft tf n ds D where D and C satisfy the hypotheses of Green’s Theorem
and the appropriate partial derivatives of f and t exist and are
continuous.
35. This exercise demonstrates a connection between the curl vector and rotations. Let B be a rigid body rotating about the E
t where c is the speed of light. Use these equations to prove the
following: D where D and C satisfy the hypotheses of Green’s Theorem
and the appropriate partial derivatives of f and t exist and are
continuous. (The quantity t n Dn t occurs in the line integral. This is the directional derivative in the direction of the
normal vector n and is called the normal derivative of t.) div H
1
c curl E 33. Use Green’s Theorem in the form of Equation 13 to prove yy f v F curl G 30. Verify each identity. (a)
(c) d B F ■ Let r w 0 t ■  F G curl F 29. curl curl F
■ G x, y, z curl G f curl F 27. div F z G x, y, z div G f div F 26. curl f F 28. div div F 1133 zaxis. The rotation can be described by the vector w
k,
where is the angular speed of B, that is, the tangential speed
of any point P in B divided by the distance d from the axis of
rotation. Let r
x, y, z be the position vector of P.
(a) By considering the angle in the ﬁgure, show that the
velocity ﬁeld of B is given by v w r.
(b) Show that v
yi
x j.
(c) Show that curl v 2w. is incompressible. f F x, y, z ❙❙❙❙ t
satisfy the equation curl F 0 and that all vector ﬁelds of the
form F curl G satisfy the equation div F 0 (assuming
continuity of the appropriate partial derivatives). This suggests
the question: Are there any equations that all functions of the
form f div G must satisfy? Show that the answer to this
question is “No” by proving that every continuous function
f on 3 is the divergence of some vector ﬁeld. [Hint: Let
G x, y, z
t x, y, z , 0, 0 ,where t x, y, z
x0x f t, y, z dt.] 5E17(pp 11301139) 1134 ❙❙❙❙ 1/19/06 2:58 PM Page 1134 CHAPTER 17 VECTOR CALCULUS  17.6 Parametric Surfaces and Their Areas
In Section 9.2 we found the area of a surface of revolution and in Section 16.6 we found
the area of a surface with the equation z f x, y . Here we discuss more general surfaces,
called parametric surfaces, and compute their areas. Parametric Surfaces
In much the same way that we describe a space curve by a vector function r t of a single
parameter t, we can describe a surface by a vector function r u, v of two parameters u and
v. We suppose that
r u, v 1 x u, v i z u, v k y u, v j is a vectorvalued function deﬁned on a region D in the u vplane. So x, y, and z, the component functions of r, are functions of the two variables u and v with domain D. The set
of all points x, y, z in 3 such that
x 2 x u, v y z y u, v z u, v and u, v varies throughout D, is called a parametric surface S and Equations 2 are called
parametric equations of S. Each choice of u and v gives a point on S; by making all
choices, we get all of S. In other words, the surface S is traced out by the tip of the position vector r u, v as u, v moves throughout the region D. (See Figure 1.)
√ z S
D r
(u, √) 0 r (u, √) u 0 FIGURE 1
x A parametric surface y EXAMPLE 1 Identify and sketch the surface with vector equation r u, v
z 0 F IGURE 2 2 sin u k 2 cos u y v z 2 sin u 4 sin 2u 4 So for any point x, y, z on the surface, we have
y (2, 0, 0) vj SOLUTION The parametric equations for this surface are (0, 0, 2) x x 2 cos u i x2 z2 4 cos 2u This means that vertical crosssections parallel to the x zplane (that is, with y constant)
are all circles with radius 4. Since y v and no restriction is placed on v, the surface is a
circular cylinder with radius 2 whose axis is the yaxis (see Figure 2). 5E17(pp 11301139) 1/19/06 2:58 PM Page 1135 S ECTION 17.6 PARAMETRIC SURFACES AND THEIR AREAS z ❙❙❙❙ 1135 In Example 1 we placed no restrictions on the parameters u and v and so we obtained
the entire cylinder. If, for instance, we restrict u and v by writing the parameter domain as
(0, 3, 2) 0 u 2 0 v 3 0
x y FIGURE 3 then x 0, z 0, 0 y 3, and we get the quartercylinder with length 3 illustrated in
Figure 3.
If a parametric surface S is given by a vector function r u, v , then there are two useful
families of curves that lie on S, one family with u constant and the other with v constant.
These families correspond to vertical and horizontal lines in the uvplane. If we keep u
constant by putting u u 0, then r u 0 , v becomes a vector function of the single parameter v and deﬁnes a curve C1 lying on S. (See Figure 4.)
z √ (u¸, √¸)
√=√¸ Visual 17.6 shows animated versions of
Figures 4 and 5, with moving grid curves,
for several parametric surfaces. D r
C¡
u=u¸ 0 C™ 0 u y FIGURE 4 x Similarly, if we keep v constant by putting v v0, we get a curve C2 given by r u, v0
that lies on S. We call these curves grid curves. (In Example 1, for instance, the grid
curves obtained by letting u be constant are horizontal lines whereas the grid curves with
v constant are circles.) In fact, when a computer graphs a parametric surface, it usually
depicts the surface by plotting these grid curves, as we see in the following example. z EXAMPLE 2 Use a computer algebra system to graph the surface r u, v
√ constant sin v sin u, u cos v SOLUTION We graph the portion of the surface with parameter domain 0 u 4,
0 v 2 in Figure 5. It has the appearance of a spiral tube. To identify the grid
curves, we write the corresponding parametric equations:
x FIGURE 5 sin v cos u, 2 Which grid curves have u constant? Which have v constant?
u constant x 2 y 2 sin v cos u y 2 sin v sin u z u cos v If v is constant, then sin v and cos v are constant, so the parametric equations resemble
those of the helix in Example 4 in Section 14.1. So the grid curves with v constant are
the spiral curves in Figure 5. We deduce that the grid curves with u constant must be the
curves that look like circles in the ﬁgure. Further evidence for this assertion is that if u
is kept constant, u u 0 , then the equation z u 0 cos v shows that the zvalues vary
from u 0 1 to u 0 1.
In Examples 1 and 2 we were given a vector equation and asked to graph the corresponding parametric surface. In the following examples, however, we are given the more
challenging problem of ﬁnding a vector function to represent a given surface. In the rest of
this chapter we will often need to do exactly that. 5E17(pp 11301139) ❙❙❙❙ 1136 1/19/06 2:58 PM Page 1136 CHAPTER 17 VECTOR CALCULUS EXAMPLE 3 Find a vector function that represents the plane that passes through the
point P0 with position vector r0 and that contains two nonparallel vectors a and b.
P
√b
b P¸ a SOLUTION If P is any point in the plane, we can get from P0 to P by moving a certain
distance in the direction of a and another distance in the direction of b. So there are
scalars u and v such that P0 P
A
u a vb. (Figure 6 illustrates how this works, by
means of the Parallelogram Law, for the case where u and v are positive. See also
Exercise 38 in Section 13.2.) If r is the position vector of P, then ua OP0
A r P0 P
A r0 ua vb FIGURE 6 So the vector equation of the plane can be written as
r u, v r0 ua vb where u and v are real numbers.
If we write r
x, y, z , r0
x0 , y0 , z0 , a
a1 , a2 , a3 , and b
b1 , b2 , b3 ,
then we can write the parametric equations of the plane through the point x0 , y0 , z0 as
follows:
x x0 u a1 v b1 y y0 u a2 z v b2 z0 u a3 v b3 EXAMPLE 4 Find a parametric representation of the sphere x2 y2 z2 a2 SOLUTION The sphere has a simple representation
a in spherical coordinates, so let’s
choose the angles and in spherical coordinates as the parameters (see Section 13.7).
Then, putting
a in the equations for conversion from spherical to rectangular coordinates (Equations 13.7.3), we obtain x a sin cos y a sin sin z a cos as the parametric equations of the sphere. The corresponding vector equation is
r, a sin cos i a sin sin j a cos k We have 0
and 0
2 , so the parameter domain is the rectangle
D
0,
0, 2 . The grid curves with constant are the circles of constant latitude (including the equator). The grid curves with constant are the meridians (semicircles), which connect the north and south poles.
 One of the uses of parametric surfaces is in
computer graphics. Figure 7 shows the result of
trying to graph the sphere x 2 y 2 z 2 1
by solving the equation for z and graphing the
top and bottom hemispheres separately. Part of
the sphere appears to be missing because of the
rectangular grid system used by the computer.
The much better picture in Figure 8 was produced by a computer using the parametric
equations found in Example 4. FIGURE 7 FIGURE 8 5E17(pp 11301139) 1/19/06 2:58 PM Page 1137 S ECTION 17.6 PARAMETRIC SURFACES AND THEIR AREAS ❙❙❙❙ 1137 EXAMPLE 5 Find a parametric representation for the cylinder x2 y2 4 z 0 1 SOLUTION The cylinder has a simple representation r
2 in cylindrical coordinates, so
we choose as parameters and z in cylindrical coordinates. Then the parametric equations of the cylinder are x
where 0 2 cos
z 2 and 0 y z 2 sin z 1. EXAMPLE 6 Find a vector function that represents the elliptic paraboloid z x2 2y 2. SOLUTION If we regard x and y as parameters, then the parametric equations are simply x x y r x, y xi x2 z y 2y 2 and the vector equation is In Module 17.6 you can investigate
several families of parametric surfaces. x2 yj 2y 2 k In general, a surface given as the graph of a function of x and y, that is, with an equation of the form z f x, y , can always be regarded as a parametric surface by taking x
and y as parameters and writing the parametric equations as
x x y z y f x, y Parametric representations (also called parametrizations) of surfaces are not unique. The
next example shows two ways to parametrize a cone.
EXAMPLE 7 Find a parametric representation for the surface z top half of the cone z 2 4x 2 2 sx 2 y 2, that is, the 2 4y . SOLUTION 1 One possible representation is obtained by choosing x and y as parameters: x x y z y 2 sx 2 y2 So the vector equation is
r x, y
 For some purposes the parametric representations in Solutions 1 and 2 are equally good,
but Solution 2 might be preferable in certain
situations. If we are interested only in the part
of the cone that lies below the plane z 1,
for instance, all we have to do in Solution 2 is
change the parameter domain to
0 r 1
2 0 2 xi yj 2 sx 2 y2 k SOLUTION 2 Another representation results from choosing as parameters the polar coor dinates r and . A point x, y, z on the cone satisﬁes x r cos , y
z 2 s x 2 y 2 2r. So a vector equation for the cone is
r r,
where r 0 and 0 r cos i r sin j r sin , and 2r k 2. Surfaces of Revolution
Surfaces of revolution can be represented parametrically and thus graphed using a computer. For instance, let’s consider the surface S obtained by rotating the curve y f x ,
a x b, about the xaxis, where f x
0. Let be the angle of rotation as shown in 5E17(pp 11301139) 1138 ❙❙❙❙ 1/19/06 2:58 PM Page 1138 CHAPTER 17 VECTOR CALCULUS Figure 9. If x, y, z is a point on S, then z x 3 x y z f x cos f x sin 0
y y=ƒ
ƒ
x
x EXAMPLE 8 Find parametric equations for the surface generated by rotating the curve (x, y, z) ¨ Therefore, we take x and as parameters and regard Equations 3 as parametric equations
of S. The parameter domain is given by a x b, 0
2. z y sin x, 0
revolution. ƒ x 2 , about the xaxis. Use these equations to graph the surface of SOLUTION From Equations 3, the parametric equations are x x y z sin x cos sin x sin FIGURE 9
z and the parameter domain is 0 x 2 , 0
2 . Using a computer to plot these
equations and rotate the image, we obtain the graph in Figure 10. y x We can adapt Equations 3 to represent a surface obtained through revolution about the
y or zaxis. (See Exercise 28.) Tangent Planes FIGURE 10 We now ﬁnd the tangent plane to a parametric surface S traced out by a vector function
r u, v x u, v i y u, v j z u, v k at a point P0 with position vector r u 0 , v0 . If we keep u constant by putting u u 0 , then
r u 0 , v becomes a vector function of the single parameter v and deﬁnes a grid curve C1
lying on S. (See Figure 11.) The tangent vector to C1 at P0 is obtained by taking the partial
derivative of r with respect to v :
rv 4 x
v u 0 , v0 i y u 0 , v0 j v z
v u 0 , v0 k z √ P¸
(u¸, √¸)
√=√¸
D ru r√ 0 C¡ r u=u¸ 0 u y x FIGURE 11 C™ Similarly, if we keep v constant by putting v v0 , we get a grid curve C2 given by
r u, v0 that lies on S, and its tangent vector at P0 is
5 ru x
u 0 , v0 i
u y
u 0 , v0 j
u z
u 0 , v0 k
u 5E17(pp 11301139) 1/19/06 2:58 PM Page 1139 S ECTION 17.6 PARAMETRIC SURFACES AND THEIR AREAS ❙❙❙❙ 1139 If ru rv is not 0, then the surface S is called smooth (it has no “corners”). For a smooth
surface, the tangent plane is the plane that contains the tangent vectors ru and rv , and the
vector ru rv is a normal vector to the tangent plane.
 Figure 12 shows the selfintersecting surface
in Example 9 and its tangent plane at 1, 1, 3 .
z EXAMPLE 9 Find the tangent plane to the surface with parametric equations x
z u 2v at the point 1, 1, 3 . u 2, y v 2, SOLUTION We ﬁrst compute the tangent vectors:
(1, 1, 3) x
i
u y rv x y
j
u z
k
u x ru y z v i v j v 2u i
2v j k k
2k Thus, a normal vector to the tangent plane is
F IGURE 12 ru i
2u
0 rv jk
01
2v 2 2v i 4u j 4uv k Notice that the point 1, 1, 3 corresponds to the parameter values u
the normal vector there is
2i 4j 1 and v 1, so 4k Therefore, an equation of the tangent plane at 1, 1, 3 is
2x 1 4y or 1
x 4z 3 2z 2y 3 0
0 Surface Area
Now we deﬁne the surface area of a general parametric surface given by Equation 1. For
simplicity we start by considering a surface whose parameter domain D is a rectangle, and
we divide it into subrectangles Rij . Let’s choose u i*, vj* to be the lower left corner of Rij.
(See Figure 13.) The part Sij of the surface S that corresponds to Rij is called a patch and
has the point Pij with position vector r u i*, vj* as one of its corners. Let
*
ru ru u i*, vj* and *
rv rv u i*, vj* be the tangent vectors at Pij as given by Equations 5 and 4.
√ z R ij r Î√ Pij Sij Îu (u *, √ *)
ij FIGURE 13 The image of the
subrectangle Rij is the patch Sij . 0 0 u
x y 5E17(pp 11401149) 1140 ❙❙❙❙ 1/19/06 3:00 PM Page 1140 CHAPTER 17 VECTOR CALCULUS Sij Pij Figure 14(a) shows how the two edges of the patch that meet at Pij can be approximated
by vectors. These vectors, in turn, can be approximated by the vectors u ru and v r*
*
v
because partial derivatives can be approximated by difference quotients. So we approximate Sij by the parallelogram determined by the vectors u ru and v r*. This parallelo*
v
gram is shown in Figure 14(b) and lies in the tangent plane to S at Pij. The area of this
parallelogram is (a) u ru
* v r*
v ru
* r*
v uv and so an approximation to the area of S is
m n r*
u rv
* uv i 1j 1 *
Î√ r √
*
Îu r u Our intuition tells us that this approximation gets better as we increase the number of subrectangles, and we recognize the double sum as a Riemann sum for the double integral
xxD ru rv du dv. This motivates the following deﬁnition. (b)
FIGURE 14 6 Definition If a smooth parametric surface S is given by the equation Approximating a patch
by a parallelogram r u, v x u, v i z u, v k y u, v j u, v D and S is covered just once as u, v ranges throughout the parameter domain D,
then the surface area of S is yy AS ru rv d A D where x
i
u ru z
k
u y
j
u x rv v y i v z j v k EXAMPLE 10 Find the surface area of a sphere of radius a.
SOLUTION In Example 4 we found the parametric representation x a sin cos y a sin z sin a cos where the parameter domain is
D , 0 ,0 2 We ﬁrst compute the cross product of the tangent vectors:
i
x
r j
y k
z x y z r a 2 sin 2 cos i i
a cos
a sin
a 2 sin2 j
cos
sin sin j a cos
a sin
a 2 sin k
a sin
0 sin
cos
cos k 5E17(pp 11401149) 1/19/06 3:01 PM Page 1141 S ECTION 17.6 PARAMETRIC SURFACES AND THEIR AREAS ❙❙❙❙ 1141 Thus
r sa 4 sin 4 r cos 2 sa 4 sin 4
since sin 0 for 0 a 4 sin 4 a 4 sin 2 sin 2 cos 2 a 4 sin 2 a 2ssin 2 cos 2
a 2 sin . Therefore, by Deﬁnition 6, the area of the sphere is yy A r 2 yy r dA 0 0 a 2 sin dd D a2 y 2 0 d y 0 sin a2 2 2 d 4 a2 Surface Area of the Graph of a Function
For the special case of a surface S with equation z f x, y , where x, y lies in D and f
has continuous partial derivatives, we take x and y as parameters. The parametric equations
are
xx
yy
z f x, y
so rx f
x i j f
y f
i
x k f
j
y ry k and
i
1 ry 0 0 rx 7 j k
f
x
f
y 1 k Thus, we have
8
 Notice the similarity between the surface
area formula in Equation 9 and the arc length
formula y L b a 1 dy
dx from Section 9.1. rx f
x ry 2 f
y 2 1 2 z
x 1 z
y 2 and the surface area formula in Deﬁnition 6 becomes 2 dx yy AS 9 z
x 1 2 z
y
2 dA D z EXAMPLE 11 Find the area of the part of the paraboloid z 9 plane z x2 y 2 that lies under the 9. SOLUTION The plane intersects the paraboloid in the circle x 2 y 2 9, z 9. Therefore, the given surface lies above the disk D with center the origin and radius 3. (See
Figure 15.) Using Formula 9, we have
A D
x 3 FIGURE 15 y yy z
x 1 2 D yy s1
D 4 x2 y 2 dA z
y 2 dA yy s1
D 2x 2 2y 2 d A 5E17(pp 11401149) ❙❙❙❙ 1142 1/19/06 3:01 PM Page 1142 CHAPTER 17 VECTOR CALCULUS Converting to polar coordinates, we obtain
2 yy A 0 2 3 0 (1) 2
83 y 4r 2 r dr d s1 4r 2 1 2 0 32 3
0 6 y d 3 0 4r 2 dr r s1 (37s37 1) The question remains whether our deﬁnition of surface area (6) is consistent with the
surface area formula from singlevariable calculus (9.2.4).
We consider the surface S obtained by rotating the curve y f x , a x b, about
the xaxis, where f x
0 and f is continuous. From Equations 3 we know that parametric equations of S are
x x y z f x cos f x sin a x b 0 2 To compute the surface area of S we need the tangent vectors
rx i r f x cos
f x sin j j f x sin k
f x cos k Thus
rx i
1
0 r j
f x cos
f x sin fxf x i
rx because f x sfx r 2 sfx and so k
f x sin
f x cos 2 f x cos j f x sin k 1 2 fx 2 fx fx cos 2 2 f x s1 fx
fx 2 sin 2
2 0. Therefore, the area of S is yy A rx 2 yy r dA b 0 a fx 2 f x s1 fx 2 dx d D 2 y b a f x s1 dx This is precisely the formula that was used to deﬁne the area of a surface of revolution in
singlevariable calculus (9.2.4).  17.6
1–4  Exercises Identify the surface with the given vector equation. 1. r u, v u cos v i 2. r u, v 1 2u i u 3. r x, ■  Use a computer to graph the parametric surface. Get a
printout and indicate on it which grid curves have u constant and
which have v constant. 3v j 2 4u 5v k x, x cos , x sin ■ ■ ■ ■ 1
■ u u2
1, u u v, u 2, v 2 ,
1, 1 v 1 5. r u, v x, cos , sin 4. r x, ; 5–10 u2 k u sin v j 6. r u, v
■ ■ ■ ■ ■ ■ 1 1, v 3 1, u
1v1 v, 5E17(pp 11401149) 1/19/06 3:01 PM Page 1143 SECTION 17.6 PARAMETRIC SURFACES AND THEIR AREAS cos 3u cos 3v, sin 3u cos 3v, sin 3v ,
,0 v 2 7. r u, v 0 u 8. r u, v 0 cos u sin v, sin u sin v, cos v
2 , 0.1 v 6.2 u 9. x cos u sin 2v, 10. x u sin u cos v, ■ ■ ■ ■ y
y ■ ■ the vectors i sin v
z j k and i ■ ■ ■  Match the equations with the graphs labeled I–VI and
give reasons for your answers. Determine which families of grid
curves have u constant and which have v constant. 12. r u, v u cos v i 13. r u, v u 3, 15. x u 16. x 1
1
3u u sin v, z sin u cos v, y y cone z
planes z uk y x2
z cos u sin v, z2 1 1 that lies to the y2 0 and x u ■ CAS y2 2z 2 4 that lies y2 z2 4 that lies above the y2 z2 16 that lies between the 2
z2 16 that lies between the 5 24. The part of the plane z u cos v z 2 and z planes x vk 1 z 2 2 23. The part of the cylinder y 2 u 3 cos v cos 4 u,
u 3 cos v sin 4 u,
1 u sin v I sx 2 22. The part of the sphere x 2 u sin v j y 4y2
2 0 21. The part of the sphere x 2 uk u sin v j u cos v i 14. x y
z sin v j 2 20. The part of the elliptic paraboloid x ■ 11–16 cos v i 3 and contains k. right of the xzplane
in front of the plane x 11. r u, v j 19. The part of the hyperboloid x u sin v ■ 1143 Find a parametric representation for the surface. 18. The lower half of the ellipsoid 2 x 2 u cos u cos v, ■  17. The plane that passes through the point 1, 2, ln tan v 2 , z sin u sin 2v, 17–24 ❙❙❙❙ 3 that lies inside the cylinder x 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 25–26  Use a computer algebra system to produce a graph that
looks like the given one. 25.
3 z II z0 x x
z III _3
_3 y y x 05 y 0 26.
z IV z 0 x
_1
_1 y
y x ■ z V ■ ■ 0 0
11
■ ■ ■ x
■ ■ ■ ■ ■ ■ ; 27. Find parametric equations for the surface obtained by rotating z VI _1
y the curve y e x, 0
graph the surface. x 3, about the xaxis and use them to ; 28. Find parametric equations for the surface obtained by rotating
the curve x 4y 2 y 4, 2
them to graph the surface. y x y x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ y 2, about the yaxis and use ; 29. (a) What happens to the spiral tube in Example 2 (see Figure 5) if we replace cos u by sin u and sin u by cos u ?
(b) What happens if we replace cos u by cos 2u and sin u
by sin 2u ? 5E17(pp 11401149) 1144 ❙❙❙❙ 1/19/06 3:02 PM Page 1144 CHAPTER 17 VECTOR CALCULUS 46–47 x 2 cos r cos 2  Find the area of the surface correct to four decimal places
by expressing the area in terms of a single integral and using your
calculator to estimate the integral. y 2 sin r cos 2 46. The part of the surface z ; 30. The surface with parametric equations z r sin cylinder x 2 2 x
■ C AS Find an equation of the tangent plane to the given parametric surface at the speciﬁed point. If you have software that
graphs parametric surfaces, use a computer to graph the surface
and the tangent plane.
 31. x u v, 32. x u 2, y v 2,
2 33. r u, v ui 34. r u, v
■ 3u 2, y ■ uv ; u ■ v cos u k;
■ ■ u CAS 0 1, v
0, v ■ ■ ■ ■ y
■ 2 y 2 that lies inside the e x2 y2 that lies above the disk 4
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 48. Find, to four decimal places, the area of the part of the 1 x 2 1 y 2 that lies above the square
1. Illustrate by graphing this part of the surface. 49. (a) Use the Midpoint Rule for double integrals (see Sec 1 1, v 2 surface z
x
y 2, 3, 0 u cos v k; u u sin v j
■ v; u 2 u sin v j uv i ■ z z cos x 2 1 47. The part of the surface z where 1 r 1 and 0
2 , is called a Möbius strip.
2
2
Graph this surface with several viewpoints. What is unusual
about it?
31–34 y2 tion 16.1) with four squares to estimate the surface area
of the portion of the paraboloid z x 2 y 2 that lies above
the square 0, 1
0, 1 .
(b) Use a computer algebra system to approximate the surface
area in part (a) to four decimal places. Compare with the
answer to part (a). ■ C AS 35–45  50. Find the area of the surface with vector equation
r u, v
cos 3u cos 3v, sin 3u cos 3v, sin 3v , 0 u
,
0 v 2 . State your answer correct to four decimal places. CAS 51. Find the exact area of the surface z Find the area of the surface. 35. The part of the plane x cylinder x 2 y2 z 2y 4 that lies inside the 4 1 36. The part of the plane with vector equation
1 v, u 2v, 3 5u v that is given by
r u, v
0 u 1, 0 v 1
37. The part of the surface z x2 y2 x y that lies within the cylinder 3x 2 y 2 that lies above the
triangle with vertices 0, 0 , 0, 1 , and 2, 1
1 39. The part of the hyperbolic paraboloid z between the cylinders x 2 y2 cylinder y 2 z2 1, z 0, x cylinder x y 2 cylinder x y 2 x that lies
4 z 2 that lies inside the a z2
y2 z2 ■ ■ ■ ; a 2 that lies inside the ■ ■ ■ ■ u v, y
■ ■ ; v u v, ■ 1 2x 3y 4 y 2, 1. (b) Eliminate the parameters to show that the surface is an
elliptic paraboloid and set up another double integral for
the surface area.
(c) Use the parametric equations in part (a) with a 2 and
b 3 to graph the surface.
(d) For the case a 2, b 3, use a computer algebra system
to ﬁnd the surface area correct to four decimal places.
a sin u cos v,
,0 v 2 , represent an ellipsoid.
(b) Use the parametric equations in part (a) to graph the ellipsoid for the case a 1, b 2, c 3.
(c) Set up, but do not evaluate, a double integral for the surface
area of the ellipsoid in part (b).
a cosh u cos v,
54. (a) Show that the parametric equations x
y b cosh u sin v, z c sinh u, represent a hyperboloid of ax 45. The surface with parametric equations x
z u v, u 2 v 2 1 y 53. (a) Show that the parametric equations x
y b sin u sin v, z c cos u, 0 u a 2 that lies inside the 44. The helicoid (or spiral ramp) with vector equation
r u, v
u cos v i u sin v j v k, 0 u 1, 0 ■ CAS 2 43. The part of the sphere x 2
2 y2 4 x z 2 that lies between the
0, and z 1 42. The part of the cylinder x 2
2 ; 2 9 41. The part of the surface y planes x 1 and x 2
y2 40. The part of the paraboloid x y 2 4, 0 52. (a) Set up, but do not evaluate, a double integral for the area
of the surface with parametric equations x au cos v,
y bu sin v, z u 2, 0 u 2, 0 v 2 . 1 38. The part of the surface z x ■ one sheet.
(b) Use the parametric equations in part (a) to graph the hyperboloid for the case a 1, b 2, c 3.
(c) Set up, but do not evaluate, a double integral for the surface
area of the part of the hyperboloid in part (b) that lies
between the planes z
3 and z 3. 5E17(pp 11401149) 1/19/06 3:02 PM Page 1145 ❙❙❙❙ S ECTION 17.7 SURFACE INTEGRALS CAS 1145 z 55. Find the area of the surface in Exercise 7 correct to four deci mal places. (x, y, z) 56. (a) Find a parametric representation for the torus obtained by rotating about the z axis the circle in the x zplane with
center b, 0, 0 and radius a b. [Hint: Take as parameters the angles and shown in the ﬁgure.]
(b) Use the parametric equations found in part (a) to graph the
torus for several values of a and b.
(c) Use the parametric representation from part (a) to ﬁnd the
surface area of the torus. ;  17.7 0 å ¨ x y (b, 0, 0) Surface Integrals
The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length. Suppose f is a function of three variables
whose domain includes a surface S. We divide S into patches Sij with area Sij . We evalu*
ate f at a point Pij in each patch, multiply by the area Sij , and form the sum
m n *
f Pij Sij i 1j 1 Then we take the limit as the patch size approaches 0 and deﬁne the surface integral of f
over the surface S as
m yy 1 f x, y, z dS n *
f Pij lim m, n l S Sij i 1j 1 Notice the analogy with the deﬁnition of a line integral (17.2.2) and also the analogy with
the deﬁnition of a double integral (16.1.5).
To evaluate the surface integral in Equation 1 we approximate the patch area Sij by the
area Tij of an approximating parallelogram in the tangent plane, and the limit becomes a
double integral. We now explain the details for two types of surfaces, graphs and parametric surfaces.
Graphs If the surface S is a graph of a function of two variables, then it has an equation of the form z t x, y , x, y
D, and is illustrated in Figure 1. We ﬁrst assume that
the parameter domain D is a rectangle and we divide it into smaller rectangles Rij of equal
*
size. The patch Sij lies directly above the rectangle Rij and the point Pij in Sij is of the
form x i*, yj*, t x i*, yj* . As with surface area in Section 16.6, we use the approximation z P*
ij ÎTij ÎS ij
S Sij y D R ij m yy
ÎA f x, y, z dS 2 ty xi, yj 2 1A n f x i*, yj*, t x i*, yj* s tx xi, yj lim m, n l S yy f
FIGURE 1 s tx xi, yj In fact, it can be shown that, if f is continuous on S and t has continuous derivatives, then
Deﬁnition 1 becomes 0 x Tij D 2 ty xi, yj i 1j 1 x, y, t x, y s tx x, y 2 ty x, y 2 1 dA 2 1A 5E17(pp 11401149) 1146 ❙❙❙❙ 1/19/06 3:03 PM Page 1146 CHAPTER 17 VECTOR CALCULUS This formula is true even when D isn’t a rectangle and is usually written in the following
form. yy f 2 yy f x, y, z dS S 2 z
x x, y, t x, y D 2 z
y 1 dA Similar formulas apply when it is more convenient to project S onto the y zplane or
x zplane. For instance, if S is a surface with equation y h x, z and D is its projection on
the x zplane, then yy f yy f x, y, z dS S z D EXAMPLE 1 Evaluate 2 y
x x, h x, z , z xxS y dS, where S is the surface z 2 y
z 1 dA y 2, 0 x x 1, 0 y 2. (See Figure 2.)
z
x SOLUTION Since
y 1 z
y and 2y Formula 2 gives
x yy y dS yy y FIGURE 2 S z
x 1 D
1 yy
0 y 1 0 2 0 y s1 2 z
y dA 4y 2 d y d x 1
2 d x s2 y y s1
0 12
s2 ( 4 ) 3 1 2 2y 2 2y 2 d y 322
0 13 s2
3 Surface integrals have applications similar to those for the integrals we have previously
considered. For example, if a thin sheet (say, of aluminum foil) has the shape of a surface
S and the density (mass per unit area) at the point x, y, z is x, y, z , then the total mass
of the sheet is yy m x, y, z dS S and the center of mass is x, y, z , where
x 1
m yy x x, y, z dS y S 1
m yy y x, y, z dS 1
m z S yy z
S Moments of inertia can also be deﬁned as before (see Exercise 37).
Parametric Surfaces Suppose that a surface S has a vector equation
r u, v x u, v i y u, v j z u, v k u, v D x, y, z dS 5E17(pp 11401149) 1/19/06 3:03 PM Page 1147 SECTION 17.7 SURFACE INTEGRALS ❙❙❙❙ 1147 We ﬁrst assume that the parameter domain D is a rectangle and we divide it into subrectangles Rij with dimensions u and v. Then the surface is divided into corresponding
patches Sij with areas Sij , as in Figure 3.
z
√ P*
ij S R ij Sij
D Î√ r Îu 0 0 y u x FIGURE 3 In our discussion of surface area in Section 17.6 we made the approximation
Sij where ru x
i
u ru rv z
k
u y
j
u uv
x rv y i v v j z
v k are the tangent vectors at a corner of Sij . If the components are continuous and ru and rv
are nonzero and nonparallel in the interior of D, it can be shown from Deﬁnition 1, even
when D is not a rectangle, that
 We assume that the surface is covered only
once as u, v ranges throughout D. The value
of the surface integral does not depend on the
parametrization that is used. 3 yy f yy f x, y, z dS S r u, v ru rv d A D This should be compared with the formula for a line integral: y C y f x, y, z ds b a f rt r t dt Observe also that yy 1 dS yy
S ru rv d A AS D Formula 3 allows us to compute a surface integral by converting it into a double integral over the parameter domain D. When using this formula, remember that f r u, v is
evaluated by writing x x u, v , y y u, v , and z z u, v in the formula for f x, y, z .
Any surface S with equation z t x, y can be regarded as a parametric surface with
parametric equations
xx
yy
z t x, y
and from Equation 17.6.8 we have
rx ry 1 z
x 2 Therefore, in this case, Formula 3 becomes Formula 2. z
y 2 5E17(pp 11401149) 1148 ❙❙❙❙ 1/19/06 3:03 PM Page 1148 CHAPTER 17 VECTOR CALCULUS EXAMPLE 2 Compute the surface integral x 2 y 2 z 2 xxS x 2 dS, where S is the unit sphere 1. SOLUTION As in Example 4 in Section 17.6, we use the parametric representation x sin cos y sin sin r, that is, z sin cos i cos sin 0 sin j 0
cos 2 k As in Example 10 in Section 17.6, we can compute that
r r sin Therefore, by Formula 3, yy x 2 yy dS S sin 2 cos r r dA D
2 yy
0 y 2 0 1
2 0 1
2 [ sin 2
1
1
2 cos 2 sin cos 2
sin 2 d
[
2
0 2 dd y 0 y sin cos 2 d y cos 2 d sin
1
3 cos 0 cos 3 0 sin 3 d 4
3 0 If S is a piecewisesmooth surface, that is, a ﬁnite union of smooth surfaces S1 , S2 , . . . ,
Sn that intersect only along their boundaries, then the surface integral of f over S is deﬁned
by yy f x, y, z dS S yy f yy f x, y, z dS S1 x, y, z dS Sn EXAMPLE 3 Evaluate xxS z dS, where S is the surface whose sides S1 are given by the
cylinder x 2 y 2 1, whose bottom S2 is the disk x 2 y 2 1 in the plane z 0, and
whose top S3 is the part of the plane z 1 x that lies above S2 .
z SOLUTION The surface S is shown in Figure 4. (We have changed the usual position of
the axes to get a better look at S.) For S1 we use and z as parameters (see Example 5
in Section 17.6) and write its parametric equations as S£ (z=1+x) y x S¡ (≈+¥=1)
x where 0 cos 2 and y z sin
0 z 1 z x 1 cos 0 Therefore
S™
FIGURE 4 r and i
sin
0 rz r rz j
cos
0
scos 2 k
0
1 cos i sin 2 1 sin j 5E17(pp 11401149) 1/19/06 3:04 PM Page 1149 SECTION 17.7 SURFACE INTEGRALS ❙❙❙❙ 1149 Thus, the surface integral over S1 is yy z dS yy z
S1 r rz d A D
2 yy
0 1
2 1 cos y 2 cos 1
2 0 y 2 1 0 [ 13
22 Since S2 lies in the plane z z dz d 2 0 1
4 2 sin 1
2 1 1 sin 2 cos cos 2 2 d d 3
2 2
0 0, we have yy z dS yy 0 dS
S2 0 S2 The top surface S3 lies above the unit disk D and is part of the plane z 1 x. So,
taking t x, y
1 x in Formula 2 and converting to polar coordinates, we have yy z dS yy
S3 1 x D
2 yy
0 1 0 s2 y 2 0 s2 y 2 0 Therefore 1 y r cos
1 0 ( 1
2 2 z
x 1 s1 1 r 2 cos r
1
3 cos z
y 2 dA 0 r dr d dr d )d s2 2 sin
3 2 s2
0 yy z dS yy z dS yy z dS yy z dS
S S1 3
2 S2 0 S3 (3
2 s2 s2 ) Oriented Surfaces
P FIGURE 5 In order to deﬁne surface integrals of vector ﬁelds, we need to rule out nonorientable surfaces such as the Möbius strip shown in Figure 5. [It is named after the German geometer
August Möbius (1790–1868).] You can construct one for yourself by taking a long rectangular strip of paper, giving it a halftwist, and taping the short edges together as in Figure 6. If an ant were to crawl along the Möbius strip starting at a point P, it would end
up on the “other side” of the strip (that is, with its upper side pointing in the opposite
direction). Then, if the ant continued to crawl in the same direction, it would end up back A Möbius strip
Visual 17.7 shows a Möbius strip with
a normal vector that can be moved
along the surface. B C A D F IGURE 6 Constructing a Möbius strip B D A C 5E17(pp 11501159) 1150 ❙❙❙❙ 1/19/06 3:06 PM Page 1150 CHAPTER 17 VECTOR CALCULUS z n¡ n™
0
y x at the same point P without ever having crossed an edge. (If you have constructed a
Möbius strip, try drawing a pencil line down the middle.) Therefore, a Möbius strip really
has only one side. You can graph the Möbius strip using the parametric equations in Exercise 30 in Section 17.6.
From now on we consider only orientable (twosided) surfaces. We start with a surface
S that has a tangent plane at every point x, y, z on S (except at any boundary points).
There are two unit normal vectors n1 and n 2
n1 at x, y, z . (See Figure 7.) If it is possible to choose a unit normal vector n at every such point x, y, z so that n varies continuously over S, then S is called an oriented surface and the given choice of n provides
S with an orientation. There are two possible orientations for any orientable surface (see
Figure 8). FIGURE 7 n n n n n
n n F IGURE 8 n
n
n The two orientations of
an orientable surface For a surface z t x, y given as the graph of t, we use Equation 17.6.7 and see that
the induced orientation is given by the unit normal vector
t
i
x n 4 t
j
y
2 t
x 1 k
t
y 2 Since the kcomponent is positive, this gives the upward orientation of the surface.
If S is a smooth orientable surface given in parametric form by a vector function
r u, v , then it is automatically supplied with the orientation of the unit normal vector
z and the opposite orientation is given by
found the parametric representation 0
y
x r,
for the sphere x 2 FIGURE 9 Positive orientation r y2 a sin
z2 r So the orientation induced by r , x FIGURE 10 Negative orientation r
r r
r sin a 2 sin 2 cos i and n a sin sin j a cos k a 2. Then in Example 10 in Section 17.6 we found that a 2 sin 2 r rv
rv n. For instance, in Example 4 in Section 17.6 we cos i z y ru
ru n 5 sin j a 2 sin cos k a 2 sin r is deﬁned by the unit normal vector cos i sin sin j cos k 1
r,
a Observe that n points in the same direction as the position vector, that is, outward from the
sphere (see Figure 9). The opposite (inward) orientation would have been obtained (see
r
r
r.
Figure 10) if we had reversed the order of the parameters because r 5E17(pp 11501159) 1/19/06 3:06 PM Page 1151 SECTION 17.7 SURFACE INTEGRALS ❙❙❙❙ 1151 For a closed surface, that is, a surface that is the boundary of a solid region E, the
convention is that the positive orientation is the one for which the normal vectors point
outward from E, and inwardpointing normals give the negative orientation (see Figures 9
and 10). Surface Integrals of Vector Fields
z F=∏ v n
S ij
S
0
x FIGURE 11 Suppose that S is an oriented surface with unit normal vector n, and imagine a ﬂuid with
density x, y, z and velocity ﬁeld v x, y, z ﬂowing through S. (Think of S as an imaginary surface that doesn’t impede the ﬂuid ﬂow, like a ﬁshing net across a stream.) Then the
rate of ﬂow (mass per unit time) per unit area is v. If we divide S into small patches Sij ,
as in Figure 11 (compare with Figures 1 and 3), then Sij is nearly planar and so we can
approximate the mass of ﬂuid crossing Sij in the direction of the normal n per unit time by
the quantity y v n A Sij
where , v, and n are evaluated at some point on Sij . (Recall that the component of the vector v in the direction of the unit vector n is v n.) By summing these quantities and taking the limit we get, according to Deﬁnition 1, the surface integral of the function v n
over S :
6 yy yy v n dS S x, y, z v x, y, z n x, y, z dS S and this is interpreted physically as the rate of ﬂow through S.
If we write F
v, then F is also a vector ﬁeld on 3 and the integral in Equation 6
becomes yy F n dS S A surface integral of this form occurs frequently in physics, even when F is not v, and is
called the surface integral (or ﬂux integral ) of F over S.
7 Definition If F is a continuous vector ﬁeld deﬁned on an oriented surface S with
unit normal vector n, then the surface integral of F over S is yy F
S dS yy F n dS S This integral is also called the ﬂux of F across S.
In words, Deﬁnition 7 says that the surface integral of a vector ﬁeld over S is equal to
the surface integral of its normal component over S (as previously deﬁned).
Graphs In the case of a surface S given by a graph z t x, y , we can ﬁnd n by
noting that S is also the level surface f x, y, z
z t x, y
0. We know that the gradient ∇f x, y, z is normal to this surface at x, y, z and so a unit normal vector is
n f x, y, z
f x, y, z t x x, y i
s t x x, y 2 t y x, y j k
t y x, y 2 1 5E17(pp 11501159) 1152 ❙❙❙❙ 1/19/06 3:07 PM Page 1152 CHAPTER 17 VECTOR CALCULUS Since the kcomponent is positive, this is the upward unit normal. If we now use Formula 2
to evaluate the surface integral (7) with
F x, y, z P x, y, z i Q x, y, z j R x, y, z k we get yy F dS S yy F n dS S yy Pi Qj t
i
x Rk 2 t
x D t
j
y k t
y 2 t
x 2 1 t
y 2 1 dA or yy F 8 yy dS S P t
x Q t
y R dA D For a downward unit normal we multiply by 1. Similar formulas can be worked out if S
is given by y h x, z or x k y, z . (See Exercises 33 and 34.)
EXAMPLE 4 Evaluate xxS F d S, where F x, y, z
yi xj
of the solid region E enclosed by the paraboloid z 1 x 2
z z k and S is the boundary
y 2 and the plane z 0. SOLUTION S consists of a parabolic top surface S1 and a circular bottom surface S2. (See Figure 12.) Since S is a closed surface, we use the convention of positive (outward) orientation. This means that S1 is oriented upward and we can use Equation 8 with D being
the projection of S1 on the xyplane, namely, the disk x 2 y 2 1. Since S¡ P x, y, z S™ Q x, y, z y R x, y, z x z x2 1 y
x FIGURE 12 t
x on S1 and t
y 2x 2y we have yy F
S1 dS yy P t
x Q t
y R dA D yy y 2x x 4 xy x2 2y x2 1 y2 dA D yy 1 y 2 dA D
2 yy
0 2 0 yy
0 2 y(
0 1 1 0 1
4 1 4r 2 cos sin r r3
cos r 2 r dr d 4r 3 cos sin
sin )d 1
4 2 dr d
0 2 y2 5E17(pp 11501159) 1/19/06 3:07 PM Page 1153 SECTION 17.7 SURFACE INTEGRALS The disk S2 is oriented downward, so its unit normal vector is n yy F yy F dS S2 yy k dS S2 1153 k and we have yy 0 d A z dA D ❙❙❙❙ 0 D since z 0 on S2 . Finally, we compute, by deﬁnition, xxS F d S as the sum of the surface integrals of F over the pieces S1 and S2 : yy F yy F dS S yy F dS S1 dS 0 2 S2 2 Parametric Surfaces If S is given by a vector function r u, v , then n is given by Equation 5, and from Deﬁnition 7 and Equation 3 we have yy F ru
ru yy F dS S S yy rv
dS
rv
ru
ru F r u, v rv
rv ru rv d A D where D is the parameter domain. Thus, we have
 Compare Equation 9 to the similar expression for evaluating line integrals of vector ﬁelds
in Deﬁnition 17.2.13: y C F dr y b a Frt yy F 9 yy F dS S r t dt ru rv d A D Notice that, in view of Equation 17.6.7, we have
rx t
i
x ry t
j
y k and so Formula 8 is just a special case of Formula 9.
 Figure 13 shows the vector ﬁeld F in
Example 5 at points on the unit sphere.
z EXAMPLE 5 Find the ﬂux of the vector ﬁeld F x, y, z sphere x 2 y2 z2 zi yj x k across the unit 1. SOLUTION Using the parametric representation r, sin cos i sin sin j cos k 0 0 2 we have
Fr , cos i sin sin j sin cos k y and, from Example 10 in Section 17.6,
x r
FIGURE 13 sin 2 r cos i sin 2 sin j sin cos sin 3 sin 2 cos k Therefore
Fr , r r cos sin 2 sin 2 cos cos 5E17(pp 11501159) 1154 ❙❙❙❙ 1/19/06 3:08 PM Page 1154 CHAPTER 17 VECTOR CALCULUS and, by Formula 9, the ﬂux is yy F
S dS yy F r r dA D
2 2 sin 2 yy
0 0 2 y sin2
0 0 y 0 cos cos y d sin 3 d sin 3 cos
2 cos d 0 y 2 0 sin 2 y 0 dd sin3 d
2 sin 2 d since y cos d
0 y 2 0 sin2 d 0 4
3
by the same calculation as in Example 2.
If, for instance, the vector ﬁeld in Example 5 is a velocity ﬁeld describing the ﬂow of a
ﬂuid with density 1, then the answer, 4 3, represents the rate of ﬂow through the unit
sphere in units of mass per unit time.
Although we motivated the surface integral of a vector ﬁeld using the example of ﬂuid
ﬂow, this concept also arises in other physical situations. For instance, if E is an electric
ﬁeld (see Example 5 in Section 17.1), then the surface integral yy E dS S is called the electric ﬂux of E through the surface S. One of the important laws of electrostatics is Gauss’s Law, which says that the net charge enclosed by a closed surface S is
Q 10 0 yy E dS S where is a constant (called the permittivity of free space) that depends on the units used.
(In the SI system, 0 8.8542 10 12 C 2 N m 2.) Therefore, if the vector ﬁeld
F in Example 5 represents an electric ﬁeld, we can conclude that the charge enclosed by S
is Q 4 0 3.
Another application of surface integrals occurs in the study of heat ﬂow. Suppose the
temperature at a point x, y, z in a body is u x, y, z . Then the heat ﬂow is deﬁned as the
vector ﬁeld
F
K ∇u
0 where K is an experimentally determined constant called the conductivity of the substance. The rate of heat ﬂow across the surface S in the body is then given by the surface
integral yy F
S dS K yy ∇u d S
S EXAMPLE 6 The temperature u in a metal ball is proportional to the square of the distance
from the center of the ball. Find the rate of heat ﬂow across a sphere S of radius a with
center at the center of the ball. 5E17(pp 11501159) 1/19/06 3:08 PM Page 1155 S ECTION 17.7 SURFACE INTEGRALS ❙❙❙❙ 1155 SOLUTION Taking the center of the ball to be at the origin, we have C x2 u x, y, z y2 z2 where C is the proportionality constant. Then the heat ﬂow is
F x, y, z Ku KC 2 x i 2z k 2y j where K is the conductivity of the metal. Instead of using the usual parametrization
of the sphere as in Example 5, we observe that the outward unit normal to the sphere
x 2 y 2 z 2 a 2 at the point x, y, z is
1
xi
a n and so 2KC 2
x
a Fn But on S we have x 2
ﬂow across S is yy F
S y2 dS z2 yy F n dS 1, 1, 1 . Approximate
2y 2 3z 2 dS by using a Riemann sum as in Deﬁnition 1, taking the patches Sij to be the squares that are the faces
*
of the cube and the points Pij to be the centers of the squares. 2 aKC yy dS
S 2 aKC 4 a 2 8KC a 3 6. xxS x y dS,
S is the triangular region with vertices (1, 0, 0), (0, 2, 0),
and (0, 0, 2) 7. xxS y z dS, 2. A surface S consists of the cylinder x 2 y 2 1, 1 z 1,
together with its top and bottom disks. Suppose you know
that f is a continuous function with f 1, 0, 0
2,
f 0, 1, 0
3, and f 0, 0, 1
4. Estimate the value of
xxS f x, y, z dS by using a Riemann sum, taking the patches Sij
to be four quartercylinders and the top and bottom disks.
y 2 z 2 50, z 0, and
suppose f is a continuous function with f 3, 4, 5
7,
f 3, 4, 5
8, f 3, 4, 5
9, and f 3, 4, 5
12.
By dividing H into four patches, estimate the value of
xxH f x, y, z dS. S is the part of the plane x
ﬁrst octant
8. t(s x 2 y 2 z 2 ), where t is a
function of one variable such that t 2
5. Evaluate
xxS f x, y, z dS, where S is the sphere x 2 y 2 z 2 4. 4. Suppose that f x, y, z 9. 10.
11. x3 2 1 that lies in the y3 2 , 0 z dS,
S is the part of the cone z 2
planes z 1 and z 3 xxS z dS,
y x2 2z 2, 0 x 1, 0 y xxS y dS, z 1, 0 x2 1 z 2 that lies inside the xxS x y dS,
S is the boundary of the region enclosed by the cylinder
x 2 z 2 1 and the planes y 0 and x y 2 xxS x 2 y z dS,
2x 3y that lies above the 13. xxS 1 y 2 that lies between the y S is the part of the paraboloid y
cylinder x 2 z 2 4
12. 1 xxS x 2
3 z y 22 S is the surface x Evaluate the surface integral. S is the part of the plane z
rectangle 0, 3
0, 2 xxS y dS,
S is the surface z 3. Let H be the hemisphere x 2 5. 2 aKC. Therefore, the rate of heat S xxS s x 2  z2 Exercises 1. Let S be the cube with vertices 5–18 y2 a 2, so F n 2 aKCA S  17.7 zk yj x 2 z y 2 z dS,
S is the hemisphere x 2 y2 z2 4, z 0 5E17(pp 11501159) 1156 14. ❙❙❙❙ 1/19/06 3:09 PM CHAPTER 17 VECTOR CALCULUS xxS x y z dS, CAS S is the part of the sphere x
cone z s x 2 y 2
15. Page 1156 2 y xxS x 2 y z 2 dS,
S is the part of the cylinder x 2
z 0 and z 2 2 z 2 1 that lies above the 9 between the planes 16. xxS x 2 y 2 z 2 dS,
S consists of the cylinder in Exercise 15 together with its top
and bottom disks 17. xxS y z dS, u 2, y S is the surface with parametric equations x
z u cos v, 0 u 1, 0 v
2
18. x 2 y 2 dS,
S is the helicoid with vector equation
r u, v
u cos v i u sin v j v k, 0
■ ■ ■ ■ ■ ■ 1, 0 u ■ ■ v ■ ■ x y i 4 x 2 j y z k, S is the surface z
1, 0 y 1, with upward orientation 20. F x, y, z 0 x ■ x z e y i x z e y j z k,
S is the part of the plane x y z
has downward orientation z 1 in the ﬁrst octant and 1 z2 4 in the ﬁrst octant, x 25, y 0, oriented in the 2 2 z,0 CAS 1, 1 x y, 0 x ■ ■ ■ ■ ■ 1, 0 y 1. 30. Find the exact value of xxS x 2 y z dS, where S is the surface in Exercise 29. v
y i x j 2 z k. Find the rate of ﬂow outward through
the sphere x 2 y 2 z 2 25.
y 2 z 2 a 2, z
y j 2 z k. 0, if the electric ﬁeld is 43. The temperature at the point x, y, z in a substance with con 29. Evaluate xxS x y z dS correct to four decimal places, where S is the surface z
CAS ■ z k. Find the rate of ﬂow upward through the
9 1 x 2 y 2 , x 2 y 2 36.
4 with vertices 1, 1, 1 if the electric ﬁeld is
E x, y, z
x i y j z k. y i x j z k,
S is the helicoid of Exercise 18 with upward orientation
■ v yi j
paraboloid z 42. Use Gauss’s Law to ﬁnd the charge enclosed by the cube 2 ■ x 2 y 2, 0 z a, has constant
density k. Find (a) the center of mass and (b) the moment of
inertia about the z axis. hemisphere x 2
E x, y, z
xi 1, and the y S is the surface of Exercise 12 x i 2y j 3z k,
S is the cube with vertices 1, ■ 4, if its density function is 40. A ﬂuid has density 1500 and velocity ﬁeld 27. F x, y, z ■ y 2, 1 z
10 z. 41. Use Gauss’s Law to ﬁnd the charge contained in the solid y j z k,
S consists of the paraboloid y
disk x 2 z 2 1, y 1 ■ sx 2
x, y, z 38. The conical surface z 2 25. F x, y, z 28. F x, y, z 0, if it has constant density. 39. A ﬂuid with density 1200 ﬂows with velocity x z i x j y k,
S is the hemisphere x 2 y 2 z 2
direction of the positive yaxis 5 k, a 2, about the z axis of a thin sheet in the shape of a surface S
if the density function is .
(b) Find the moment of inertia about the z axis of the funnel in
Exercise 36. 24. F x, y, z yj z2
37. (a) Give an integral expression for the moment of inertia I z x e y, y 2 beneath the plane z x i z j y k,
S is the part of the sphere x 2 y 2
with orientation toward the origin xi y2 36. Find the mass of a thin funnel in the shape of a cone 23. F x, y, z 26. F x, y, z h x, z and n is the unit normal that where S is given by x k y, z and n is the unit normal that
points forward (that is, toward the viewer when the axes are
drawn in the usual way).
z 4 x i y j z k,
S is the part of the cone z sx 2
with downward orientation sin x y z i x 2 y j z 2e x 5 k
across the part of the cylinder 4y 2 z 2 4 that lies above
the xyplane and between the planes x
2 and x 2 with
upward orientation. Illustrate by using a computer algebra
system to draw the cylinder and the vector ﬁeld on the same
screen. 35. Find the center of mass of the hemisphere x 2 21. F x, y, z 22. F x, y, z y 2 that 34. Find a formula for xxS F d S similar to Formula 8 for the case x y i y z j z x k, S is the part of the parabo4 x 2 y 2 that lies above the square 0 x 1,
1, and has upward orientation loid z
0y 2x 2 32. Find the ﬂux of F x, y, z where S is given by y
points toward the left. 19–28  Evaluate the surface integral xxS F d S for the given
vector ﬁeld F and the oriented surface S. In other words, ﬁnd the
ﬂux of F across S. For closed surfaces, use the positive (outward)
orientation. 19. F x, y, z 3 33. Find a formula for xxS F d S similar to Formula 8 for the case u sin v, xxS s1 ■ where S is the part of the paraboloid z
lies above the x yplane.
CAS y2 31. Find the value of xxS x 2 y 2z 2 dS correct to four decimal places, ■ ductivity K 6.5 is u x, y, z
2 y 2 2 z 2. Find the rate of
heat ﬂow inward across the cylindrical surface y 2 z 2 6,
0 x 4.
44. The temperature at a point in a ball with conductivity K is inversely proportional to the distance from the center of the
ball. Find the rate of heat ﬂow across a sphere S of radius a
with center at the center of the ball. 5E17(pp 11501159) 1/19/06 3:09 PM Page 1157 SECTION 17.8 STOKES’ THEOREM  17.8 n
S C
0
x 1157 Stokes’ Theorem z n ❙❙❙❙ y FIGURE 1  Stokes’ Theorem is named after the Irish
mathematical physicist Sir George Stokes
(1819–1903). Stokes was a professor at Cambridge University (in fact he held the same
position as Newton, Lucasian Professor of
Mathematics) and was especially noted for his
studies of ﬂuid ﬂow and light. What we call
Stokes’ Theorem was actually discovered by
the Scottish physicist Sir William Thomson
(1824–1907, known as Lord Kelvin). Stokes
learned of this theorem in a letter from Thomson
in 1850 and asked students to prove it on an
examination at Cambridge University in 1854.
We don’t know if any of those students was
able to do so. Stokes’ Theorem can be regarded as a higherdimensional version of Green’s Theorem.
Whereas Green’s Theorem relates a double integral over a plane region D to a line integral
around its plane boundary curve, Stokes’ Theorem relates a surface integral over a surface
S to a line integral around the boundary curve of S (which is a space curve). Figure 1 shows
an oriented surface with unit normal vector n. The orientation of S induces the positive
orientation of the boundary curve C shown in the ﬁgure. This means that if you walk in
the positive direction around C with your head pointing in the direction of n, then the surface will always be on your left.
Stokes’ Theorem Let S be an oriented piecewisesmooth surface that is bounded by a simple, closed, piecewisesmooth boundary curve C with positive orientation.
Let F be a vector ﬁeld whose components have continuous partial derivatives on an
open region in 3 that contains S. Then y C F dr yy curl F dS S Since y C F dr y C F T ds yy curl F and dS S yy curl F n dS S Stokes’ Theorem says that the line integral around the boundary curve of S of the tangential component of F is equal to the surface integral of the normal component of the curl
of F.
The positively oriented boundary curve of the oriented surface S is often written as
S, so Stokes’ Theorem can be expressed as yy curl F 1 dS y S F dr S There is an analogy among Stokes’ Theorem, Green’s Theorem, and the Fundamental
Theorem of Calculus. As before, there is an integral involving derivatives on the left side
of Equation 1 (recall that curl F is a sort of derivative of F ) and the right side involves the
values of F only on the boundary of S.
In fact, in the special case where the surface S is ﬂat and lies in the xyplane with
upward orientation, the unit normal is k, the surface integral becomes a double integral,
and Stokes’ Theorem becomes y C F dr yy curl F
S dS yy curl F k dA S This is precisely the vector form of Green’s Theorem given in Equation 17.5.12. Thus, we
see that Green’s Theorem is really a special case of Stokes’ Theorem.
Although Stokes’ Theorem is too difﬁcult for us to prove in its full generality, we can
give a proof when S is a graph and F, S, and C are well behaved. 5E17(pp 11501159) 1158 ❙❙❙❙ 1/19/06 3:10 PM Page 1158 CHAPTER 17 VECTOR CALCULUS z Proof of a Special Case of Stokes’ Theorem We assume that the equation of S is z t x, y ,
x, y
D, where t has continuous secondorder partial derivatives and D is a simple
plane region whose boundary curve C1 corresponds to C. If the orientation of S is
upward, then the positive orientation of C corresponds to the positive orientation of C1.
(See Figure 2.) We are given that F P i Q j R k, where the partial derivatives of
P, Q, and R are continuous.
Since S is a graph of a function, we can apply Formula 17.7.8 with F replaced by
curl F. The result is n
z=g(x, y)
S
0
x C D y 2 C¡ yy curl F dS S FIGURE 2 R
y yy z
x Q
z P
z z
y R
x Q
x P
y dA D where the partial derivatives of P, Q, and R are evaluated at x, y, t x, y . If
x xt y yt a t b is a parametric representation of C1, then a parametric representation of C is
x xt y z yt t x t ,y t a t b This allows us, with the aid of the Chain Rule, to evaluate the line integral as follows: y C F dr y P dx
dt Q dy
dt P dx
dt Q dy
dt R a y b a y z
x b P R a y dz
dt R b P R C1 yy x Q z dx
x dt dx
dt z
x Q dx Q z
y R dt
z dy
y dt
z
y R
R z
y dy
dt dt dy P y dt z
x R dA D where we have used Green’s Theorem in the last step. Then, using the Chain Rule again
and remembering that P, Q, and R are functions of x, y, and z and that z is itself a function of x and y, we get y C F dr yy Q
x Q
z z
x R
x z
y R
z z
x z
y 2 R z xy D P
y P
z z
y R
y z
x R
z z
y z
x 2 R z yx Four of the terms in this double integral cancel and the remaining six terms can be
arranged to coincide with the right side of Equation 2. Therefore y C F dr yy curl F
S dS dA 5E17(pp 11501159) 1/19/06 3:11 PM Page 1159 S ECTION 17.8 STOKES’ THEOREM ❙❙❙❙ 1159 xC F d r, where F x, y, z
y 2 i x j z 2 k and C is the curve
of intersection of the plane y z 2 and the cylinder x 2 y 2 1. (Orient C to be
counterclockwise when viewed from above.)
EXAMPLE 1 Evaluate z S SOLUTION The curve C (an ellipse) is shown in Figure 3. Although xC F d r could be
evaluated directly, it’s easier to use Stokes’ Theorem. We ﬁrst compute
C i j k x
y2 y+z=2 y
x z curl F D0
x FIGURE 3 y C F dr yy curl F z2 yy dS S
2 0 y
z ≈+¥+z@=4
S
C 1 1 0 2 0
1
2 1 2y d A D yy 2r sin r2
2 1 r3
sin
3 2 2 r dr d
d
0 2 y(
0 1
2 2
3 sin )d 0 EXAMPLE 2 Use Stokes’ Theorem to compute the integral xxS curl F d S, where
F x, y, z
x z i y z j x y k and S is the part of the sphere x 2 y 2 z 2 4 that
lies inside the cylinder x 2 y 2 1 and above the xyplane. (See Figure 4.)
SOLUTION To ﬁnd the boundary curve C we solve the equations x 2
2 2 y 2 z 2 4 and
3 and so z s3 (since z 0). Thus, C is the
1, z s3. A vector equation of C is 2 x
y
1. Subtracting, we get z
circle given by the equations x 2 y 2 0 y FIGURE 4 2y k Although there are many surfaces with boundary C, the most convenient choice is the
elliptical region S in the plane y z 2 that is bounded by C. If we orient S upward,
then C has the induced positive orientation. The projection D of S on the xyplane is the
2 y, we have
disk x 2 y 2 1 and so using Equation 17.7.8 with z t x, y y x 1 rt ≈+¥=1 cos t i rt so sin t j sin t i s3 k 0 t 2 cos t j Also, we have
Frt s3 cos t i s3 sin t j cos t sin t k Therefore, by Stokes’ Theorem, yy curl F dS y C F dr y 2 Frt 0 r t dt S
2 y( s3 cos t sin t 0 s3 y 2 0 0 dt 0 s3 sin t cos t) dt 5E17(pp 11601169) 1160 ❙❙❙❙ 1/19/06 3:14 PM Page 1160 CHAPTER 17 VECTOR CALCULUS Note that in Example 2 we computed a surface integral simply by knowing the values
of F on the boundary curve C. This means that if we have another oriented surface with
the same boundary curve C, then we get exactly the same value for the surface integral!
In general, if S1 and S2 are oriented surfaces with the same oriented boundary curve C
and both satisfy the hypotheses of Stokes’ Theorem, then yy curl F 3 y dS C yy curl F F dr S1 This fact is useful when it is difﬁcult to integrate over one surface but easy to integrate over
the other.
We now use Stokes’ Theorem to throw some light on the meaning of the curl vector.
Suppose that C is an oriented closed curve and v represents the velocity ﬁeld in ﬂuid ﬂow.
Consider the line integral T y v C dS S2 (a) jC v d r>0, positive circulation T C C y v dr C v T ds and recall that v T is the component of v in the direction of the unit tangent vector T.
This means that the closer the direction of v is to the direction of T, the larger the value of
v T. Thus, xC v d r is a measure of the tendency of the ﬂuid to move around C and is
called the circulation of v around C. (See Figure 5.)
Now let P0 x 0 , y0 , z0 be a point in the ﬂuid and let Sa be a small disk with radius a and
center P0. Then (curl F P
curl F P0 for all points P on Sa because curl F is continuous. Thus, by Stokes’ Theorem, we get the following approximation to the circulation
around the boundary circle Ca : v y (b) jC v d r<0, negative circulation Ca yy curl v v dr dS FIGURE 5 0 curl v FIGURE 6 n dS Sa yy curl v P
 Imagine a tiny paddle wheel placed in the
ﬂuid at a point P, as in Figure 6; the paddle
wheel rotates fastest when its axis is parallel
to curl v. yy curl v Sa n P0 dS curl v P0 n P0 a2 Sa This approximation becomes better as a l 0 and we have
curl v P0 4 n P0 lim al0 1
a2 y Ca v dr Equation 4 gives the relationship between the curl and the circulation. It shows that
curl v n is a measure of the rotating effect of the ﬂuid about the axis n. The curling effect
is greatest about the axis parallel to curl v.
Finally, we mention that Stokes’ Theorem can be used to prove Theorem 17.5.4 (which
states that if curl F 0 on all of 3, then F is conservative). From our previous work
(Theorems 17.3.3 and 17.3.4), we know that F is conservative if xC F d r 0 for every
closed path C. Given C, suppose we can ﬁnd an orientable surface S whose boundary is
C. (This can be done, but the proof requires advanced techniques.) Then Stokes’ Theorem
gives y C F dr yy curl F
S dS yy 0 dS 0 S A curve that is not simple can be broken into a number of simple curves, and the integrals
around these simple curves are all 0. Adding these integrals, we obtain xC F d r 0 for
any closed curve C. 5E17(pp 11601169) 1/19/06 3:15 PM Page 1161 SECTION 17.8 STOKES’ THEOREM  17.8 xi yj
x 2 y 2 k,
C is the boundary of the part of the paraboloid
z 1 x 2 y 2 in the ﬁrst octant 10. F x, y, z 3 Suppose F is a vector ﬁeld on
whose components have continuous partial derivatives. Explain why
■ yy curl F yy curl F dS H z
4 ; ;
2 x y 2 2 2. F x, y, z x 2e y z i y 2e x z j
S is the hemisphere x 2 y 2
oriented upward
2 x2 z 2e x y k,
z 2 4, z 3. F x, y, z ; ■ ■ x 2z i xy2 j z2 k y 2 i x j z 2 k,
S is the part of the paraboloid z
plane z 1, oriented upward 13. F x, y, z ■ ■ ■ ■ ■ y 2 that lies below the z x i y j x y z k,
S is the part of the plane 2 x y
octant, oriented upward 2 that lies in the ﬁrst 15. F x, y, z y i z j x k,
S is the hemisphere x 2 y 2 z 2
direction of the positive yaxis ■ ■ x2 14. F x, y, z e xy cos z i x 2 z j x y k,
S is the hemisphere x s1 y 2 z 2, oriented in the direction of the positive xaxis [Hint: Use Equation 3.]
■ ■  Verify that Stokes’ Theorem is true for the given vector
ﬁeld F and surface S. 6. F x, y, z ■ ■ 13–15 3 x y z i x y j x 2 y z k,
S consists of the top and the four sides (but not the bottom)
of the cube with vertices 1, 1, 1 , oriented outward
[Hint: Use Equation 3.] ■ ■ 0, 5. F x, y, z ■ ■ F x, y, z
x 2 y i 1 x 3 j x y k and C is the curve of
3
intersection of the hyperbolic paraboloid z y 2 x 2 and
the cylinder x 2 y 2 1 oriented counterclockwise as
viewed from above.
(b) Graph both the hyperbolic paraboloid and the cylinder with
domains chosen so that you can see the curve C and the
surface that you used in part (a).
(c) Find parametric equations for C and use them to graph C. ; y 2 that lies above x y z i sin x y z j x y z k,
S is the part of the cone y 2 x 2 z 2 that lies between the
planes y 0 and y 3, oriented in the direction of the
positive yaxis ■ ■ 12. (a) Use Stokes’ Theorem to evaluate xC F d r, where y Use Stokes’ Theorem to evaluate xxS curl F d S. y z i x z j x y k,
S is the part of the paraboloid z 9
the plane z 5, oriented upward ■ and C is the curve of intersection of the plane
x y z 1 and the cylinder x 2 y 2 9 oriented
counterclockwise as viewed from above.
(b) Graph both the plane and the cylinder with domains
chosen so that you can see the curve C and the surface
that you used in part (a).
(c) Find parametric equations for C and use them to graph C. H 2 ■ F x, y, z P 4. F x, y, z ■ 11. (a) Use Stokes’ Theorem to evaluate xC F d r, where P 4  ■ dS z 2– 6 1161 Exercises 1. A hemisphere H and a portion P of a paraboloid are shown. x ❙❙❙❙ ■ ■ ■ ■ ■ ■ 1, y 0, oriented in the ■ ■ ■ ■ ■ ■ 16. Let Use Stokes’ Theorem to evaluate xC F d r. In each case C
is oriented counterclockwise as viewed from above.
7–10  x y2 i
y z2 j
z x 2 k,
C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) 7. F x, y, z e x i e x j e z k,
C is the boundary of the part of the plane 2 x
the ﬁrst octant 8. F x, y, z 9. F x, y, z yz i
C is the circle x 2 xy 2 x z j e k,
y 2 16, z 5 y 2z 2 in ax 3 F x, y, z 3 x z 2, x 2 y b y 3, c z 3 Let C be the curve in Exercise 12 and consider all possible
smooth surfaces S whose boundary curve is C. Find the values
of a, b, and c for which xxS F d S is independent of the choice
of S.
17. Calculate the work done by the force ﬁeld F x, y, z xx z2 i yy x2 j zz y2 k when a particle moves under its inﬂuence around the edge of 5E17(pp 11601169) 1162 ❙❙❙❙ 1/19/06 3:15 PM Page 1162 CHAPTER 17 VECTOR CALCULUS the part of the sphere x 2 y 2 z 2 4 that lies in the ﬁrst
octant, in a counterclockwise direction as viewed from above.
18. Evaluate xC y sin x d x
z 2 cos y d y x 3 dz, where
C is the curve r t
sin t, cos t, sin 2 t , 0 t 2 .
[Hint: Observe that C lies on the surface z 2 x y.] 19. If S is a sphere and F satisﬁes the hypotheses of Stokes’ Theorem, show that xxS curl F d S 0. 20. Suppose S and C satisfy the hypotheses of Stokes’ Theorem and f , t have continuous secondorder partial derivatives. Use
Exercises 24 and 26 in Section 17.5 to show the following.
(a) xC f t d r xxS f
t dS
(b)
(c) xC
xC ff dr ft tf 0
dr 0 WRITING PROJECT
Three Men and Two Theorems  The photograph shows a stainedglass
window at Cambridge University in honor of
George Green. Although two of the most important theorems in vector calculus are named after George Green
and George Stokes, a third man, William Thomson (also known as Lord Kelvin), played a large
role in the formulation, dissemination, and application of both of these results. All three men
were interested in how the two theorems could help to explain and predict physical phenomena
in electricity and magnetism and ﬂuid ﬂow. The basic facts of the story are given in the margin
notes on pages 1120 and 1157.
Write a report on the historical origins of Green’s Theorem and Stokes’ Theorem. Explain the
similarities and relationship between the theorems. Discuss the roles that Green, Thomson, and
Stokes played in discovering these theorems and making them widely known. Show how both
theorems arose from the investigation of electricity and magnetism and were later used to study a
variety of physical problems.
The dictionary edited by Gillispie [2] is a good source for both biographical and scientiﬁc
information. The book by Hutchinson [5] gives an account of Stokes’ life and the book by
Thompson [8] is a biography of Lord Kelvin. The articles by GrattanGuinness [3] and Gray [4]
and the book by Cannell [1] give background on the extraordinary life and works of Green.
Additional historical and mathematical information is found in the books by Katz [6] and
Kline [7].
1. D. M. Cannell, George Green, Mathematician and Physicist 1793–1841: The Background to His Life and Work (Philadelphia: Society for Industrial and Applied Mathematics, 2001).
2. C. C. Gillispie, ed., Dictionary of Scientiﬁc Biography (New York: Scribner’s, 1974). See the article on Green by P. J. Wallis in Volume XV and the articles on Thomson by Jed Buchwald
and on Stokes by E. M. Parkinson in Volume XIII.
3. I. GrattanGuinness, “Why did George Green write his essay of 1828 on electricity and magnetism?” Amer. Math. Monthly, Vol. 102 (1995), pp. 387–396.
4. J. Gray, “There was a jolly miller.” The New Scientist, Vol. 139 (1993), pp. 24–27.
5. G. E. Hutchinson, The Enchanted Voyage and Other Studies (Westport Conn.: Greenwood Press, 1978).
6. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993), pp. 678–680.
7. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford University Press, 1972), pp. 683–685.
8. Sylvanus P. Thompson, The Life of Lord Kelvin (New York: Chelsea, 1976). 5E17(pp 11601169) 1/19/06 3:15 PM Page 1163 SECTION 17.9 THE DIVERGENCE THEOREM  17.9 ❙❙❙❙ 1163 The Divergence Theorem
In Section 17.5 we rewrote Green’s Theorem in a vector version as y C F n ds yy div F x, y dA D where C is the positively oriented boundary curve of the plane region D. If we were seeking to extend this theorem to vector ﬁelds on 3, we might make the guess that yy F 1 yyy div F x, y, z n dS S dV E where S is the boundary surface of the solid region E. It turns out that Equation 1 is true,
under appropriate hypotheses, and is called the Divergence Theorem. Notice its similarity
to Green’s Theorem and Stokes’ Theorem in that it relates the integral of a derivative of a
function (div F in this case) over a region to the integral of the original function F over the
boundary of the region.
At this stage you may wish to review the various types of regions over which we were
able to evaluate triple integrals in Section 16.7. We state and prove the Divergence Theorem for regions E that are simultaneously of types 1, 2, and 3 and we call such regions
simple solid regions. (For instance, regions bounded by ellipsoids or rectangular boxes are
simple solid regions.) The boundary of E is a closed surface, and we use the convention,
introduced in Section 17.7, that the positive orientation is outward; that is, the unit normal
vector n is directed outward from E.
The Divergence Theorem Let E be a simple solid region and let S be the boundary  The Divergence Theorem is sometimes called
Gauss’s Theorem after the great German mathematician Karl Friedrich Gauss (1777–1855), who
discovered this theorem during his investigation
of electrostatics. In Eastern Europe the Divergence Theorem is known as Ostrogradsky’s
Theorem after the Russian mathematician
Mikhail Ostrogradsky (1801–1862), who
published this result in 1826. surface of E, given with positive (outward) orientation. Let F be a vector ﬁeld
whose component functions have continuous partial derivatives on an open region
that contains E. Then yy F yyy div F dV dS S Proof Let F Pi Qj E R k. Then
P
x Q
y R
z P
dV
x yyy Q
dV
y div F yyy div F dV yyy so E E E yyy
E R
dV
z If n is the unit outward normal of S, then the surface integral on the left side of the
Divergence Theorem is yy F
S dS yy F n dS S yy P i
S yy Pi Qj Rk n dS S n dS yy Q j
S n dS yy R k
S n dS 5E17(pp 11601169) 1164 ❙❙❙❙ 1/19/06 3:16 PM Page 1164 CHAPTER 17 VECTOR CALCULUS Therefore, to prove the Divergence Theorem, it sufﬁces to prove the following three
equations: yy P i 2 n dS yyy S E yy Q j 3 n dS yyy S E yy R k 4 n dS yyy S E P
dV
x
Q
dV
y
R
dV
z To prove Equation 4 we use the fact that E is a type 1 region:
x, y, z E x, y z D, u1 x, y u 2 x, y where D is the projection of E onto the xyplane. By Equation 16.7.6, we have
R
dV
z yyy
E yy y R
x, y, z dz d A
z u 2 x, y u1 x, y D and, therefore, by the Fundamental Theorem of Calculus,
5 yyy
E z S™ { z=u™(x, y)} S£ E R
dV
z yy R x, y, u 2 x, y x y yy 0 dS n dS S3 S¡ (z=u¡(x, y))
D dA The boundary surface S consists of three pieces: the bottom surface S1 , the top surface
S2 , and possibly a vertical surface S3 , which lies above the boundary curve of D. (See
Figure 1. It might happen that S3 doesn’t appear, as in the case of a sphere.) Notice that
on S3 we have k n 0, because k is vertical and n is horizontal, and so yy R k
0 R x, y, u1 x, y D 0 S3 Thus, regardless of whether there is a vertical surface, we can write
6 yy R k n dS yy R k S FIGURE 1 yy R k n dS S1 n dS S2 D, and the outward normal n points
The equation of S2 is z u 2 x, y , x, y
upward, so from Equation 17.7.8 (with F replaced by R k) we have yy R k n dS S2 On S1 we have z
multiply by 1: yy R x, y, u x, y d A 2 D u1 x, y , but here the outward normal n points downward, so we yy R k
S1 n dS yy R x, y, u 1 D x, y d A 5E17(pp 11601169) 1/19/06 3:16 PM Page 1165 S ECTION 17.9 THE DIVERGENCE THEOREM ❙❙❙❙ 1165 Therefore, Equation 6 gives yy R k n dS yy S R x, y, u 2 x, y R x, y, u1 x, y dA D Comparison with Equation 5 shows that yy R k
S  Notice that the method of proof of the
Divergence Theorem is very similar to that of
Green’s Theorem. R
dV
z yyy n dS E Equations 2 and 3 are proved in a similar manner using the expressions for E as a type 2
or type 3 region, respectively.
EXAMPLE 1 Find the ﬂux of the vector ﬁeld F x, y, z sphere x 2 y2 z2 zi yj x k over the unit 1. SOLUTION First we compute the divergence of F : div F z x y y z x 1 The unit sphere S is the boundary of the unit ball B given by x 2
the Divergence Theorem gives the ﬂux as yy F B yy F 1. Thus, B VB
EXAMPLE 2 Evaluate z2 yyy div F dV yyy 1 dV dS S  The solution in Example 1 should be
compared with the solution in Example 5 in
Section 17.7. y2 4
3 1 3 4
3 d S, where S F x, y, z xy i (y2 e xz ) j
2 sin x y k and S is the surface of the region E bounded by the parabolic cylinder z
the planes z 0, y 0, and y z 2. (See Figure 2.) 1 x 2 and z
(0, 0, 1) y=2z 0
(1, 0, 0) (0, 2, 0) y x FIGURE 2 z=1≈
SOLUTION It would be extremely difﬁcult to evaluate the given surface integral directly.
(We would have to evaluate four surface integrals corresponding to the four pieces of S.) 5E17(pp 11601169) 1166 ❙❙❙❙ 1/19/06 3:17 PM Page 1166 CHAPTER 17 VECTOR CALCULUS Furthermore, the divergence of F is much less complicated than F itself:
div F x
y (y2
y xy
2y e xz 2 ) z sin x y 3y Therefore, we use the Divergence Theorem to transform the given surface integral into a
triple integral. The easiest way to evaluate the triple integral is to express E as a type 3
region:
E x, y, z 1 x z 1, 0 x 2, 0 1 y 2 z Then we have yy F
S dS yyy div F dV yyy 3y dV
E E 3y
3y
3
2 1 y 1 y y 1x y d y dz d x
z 2 1 x2 dx 3 1 1 3 dz d x 2 1 y 2 z 2 2 1 x2 x6 1
3x 4 0
3 8 dx
3x 2 7 dx 184
35 Although we have proved the Divergence Theorem only for simple solid regions, it can
be proved for regions that are ﬁnite unions of simple solid regions. (The procedure is similar to the one we used in Section 17.4 to extend Green’s Theorem.)
For example, let’s consider the region E that lies between the closed surfaces S1 and S2 ,
where S1 lies inside S2. Let n1 and n 2 be outward normals of S1 and S2 . Then the boundary
n1 on S1 and n n 2 on S2.
surface of E is S S1 S2 and its normal n is given by n
(See Figure 3.) Applying the Divergence Theorem to S, we get n™
n¡
S¡ 7 yyy div F dV yy F
E FIGURE 3 0 0 1 0 _n ¡ y 2z 1
1
2 S™ 2 0 1 y 1x yy F dS S n dS S yy F n1 dS yy F S1 n 2 dS S2 yy F dS S1 yy F dS S2 Let’s apply this to the electric ﬁeld (see Example 5 in Section 17.1):
Ex Q
x
x3 5E17(pp 11601169) 1/19/06 3:17 PM Page 1167 S ECTION 17.9 THE DIVERGENCE THEOREM ❙❙❙❙ 1167 where S1 is a small sphere with radius a and center the origin. You can verify that
div E 0. (See Exercise 21.) Therefore, Equation 7 gives yy E yy E dS S2 dS S1 yyy div E dV
E yy E dS S1 yy E n dS S1 The point of this calculation is that we can compute the surface integral over S1 because S1
is a sphere. The normal vector at x is x x . Therefore
Q
x
x3 En x
x Q
x2
since the equation of S1 is x Q
xx
x4 Q
a2 a. Thus, we have yy E yy E dS S2 n dS S1 Q
a2 Q
A S1
a2 yy dS
S1 Q
4 a2
a2 4 Q This shows that the electric ﬂux of E is 4 Q through any closed surface S2 that contains
the origin. [This is a special case of Gauss’s Law (Equation 17.7.10) for a single charge.
1 4 0 .]
The relationship between and 0 is
Another application of the Divergence Theorem occurs in ﬂuid ﬂow. Let v x, y, z be
v is the rate of ﬂow per
the velocity ﬁeld of a ﬂuid with constant density . Then F
unit area. If P0 x 0 , y0 , z0 is a point in the ﬂuid and Ba is a ball with center P0 and very small
div F P0 for all points in Ba since div F is continuous. We
radius a, then div F P
approximate the ﬂux over the boundary sphere Sa as follows: yy F
Sa dS yyy div F dV
Ba yyy div F P 0 dV Ba div F P0 V Ba
This approximation becomes better as a l 0 and suggests that
8 div F P0 lim al0 1
V Ba yy F dS Sa Equation 8 says that div F P0 is the net rate of outward ﬂux per unit volume at P0. (This
is the reason for the name divergence.) If div F P
0, the net ﬂow is outward near P and
P is called a source. If div F P
0, the net ﬂow is inward near P and P is called a sink. 5E17(pp 11601169) 1168 ❙❙❙❙ 1/19/06 3:18 PM Page 1168 CHAPTER 17 VECTOR CALCULUS For the vector ﬁeld in Figure 4, it appears that the vectors that end near P1 are shorter
than the vectors that start near P1. Thus, the net ﬂow is outward near P1, so div F P1
0
and P1 is a source. Near P2 , on the other hand, the incoming arrows are longer than the
outgoing arrows. Here the net ﬂow is inward, so div F P2
0 and P2 is a sink. We
can use the formula for F to conﬁrm this impression. Since F x 2 i y 2 j, we have
div F 2 x 2 y, which is positive when y
x. So the points above the line y
x
are sources and those below are sinks.
y P¡ x P™
FIGURE 4 The vector field F=≈ i+¥ j  17.9 Exercises 1. A vector ﬁeld F is shown. Use the interpretation of divergence derived in this section to determine whether div F is positive or
negative at P1 and at P2. 3–6  Verify that the Divergence Theorem is true for the vector
ﬁeld F on the region E. 3. F x, y, z 3 x i x y j 2 x z k,
E is the cube bounded by the planes x
y 1, z 0, and z 1 2 0, x x 2 i x y j z k,
E is the solid bounded by the paraboloid z
the xyplane 1, y 0, 4. F x, y, z P¡
_2 2 x2 4 y 2 and 5. F x, y, z x y i y z j z x k,
E is the solid cylinder x 2 y 2 1, 0 P™ z 1 6. F x, y, z x i y j z k,
E is the unit ball x 2 y 2 z 2 _2
■ 2. (a) Are the points P1 and P2 sources or sinks for the vector ﬁeld F shown in the ﬁgure? Give an explanation based
solely on the picture.
(b) Given that F x, y
x, y 2 , use the deﬁnition of divergence to verify your answer to part (a).
2 ■ ■ ■ ■ ■ 1
■ ■ ■ ■ ■ 7–17  Use the Divergence Theorem to calculate the surface integral xxS F d S; that is, calculate the ﬂux of F across S. e x sin y i e x cos y j y z 2 k,
S is the surface of the box bounded by the planes x
y 0, y 1, z 0, and z 2 7. F x, y, z x 2z 3 i 2 x y z 3 j x z 4 k,
S is the surface of the box with vertices 0, x 8. F x, y, z
P¡
2
P™ z 1, 2, 3 3 x 3 y i x 2 y 2 j x 2 y z k,
S is the surface of the solid bounded by the hyperboloid
x 2 y 2 z 2 1 and the planes z
2 and z 2 10. F x, y, z
_2 2 3 x y i x e j z k,
S is the surface of the solid bounded by the cylinder
y 2 z 2 1 and the planes x
1 and x 2 9. F x, y, z
_2 ■ 1, 5E17(pp 11601169) 1/19/06 3:19 PM Page 1169 S ECTION 17.9 THE DIVERGENCE THEOREM  Prove each identity, assuming that S and E satisfy the
conditions of the Divergence Theorem and the scalar functions
and components of the vector ﬁelds have continuous secondorder
partial derivatives. x 2 y i x y 2 j 2 x y z k,
S is the surface of the tetrahedron bounded by the planes
x 0, y 0, z 0, and x 2y z 2 12. F x, y, z 23. sin y x 2 z k,
cos z x y 2 i x e z j
S is the surface of the solid bounded by the paraboloid
z x 2 y 2 and the plane z 4 13. F x, y, z 14. F x, y, z 32 1169 23–28 x y sin z i cos x z j y cos z k,
S is the ellipsoid x 2 a 2 y 2 b 2 z 2 c 2 1 11. F x, y, z 4 ❙❙❙❙ yy a 0, where a is a constant vector n dS S
1
3 24. V E yy F d S, where F x, y, z xi zk yj S 2 x i x z j 4 x y z k,
S is the surface of the solid bounded by the cylinder
x 2 y 2 1 and the planes z x 2 and z 0 25. 4 x 3z i 4 y 3z j 3 z 4 k,
S is the sphere with radius R and center the origin 26. yy curl F dS 0 S 15. F x, y, z yy D n f dS yyy S y 3 z sin x j 3z k,
x 3 y sin z i
S is the surface of the solid bounded by the hemispheres
z s4 x 2 y 2, z s1 x 2 y 2 and the plane z 0 2 f dV E 16. F x, y, z CAS e y tan z i y s3 x 2 j x sin y k,
S is the surface of the solid that lies above the xyplane
and below the surface z 2 x 4 y 4, 1 x 1,
1y1
■ ■ ■ ■ ■ ■ ■ ■ ■ 28.
■ ■ ■ F x, y, z
sin x cos 2 y i sin 3 y cos 4z j sin 5z cos 6x k
in the cube cut from the ﬁrst octant by the planes x
2,
y
2, and z
2. Then compute the ﬂux across the
surface of the cube.
19. Use the Divergence Theorem to evaluate xxS F d S, where (1 y3
3 z2x i tan z) j x 2z y2 k and S is the top half of the sphere x 2 y 2 z 2 1.
[Hint: Note that S is not a closed surface. First compute
integrals over S1 and S2, where S1 is the disk x 2 y 2 1,
oriented downward, and S2 S S1.]
z tan 1 y 2 i z 3 ln x 2 1 j z k. Find the
ﬂux of F across the part of the paraboloid x 2 y 2 z 2
that lies above the plane z 1 and is oriented upward. 20. Let F x, y, z 21. Verify that div E 0 for the electric ﬁeld
Q
x
x3 Ex 2x 2y
y2 z 2 dS
z2 yyy 2 f f t t dV E yy ft tf yyy n dS f 2 t t 2 f dV E
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 29. Suppose S and E satisfy the conditions of the Divergence Theo rem and f is a scalar function with continuous partial derivatives. Prove that yy f n dS yyy
S f dV E These surface and triple integrals of vector functions are
vectors deﬁned by integrating each component function.
[Hint: Start by applying the Divergence Theorem to F f c,
where c is an arbitrary constant vector.]
30. A solid occupies a region E with surface S and is immersed in a liquid with constant density . We set up a coordinate system
so that the xyplane coincides with the surface of the liquid and
positive values of z are measured downward into the liquid.
Then the pressure at depth z is p
t z, where t is the acceleration due to gravity (see Section 9.3). The total buoyant force
on the solid due to the pressure distribution is given by the surface integral yy p n dS
S S where S is the sphere x 2 n dS F 22. Use the Divergence Theorem to evaluate yy ft S 18. Use a computer algebra system to plot the vector ﬁeld F x, y, z yy
S 17. F x, y, z ■ CAS 27. 1. where n is the outer unit normal. Use the result of Exercise 29
to show that F
W k, where W is the weight of the liquid
displaced by the solid. (Note that F is directed upward because
z is directed downward.) The result is Archimedes’ principle:
The buoyant force on an object equals the weight of the displaced liquid. 5E17(pp 11701173) 1170 ❙❙❙❙ 1/19/06 3:20 PM Page 1170 CHAPTER 17 VECTOR CALCULUS  17.10 Summary
The main results of this chapter are all higherdimensional versions of the Fundamental
Theorem of Calculus. To help you remember them, we collect them together here (without hypotheses) so that you can see more easily their essential similarity. Notice that in
each case we have an integral of a “derivative” over a region on the left side, and the right
side involves the values of the original function only on the boundary of the region. y Fundamental Theorem of Calculus b a F x dx Fb Fa a b r(b) y Fundamental Theorem for Line Integrals C f dr f rb f ra
C r(a) C Green’s Theorem yy Q
x P
y dA y C P dx Q dy D D n Stokes’ Theorem yy curl F dS y C F dr S S C n
S Divergence Theorem yyy div F dV yy F
E S dS E n 5E17(pp 11701173) 1/19/06 3:21 PM Page 1171 C HAPTER 17 REVIEW  17 Review CONCEPT CHECK ■ 1. What is a vector ﬁeld? Give three examples that have physical P i Q j, how do you test to determine whether F is
conservative? What if F is a vector ﬁeld on 3 ? 3. (a) Write the deﬁnition of the line integral of a scalar function (d)
(e) f along a smooth curve C with respect to arc length.
How do you evaluate such a line integral?
Write expressions for the mass and center of mass of a thin
wire shaped like a curve C if the wire has linear density
function x, y .
Write the deﬁnitions of the line integrals along C of a
scalar function f with respect to x, y, and z.
How do you evaluate these line integrals? 11. (a) What is a parametric surface? What are its grid curves? (b) Write an expression for the area of a parametric surface.
(c) What is the area of a surface given by an equation
z t x, y ?
12. (a) Write the deﬁnition of the surface integral of a scalar func tion f over a surface S.
(b) How do you evaluate such an integral if S is a parametric
surface given by a vector function r u, v ?
(c) What if S is given by an equation z t x, y ?
(d) If a thin sheet has the shape of a surface S, and the density
at x, y, z is x, y, z , write expressions for the mass and
center of mass of the sheet. 4. (a) Deﬁne the line integral of a vector ﬁeld F along a smooth curve C given by a vector function r t .
(b) If F is a force ﬁeld, what does this line integral represent?
(c) If F
P, Q, R , what is the connection between the line
integral of F and the line integrals of the component functions P, Q, and R ? 13. (a) What is an oriented surface? Give an example of a non orientable surface.
(b) Deﬁne the surface integral (or ﬂux) of a vector ﬁeld F over
an oriented surface S with unit normal vector n.
(c) How do you evaluate such an integral if S is a parametric
surface given by a vector function r u, v ?
(d) What if S is given by an equation z t x, y ? 5. State the Fundamental Theorem for Line Integrals.
6. (a) What does it mean to say that xC F d r is independent of path?
(b) If you know that xC F d r is independent of path, what can
you say about F ?
7. State Green’s Theorem. 14. State Stokes’ Theorem. 8. Write expressions for the area enclosed by a curve C in terms of line integrals around C.
9. Suppose F is a vector ﬁeld on 3 15. State the Divergence Theorem.
16. In what ways are the Fundamental Theorem for Line Integrals, . (a) Deﬁne curl F.
(b) Deﬁne div F. Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem similar? ■ TRUEFALSE QUIZ Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
1. If F is a vector ﬁeld, then div F is a vector ﬁeld.
3. If f has continuous partial derivatives of all orders on f f dr 0. 3 P i Q j and Py
conservative. x C f x, y ds xC Q x in an open region D, then F is f x, y ds 7. If S is a sphere and F is a constant vector ﬁeld, then , then 0. 4. If f has continuous partial derivatives on circle, then xC 3 ■ 5. If F
6. 2. If F is a vector ﬁeld, then curl F is a vector ﬁeld. div curl ■ 10. If F (b) What is a potential function? (b)
(c) 1171 (c) If F is a velocity ﬁeld in ﬂuid ﬂow, what are the physical
interpretations of curl F and div F ? meaning.
2. (a) What is a conservative vector ﬁeld? ❙❙❙❙ and C is any xxS F dS 0. 8. There is a vector ﬁeld F such that curl F xi yj zk 5E17(pp 11701173) ❙❙❙❙ 1172 1/19/06 3:22 PM Page 1172 CHAPTER 17 VECTOR CALCULUS ■ EXERCISES 1. A vector ﬁeld F, a curve C, and a point P are shown. ■ 13–14 xC F (a) Is xC F d r positive, negative, or zero? Explain.
(b) Is div F P positive, negative, or zero? Explain.  Show that F is conservative and use this fact to evaluate
d r along the given curve. 4 x 3y 2
t sin 13. F x, y y C: r t 2xy3 i
2 x 4 y 3 x 2 y 2 4 y 3 j,
ti
2 t cos t j, 0 t 1 x e y e z j ye z k,
ey i
C is the line segment from 0, 2, 0 to 4, 0, 3 14. F x, y, z
C ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 15. Verify that Green’s Theorem is true for the line integral xC x y 2 d x x 2 y d y, where C consists of the parabola y x 2
from 1, 1 to 1, 1 and the line segment from 1, 1 to
1, 1 . x P 16. Use Green’s Theorem to evaluate y
2–9 2. Evaluate the line integral.  2 sin t, y t, z 5. xC x 8. t y d x x d y,
C is the circle x 2
orientation sin x, 0 x 2 x z dz,
t4 i t2 j t 1 curl F ■ ■ ■ y
2 ■ ■ zi yk 12. F x, y, z
■ ■ sin y i
■ ■ ey x ■ t y dy ■ 0 t t 2 f 2f t 2 f 0, show that the line
fx d y is independent of path in any simple cos t y z sin t
2 x 2e 2 y sin t 0 2y cot z d y t 2 y 2 csc 2z dz. x 2 2y that lies
above the triangle with vertices 0, 0 , 1, 0 , and 1, 2 . 25. Find the area of the part of the surface z 26. (a) Find an equation of the tangent plane at the point 4, 2, 1 to the parametric surface S given by sin z k
■ 2 f (b) Find xC 2 xe 2 y d x x 2e xy j x cos y j
■ G 24. (a) Sketch the curve C with parametric equations Show that F is a conservative vector ﬁeld. Then ﬁnd a
function f such that F ∇ f .
x y e xy i ft integral x fy d x
region D.  1 F ■ in moving a particle from the point 3, 0, 0 to the point
2, 3 along
0,
(a) A straight line
(b) The helix x 3 cos t, y t, z 3 sin t 11. F x, y f x dx 23. If f is a harmonic function, that is, xj F 22. If f and t are twice differentiable functions, show that 1
■ G and t are differentiable functions, show that y k and x
t G div F 21. If C is any piecewisesmooth simple closed plane curve and f 10. Find the work done by the force ﬁeld F x, y, z F div G C d r, where F x, y, z
e z i xz j
2
3
C is given by r t
t i t j t k, 0
■ G x 2 j and C is given by xC F 11–12 xz2 k 3yz j 2x i and G,
t 3 k, 0 d r, where F x, y
xy i
sin t i
1 t j, 0 t ■ e z sin x k e y sin z j sin y i curl G 1 with counterclockwise z d y x dz,
C consists of the line segments from 0, 0, 0 to 1, 1, 2 and
from 1, 1, 2 to 3, 1, 4 ■ x e 19. Show that there is no vector ﬁeld G such that xC y d x ■ x y 2 d y, where C
4 with counterclockwise orientation. 2 20. Show that, under conditions to be stated on the vector ﬁelds F e y dy
C is given by r t ■ y F x, y, z y2 xC sxy d x xC F 2 18. Find curl F and div F if 2 3 rt
9. 2 cos t, 0 xC x y d x y d y,
C is the sine curve y 7. x from (0, 0) to (1, 1) is the circle x 4. 6. 17. Use Green’s Theorem to evaluate xC x 2 y d x 2 xC x 3z ds,
C: x 2 xy d y where C is the triangle with vertices 0, 0 , 1, 0 , and 1, 3 . xC x ds,
C is the arc of the parabola y 3. x 3 dx s1 C ■ ■ ■ ■ r u, v v2 i uv j u2 k 0 u 3, 3 v 3 5E17(pp 11701173) 1/19/06 3:23 PM Page 1173 C HAPTER 17 REVIEW (b) Use a computer to graph the surface S and the tangent
plane found in part (a).
(c) Set up, but do not evaluate, an integral for the surface area
of S.
(d) If
x2
y2
z2
i
j
k
F x, y, z
2
2
1x
1y
1 z2 ; CAS ❙❙❙❙ 1173 37. Let 3x 2 yz F x, y, z x 3z 3y i x3y 3x j 2z k Evaluate xC F d r, where C is the curve with initial point
0, 0, 2 and terminal point 0, 3, 0 shown in the ﬁgure.
z
(0, 0, 2) ﬁnd xxS F d S correct to four decimal places.
27–30 27.
28.  xxS z dS, where S is the part of the paraboloid z
lies under the plane z 4 xxS
z 29.
30. ■ 0 Evaluate the surface integral.
x 2 y that 38. Let d S, where F x, y, z
x z i 2 y j 3 x k and S is the
sphere x 2 y 2 z 2 4 with outward orientation xxS F d S, where F x, y, z x i x y j z k and S is the
part of the paraboloid z x 2 y 2 below the plane z 1 with
upward orientation
2 ■ ■ ■ ■ ■ ■ y x 4 xxS F ■ (1, 1, 0)
(3, 0, 0) x 2 z y 2 z dS, where S is the part of the plane
4 x y that lies inside the cylinder x 2 y 2 ■ (0, 3, 0) 2 ■ ■ 2x3 F x, y 2xy2 2y 3
y2 2y i
x2 F x, y, z 2 xi yj 2x j Evaluate xC F d r, where C is shown in the ﬁgure. ■ y 31. Verify that Stokes’ Theorem is true for the vector ﬁeld
2 2x2y C 2 zk where S is the part of the paraboloid z 1 x 2
above the xyplane and S has upward orientation. y 2 that lies x 0 32. Use Stokes’ Theorem to evaluate xxS curl F d S, where F x, y, z
x 2 y z i y z 2 j z 3e xy k, S is the part of the sphere
x 2 y 2 z 2 5 that lies above the plane z 1, and S is oriented upward. 33. Use Stokes’ Theorem to evaluate xC F d r, where F x, y, z
x y i y z j z x k and C is the triangle with vertices 1, 0, 0 , 0, 1, 0 , and 0, 0, 1 , oriented counterclockwise
as viewed from above. 39. Find xxS F n dS, where F x, y, z x i y j z k and S is
the outwardly oriented surface shown in the ﬁgure (the boundary surface of a cube with a unit corner cube removed).
z 34. Use the Divergence Theorem to calculate the surface integral xxS F d S, where F x, y, z
x 3 i y 3 j z 3 k and S is the
surface of the solid bounded by the cylinder x 2 y 2 1 and
the planes z 0 and z 2. (0, 2, 2)
(2, 0, 2) 1 35. Verify that the Divergence Theorem is true for the vector ﬁeld F x, y, z xi yj zk
1 where E is the unit ball x 2 y 2 z 2 through the ellipsoid 4 x 2 xi
x2
9y 2 y 1.
x 36. Compute the outward ﬂux of F x, y, z yj zk
y 2 z2 3 2
6z2 1 36. S (2, 2, 0) 40. If the components of F have continuous second partial deriva tives and S is the boundary surface of a simple solid region,
show that xxS curl F d S 0. 5E17(pp 11741175) 1/19/06 PROBLEMS
PLUS 3:25 PM Page 1174 1. Let S be a smooth parametric surface and let P be a point such that each line that starts at P intersects S at most once. The solid angle S subtended by S at P is the set of lines starting
at P and passing through S. Let S a be the intersection of S with the surface of the sphere
with center P and radius a. Then the measure of the solid angle (in steradians) is deﬁned to be
area of S a
a2 S Apply the Divergence Theorem to the part of S between S a and S to show that
rn
dS
r3 yy S S where r is the radius vector from P to any point on S, r
r , and the unit normal vector n is
directed away from P.
This shows that the deﬁnition of the measure of a solid angle is independent of the radius a
of the sphere. Thus, the measure of the solid angle is equal to the area subtended on a unit
sphere. (Note the analogy with the deﬁnition of radian measure.) The total solid angle subtended by a sphere at its center is thus 4 steradians.
S
S(a) a P 2. Find the positively oriented simple closed curve C for which the value of the line integral y C y3 y dx 2x 3 dy is a maximum.
3. Let C be a simple closed piecewisesmooth space curve that lies in a plane with unit normal vector n
a, b, c and has positive orientation with respect to n. Show that the plane area
enclosed by C is
1
2 1174 y C bz cy dx cx az dy ay b x dz Page 1175 4. The ﬁgure depicts the sequence of events in each cylinder of a fourcylinder internal combus stio
n
hau sio
n Ex plo
Ex mp res
sio
n tion engine. Each piston moves up and down and is connected by a pivoted arm to a rotating
crankshaft. Let P t and V t be the pressure and volume within a cylinder at time t, where
a t b gives the time required for a complete cycle. The graph shows how P and V vary
through one cycle of a fourstroke engine. Co 3:25 PM ke 1/19/06 Int
a 5E17(pp 11741175) P $ Water # C
% Crankshaft
Connecting rod
Flywheel !
0 @
V During the intake stroke (from ① to ②) a mixture of air and gasoline at atmospheric pressure is drawn into a cylinder through the intake valve as the piston moves downward. Then the
piston rapidly compresses the mix with the valves closed in the compression stroke (from ② to
③) during which the pressure rises and the volume decreases. At ③ the sparkplug ignites the
fuel, raising the temperature and pressure at almost constant volume to ④. Then, with valves
closed, the rapid expansion forces the piston downward during the power stroke (from ④ to
⑤). The exhaust valve opens, temperature and pressure drop, and mechanical energy stored in
a rotating ﬂywheel pushes the piston upward, forcing the waste products out of the exhaust
valve in the exhaust stroke. The exhaust valve closes and the intake valve opens. We’re now
back at ① and the cycle starts again.
(a) Show that the work done on the piston during one cycle of a fourstroke engine is
W xC P dV, where C is the curve in the PVplane shown in the ﬁgure.
[Hint: Let x t be the distance from the piston to the top of the cylinder and note that
the force on the piston is F AP t i, where A is the area of the top of the piston. Then
x t i, a t b. An alternative approach is
W xC F d r, where C1 is given by r t
to work directly with Riemann sums.]
(b) Use Formula 17.4.5 to show that the work is the difference of the areas enclosed by the
two loops of C.
1 1175 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas.
 Fall '09
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