Multivariable MA1104 Calculus Lecture 15
Dr KU Cheng Yeaw Monday March 16, 2009
Recap
Last Lecture ... we finished our study on multiple integrations!
Overview
In this lecture, we begin a study on vector calculus. We first introduce vector fields. Unlike (scalar) functions of several variables, a vector field assigns a vector to a point in plane/space. we introduce line integrals over a curve in plane/space as a generalization of single integral over an interval.
Vector Field
So far, we have developed the calculus of (scalar) functions of several variables. On many instances, we borrow intuition from Single Variable Calculus. We have also seen the role and usefulness of vector functions r(t) in the study of scalar functions. In this last chapter of MA1104, we shall study the calculus of vector fields. These are functions that assign vectors to points in plane or space.
We shall define Line integrals - integrals which can be used to find the work done by a force field in moving an object along a curve. Surface integrals - 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 we have already met are given by the higher-dimensional versions of the Fundamental Theorem of Calculus: Greens Theorem Stokes Theorem Divergence Theorem
Recall that we have learned about the following different type of functions function (scalar) f (t) (vector) r(t) (scalar) f (x, y) (scalar) f (x, y, z) Today, we will introduce function (vector field) F(x, y) (vector field) F(x, y, z) Domain D D R2 D R3 Range R R V2 R V3 Domain D DR DR D R2 D R3 Range R RR R V2 or V3 RR RR
Vector field on R2 Let D R2 . A vector field of on R2 is a function F that assigns to each point (x, y) D a 2-D vector F(x, y).
Vector field on R3 Let D R3 . A vector field of on R3 is a function F that assigns to each point (x, y, z) D a 3-D vector F(x, y, z).
Suppose F is a vector function on R2 . Since F(x, y) is a 2-D vector, we can write F(x, y) in its component functions P and Q as follows:
F(x, y) = P (x, y)i + Q(x, y)j = P (x, y), Q(x, y) or for short F = P i + Qj.
Similarly, for a vector field F on R3 , we can express F as F(x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k = or for short F = P i + Qj + Rk. P (x, y, z), Q(x, y, z), R(x, y, z)
Notice that P and Q are scalar functions of two variables. They are sometimes called scalar fields to distinguish them from vector fields.
How can we visualize a vector field? The best way to picture a vector field F(x, y) is to draw the arrow representing the vector F(x, y) starting at the point (x, y). Of course, its impossible to do this for all points (x, y).
Still, we can gain a reasonable impression of F by doing it for a few representative points in D, as shown.
For F(x, y.z), we visualize it by drawing arrows starting from (x, y, z):
We can define continuity of vector fields. We can show that F is continuous if and only if its component functions P , Q, and R are continuous.
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.
Here are some examples of computer plots of vector fields:
F(x, y) = -yi + xj
F(x, y) = yi + sin xj
F(x, y) = ln(1 + y2)i + ln(1 + x2 )j
Example 1. Match the vector fields F(x, y) = y 2 , x-1 , G(x, y) = y+1, ex/6 , H(x, y) = y 3 , x2 -1 to the graphs shown: Graph A
Graph B
Graph C
Solution. Notice the vectors F(x, y) = y 2 , x - 1 will never point to the left because y 2 0. So F is matched to Graph B. Similarly, the second component of G(x, y) = y + 1, ex/6 is ex/6 > 0, so the vectors G(x, y) will always point upwards. Graph A is the only such graph.
We are left with Graph C for H(x, y) = y 3 , x2 - 1 . Lets check: for y > 0, the vectors H(x, y) point to the right; for y < 0, the vectors H(x, y) points to the left. So Graph C is reasonable.
Most 3-D vector fields, however, are virtually impossible to sketch by hand.
The gravitational field is pictured here.
Gradient Fields
If f is a scalar function of two variables, recall that its gradient is defined by: f (x, y) = fx (x, y)i + fy (x, y)j.
f
Therefore, f is really a vector field on R2 and is called a gradient vector field.
Likewise, if f is a scalar function of three variables, its gradient is a vector field on R3 given by: f (x, y, z) = fx (x, y, z)i + fy (x, y, z)j + fz (x, y, z)k.
Conservative Vector Field
We shall discover later that many calculations involving vector fields simplify dramatically if the vector field F is a gradient field of some scalar function, that is, if there exists a function f such that F = f. In this situation, F is called a conservative vector field and 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 is conservative. However, we will not go into the details here.
Example 2. Determine whether F(x, y) = 2xy - 3, x2 + cos y is conservative. Solution. The idea is to try to construct a potential function f such that f = F. Suppose such f exists. Then f = 2xy - 3, x f = x2 + cos y. y
Integrating the first of these equations with respect to x (treating y as a constant), we get f (x, y) = (2xy - 3)dx = x2 y - 3x + g(y).
Here, we have added an arbitrary function of y alone, g(y), rather than a constant of integration, because any function of y is treated as a constant when integrating with respect to x. Differentiating the expression for f (x, y) with respect to y gives f = x2 + g (y) = x2 + cos y. y Thus yields g (y) = cos y, so that g(y) = cos y dy = sin y + c.
Therefore, f (x, y) = x2 y - 3x + sin y + c where c is an arbitrary constant. Since we have been able to construct a potential function, the vector field F(x, y) is conservative.
Line Integrals
We shall 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.
They were invented in the early 19th century to solve problems involving: Fluid flow Forces Electricity Magnetism
Line integrals generalizes the idea of integrating a single-variable real function over an interval [a, b] in two ways: as the integral of a function of two or three variables over a curve in two- or three-dimensional space. as the integral of a vector field over a curve.
Line Integrals in Plane
We start with a plane curve C given by the parametric equations:
x = x(t), y = y(t), a t b.
Equivalently, C can be given by the vector equation r(t) = x(t)i + y(t)j. We assume that r(t) is smooth, that is r (t) is continuous and r (t) = 0 for all a t b.
Lets divide the parameter interval [a, b] into n subintervals [ti-1 , ti ] of equal width. We let xi = x(ti ) and yi = y(ti ). Then, the corresponding points Pi (xi , yi ) divide C into n subarcs with lengths s1 , . . ., sn .
We choose any point Pi (x , yi ) in the i-th subarc. This i in [t corresponds to a point ti i-1 , ti ].
Now, if f is any function of two variables whose domain includes the curve C, we:
1. Evaluate f at the point (x , yi ). i
2. Multiply by the length 3. Form the sum
n
si of the subarc.
f (x , yi ) si i i
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. Line Integrals If f is defined on a smooth curve C given by a smooth parametrization r(t) = x(t)i + y(t)j, t a b. Then the line integral of f along C is
n
f (x, y) ds = lim
C
n
f (x , yi ) si , i i=1
provided this limit exists and is the same for every choice of (x , yi ). i
It can be shown that, if f is a continuous function, then the limit in the above definition always exists. Notice si (xi - xi-1 )2 + (yi - yi-1 )2 .
Since x(t) and y(t) have continuous first derivatives, by the Mean Value Theorem (as in the derivation of the arc length formula), we have si for some t (ti-1 , ti ). i x (t )2 + y (t )2 ti i i
Therefore, we have
n
f (x, y) ds =
C
n
lim
f (x , yi ) si i i=1 n f (x , yi ) i i=1
=
n b
lim
x (t )2 + y (t )2 ti i i dx dt
2
=
a
f (x(t), y(t))
+
dy dt
2
dt.
This formula can be used to evaluate the line integral: Formula for Line Integrals
b
f (x, y) ds =
C a
f (x(t), y(t))
dx dt
2
+
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.
The preceding formula says essentially that the arc length element ds can be replaced by
ds =
dx dt
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:
x = x,
y = 0,
a x b.
Then, our formula reduces to
b
f (x, y) ds =
C a
f (x, 0) (
dx 2 ) + 0 dx = dx
b
f (x, 0) dx
a
which is an ordinary single integral.
Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. In fact, if f (x, y) 0, then C f (x, y) ds represents the area of one side of the `fence' or `curtain' shown below, whose: Base is C. Height above the point (x, y) is f (x, y).
Example 3. Evaluate C (2 + x2 y) ds where C is the upper half of the unit circle x2 + y 2 = 1. Solution. The upper half of the unit circle has the following smooth parametrization x = cos t, y = sin t, 0 t .
Therefore
2 2
(2 + x2 y) ds =
C 0
(2 + cos2 t sin t) (2 + cos2 t sin t)
0
dx dt
+
dy dt
dt
= =
0
sin2 t + cos2 t dt
(2 + cos2 t sin t) dt
0
=
cos3 t 3 2 = 2 + . 3 2t -
So far, we have defined line integral only for smooth curves. In fact, we can consider any piecewise smooth curve C. That is, C is a union of a finite number of smooth curves C1 , C2 , . . ., Cn , where the initial point of Ci+1 is the terminal point of Ci .
Then, we define the line integral f along C as the sum of the integrals of f along each of the smooth pieces:
f (x, y) ds =
C C1
f (x, y) ds + +
Cn
f (x, y) ds.
Very often, we need to parametrize a line segment, so it is useful to remember that a vector representation of the line segment that starts from r0 and ends at r1 is given by
r(t) = (1 - t)r0 + tr1 ,
0 t 1.
Example 4. Give a parametrization of the line segment from (-5, -3) to (0, 2). Solution. Take r0 = -5, -3 and r1 = 0, 2 . Then, a vector equation of the line is
r(t) = (1 - t) -5, -3 + t 0, 2 = = So, a parametrization is x = 5t - 5, y = 5t - 3, 0 t 1. (1 - t)(-5), (1 - t)(-3) + 2t 5t - 5, 5t - 3 , 0 t 1.
Example 5. Evaluate C 2x ds where C consists of the arc C1 of the parabola y = x2 from (0, 0) to (1, 1) followed by the vertical line segment C2 from (1, 1) to (1, 2). Solution.
We can parametrize C1 as x = x, Therefore,
2 2
y = x2 ,
0 x 1.
1
2x ds =
C1 0 1
dx dx 2x
+
dy dx
=
0
1 + 4x2 dx
=
5 5-1 . 6
On the other hand, on C2 , we choose y as the parameter: x = 1, So
2 2
y = y,
1 y 2.
2
2x ds =
C2 1 2
2(1) 2 dy
1
dx dy
+
dy dy
= = 2.
Combining the results,
2x ds =
C C1
2x ds + 5 5-1 + 2. 6
C2
2x ds
=
Two other line integrals are obtained by replacing definition, by either: xi = xi - xi-1 . yi = yi - yi-1 .
si the
Then, they are called the line integral of f along C with respect to x and y respectively:
n
f (x, y) dx = lim
C
n
f (x , yi ) xi i i
n
f (x, y) dy = lim
C
n
f (x , yi ) yi . i i
Formula for line integral with respect to x
b
f (x, y) dx =
C a
f (x(t), y(t))x (t) dt.
Formula for line integral with respect to y
b
f (x, y) dy =
C a
f (x(t), y(t))y (t) dt.
It frequently happens that line integrals with respect to x and y occur together. When this happens, its customary to abbreviate by writing
P (x, y) dx +
C C
Q(x, y) dy =
C
P (x, y) dx + Q(x, y) dy.
Example 6. Evaluate from (5, 3) to (0, 2).
C
y 2 dx + x dy where C is the line segment
Solution. A parametrization of C was given in Example 4: x = 5t - 5, So, x (t) = 5, y (t) = 5. y = 5t - 3, 0 t 1.
y 2 dx + x dy =
C
y 2 x (t) dt + intC xy (t) dt
C 1 1
=
0
(5t - 3)3 5 dt +
0 1
(5t - 5)5 dt
= 5
0
(25t2 - 25t + 4) dt
1 0
25t3 25t2 - + 4t 3 2 5 = - . 6 = 5
Curve Orientation
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.
For instance, here, the initial point A corresponds to t = a; the terminal point B corresponds to t = b.
Let -C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point B to terminal point A):
Then we have
f (x, y) dx = -
-C C
f (x, y) dx,
-C
f (x, y) dy = -
C
f (x, y) dy
This is because orientation.
xi and
yi changes sign when we reverse the
However, if we integrate with respect to the arc length, the value of the line integral does NOT change when we reverse the orientation of the curve:
f (x, y) ds =
-C C
f (x, y) ds
This is because
si is always positive in both orientations.
Line Integrals in Space
We now consider line integrals in space. Suppose that C is a smooth space curve given by the parametric equations x = x(t), or by a vector equation r(t) = x(t)i + y(t)j + z(t)k. y = y(t), atb
Suppose 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
f (x, y, z) ds = lim
C
n
f (x , yi , zi ) si i i=1
provided this limit exists and does not depend on the choice of the sample points (x , yi , zi ). i
We evaluate line integrals in space using a formula similar to that for line integrals in plane: Formula for line integrals in space
b
f (x, y, z) ds =
C a
f (x(t), y(t), z(t))
dx dt
2
+
dy dt
2
+
dz dt
2
dt
Similarly, we have the following formula for line integrals with respect to x, y and z respectively: Formula for line integrals in space
b
f (x, y, z) dx =
C a b
f (x(t), y(t), z(t))x (t) dt f (x(t), y(t), z(t))y (t) dt
a b
f (x, y, z) dy =
C
f (x, y, z) dz =
C a
f (x(t), y(t), z(t))z (t) dt
Therefore, as with line integrals in plane, we evaluate integrals of the form P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz
C
by expressing x, y, z, dx, dy, dz in terms of the parameter t.
Example 7. Evaluate y dx + z dy + x dz
C
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). Solution. We can write C1 as
r(t) = (1 - t) 2, 0, 0 + t 3, 4, 5 = 2 + t, 4t, 5t . So, a parametrization of C1 is x = 2 + t, y = 4t, z = 5t, 0 t 1.
Since x (t) = 1, y (t) = 4, z (t) = 5, we have
1
y dx + z dy + x dz =
C1 0 1
(4t + (5t)4 + (2 + t)5) dt (10 + 29t) dt
0
=
= 24.5
Similarly, we can write C2 as w(t) = (1 - t) 3, 4, 5 + t 3, 4, 0 = 3, 4, 5 - 5t . So a parametrization of C2 is x = 3, y = 4, z = 5 - 5t, 0 t 1. such that x (t) = 0, y (t) = 0, z (t) - 5.
So
1
y dx + z dy + x dz =
C2 0
3(-5) dt
= -15.
Adding the values of these integrals, we obtain y dx + z dy + x dz = 24.5 - 15 = 9.5
C