**Unformatted text preview: **Section 20.1 – The Curl of a Vector Field
Conceptual Description of Circulation. Let F~ (x, y, z) be a vector field,
and imagine placing a tiny paddlewheel (like the one shown to the right) in the
vector field at the point (x, y, z) so that the handle points in the direction of the
axis that you want to measure the circulation of F~ around. Your eye Notes.
angular velocity
• The
the circulation. of the paddlewheel measures the strength of (x,y,z) • The circulation is
positive
if the rotation is counterclockwise and
negative
if the rotation is clockwise (as you look down the handle of
the paddlewheel). Preliminary Example. Let F~ be the vector field
shown to the right. Decide whether the circulation of the
vector field at (0, 0, 0) around each of the following axes
would be positive, negative, or zero: (a) ~k (b) −~k (c) ~j
(a) Since the handle of the paddlewheel points
straight up, the circulation is positive. y
0 -2 2 2
z0
-2 (b) Since the handle of the paddlewheel points
straight down, the circulation is negative. 0
2 x -2 (c) Since the handle of the paddlewheel points straight to the right, that is, along the
positive y-axis, the flow of the vector field cannot "push" the paddles. Therefore, the
circulation is zero. Definition. The circulation density of a smooth vector field F~ at P = (x, y, z) around the direction of a unit vector
~n is defined as
R
~
r
C F · d~
~
,
circ~n F (x, y, z) =
lim
Area → 0 Area inside C
where C is a circle centered at P and perpendicular to ~n. Note. The orientation of the circle C is determined by the right- n
P C P C n
~n : handle of paddlewheel hand rule: If you place your right thumb in the direction of ~n, your
fingers curl in the direction of C.
Comments:
• The line integral R C F~ · d~r gives the circulation of F~ around ~n. • The quantity circ~n F~ measures the circulation per unit area around ~n. Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 1 2 Example 1. Let F~ (x, y, z) = −y~i + x~j + 0~k. The cross-section
of this vector field for z = 0 is shown to the right. Calculate each
of the following. 1
y (a) circ~k F~ (0, 0, 0) C1 0 (b) circ−~k F~ (0, 0, 0)
-1 (c) circ~j F~ (0, 0, 0)
-1 0
x 1 (a) Let C1 be a circle of radius a, centered at (0, 0) and oriented as shown
to the right. Observe that, along C1 , we have
F~ · d~r =
= kF~ k kd~rk cos 0◦
p
x2 + y 2 kd~rk = a kd~rk.
Therefore, we have ←− k since F~ and d~r are parallel ←− 2 a C1
p
since x2 + y 2 = a on C1 circ~k F~ (0, 0, 0) = lim a→0 = lim a→0 = lim a→0 R F~ · d~r C1 a ←− area inside C1 πa2
R C1 kd~rk πa2 a(2πa)
πa2 = 2. R
(b) In this case, C2 F~ · d~r = −2πa2 (the negative of the corresponding
integral from part (a)), because C2 is oriented in the opposite
direction. Therefore,
−2πa2
= −2.
a→0 πa2 a circ−~k F~ (0, 0, 0) = lim (c) RIn this case, F~ and d~r are orthogonal everywhere along C3 . Therefore,
~
r = 0, so we obtain
C3 F · d~
circ~j F~ (0, 0, 0) = lim a→0 2 R F~ · d~r
C3
πa2 = 0. C2
−k a C3 j Developed by Jerry Morris Example 2. To the right, you are given the graph of a vector field -2 F~ (x, y, z). For approximately what value of ~n does circn~ F~ (0, 0, 0) appear to
be the greatest? Explain. y
0 2 ÖÓ
n 2 Choosing ~n = ~k would produce the greatest circulation
because it appears that placing the paddlewheel so that
the handle points straight up would result in the fastest
counterclockwise rotation. z0 C -2
0
2 x -2 Geometric Definition of Curl. The curl of a smooth vector field F~ , written curl F~ , is the vector
field with the following properties:
1. The direction of curl F~ (x, y, z) is the direction of ~n for which circ~n F~ (x, y, z) is the greatest.
2. The magnitude of curl F~ (x, y, z) is the circulation density of F~ around that direction.
If the circulation density is zero around every direction, then we define the curl to be ~0. Coordinate Definition of Curl. If F~ = F1~i + F2~j + F3~k, then
curl F~ = =
∂F3
∂F2 ~
∂F3 ~
∂F1 ~
∂F1
∂F2
k
−
−
−
i+
j+
∂y
∂z
∂z
∂x
∂x
∂y ~ ~i
~ ∂ j∂ k∂ ∂x ∂y ∂z F
F F 1 = 2 3 ∇ × F~ Notes:
1. The geometric and coordinate definitions of curl are equivalent (see text, pp. 957-958). 2. Recall the "Curl Test": Under "nice" conditions, a vector field F~ is conservative if and
only if curl F~ = ~0 everywhere. See page ?? in the workbook for a more precise statement of
this result. Examples and Exercises
1. Consider the two vector fields shown below. For each field, draw in a possible vector representing
curl F~ (0, 0, 0).
F~ (x, y, z)
F~ (x, y, z) -2
2 y
0 2 -2 ÖÓ
curl F 2 2 z0 z0 -2 -2
0
2 x y
0 -2 Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 ÖÓ
curl F
0
2 x -2 3 2. Below, you are given a formula for three vector fields and a picture of the cross section of that field that lies
in the plane z = 0.
F~2 (x, y, z) = |y|~i F~3 (x, y, z) = −y ~i + x ~j 2 2 1 1 1 0 0 0 -1 y 2 y y F~1 (x, y, z) = ~i + 2 ~j -1 -1 0
x 1 2 -1 -1 0
x 1 2 -1 0
x 1 2 (a) For each vector field, use graphical reasoning to describe the vectors curl F~ (0, 1, 0) and curl F~ (0, −1, 0).
In particular, are they zero or nonzero? If they are nonzero, do they point in the ~k or the −~k direction?
(i) For the vector field F~1 , we have curl F~1 (0, 1, 0) = curl F~1 (0, −1, 0) = ~0, because
objects placed at these two points in F~1 would experience no rotation.
(ii) The vector curl F~2 (0, 1, 0) should point in the "−~k" direction because, for a square
placed at (0, 1, 0) in F~2 , there is more left to right flow at the top of the square
than at the bottom, resulting in a clockwise rotation (see above diagram). By a
similar argument, we see that curl F~2 (0, −1, 0) should point in the "~k" direction.
(iii) At (0, 1, 0), there is more right to left flow at the top of a square than at the
bottom, resulting in a counterclockwise rotation. At (0, −1, 0), there is more left
to right flow at the bottom of a square than at the top, also resulting in a
counterclockwise rotation. Therefore, curl F~ (0, 1, 0) and curl F~ (0, −1, 0) both point in
the "~k" direction.
(b) Confirm your answers to part (a) by calculating curl F~ (0, 1, 0) and curl F~ (0, −1, 0) for each vector field.
(i) Since F~1 is constant, its partial derivatives are constant everywhere, so
curl F~1 = ~0 everywhere, including at (0, 1, 0) and (0, −1, 0).
(ii) First, to aid in taking derivatives, we observe that |y| = p
y 2 . We therefore have y
1
y
curl F~2 (x, y, z) = 0~i + 0 ~j − (y 2 )−1/2 (2y) ~k = − p ~k = − ~k.
2
|y|
y2 Therefore, curl F~2 (0, 1, 0) = −~k and curl F~2 (0, −1, 0) = ~k.
(iii) We have curl F~3 (x, y, z) = 0~i + 0 ~j + (1 − (−1)) ~k = 2~k, so curl F~3 = 2~k everywhere, including at (0, 1, 0) and (0, −1, 0). 4 Developed by Jerry Morris 3. Using Green’s Theorem and the definition of curl, it can be shown that for any smooth vector field F~ and any
unit vector ~n, we have
curl F~ · ~n = circ~n F~ , (∗) a fact that you may use freely from here on out.
Let S represent a small, two-dimensional solid square or
disk centered at the point (x, y, z) and having area vector
~ Let C be the curve along the boundary of S, oriented
∆A.
~ according to the right hand rule (see
to correspond to ∆A
diagram to the right). Use equation (*) above to show that C
S (∗∗) ~
curl F~ (x, y, z) · ∆A ≈ Z ∆A ∆A
S C F~ · d~r.
C We have
~
curl F~ · ∆A = curl F~ · (~n ∆A) ←− where ~n is a unit normal vector to S
and ∆A is the scalar area of S = (curl F~ · ~n) ∆A ←− by properties of dot products = (circ~n F~ ) ∆A ←− by Equation (∗) above R C ≈
= Z F~ · d~r
∆A ! ∆A ←− by the definition of circulation F~ · d~r. C Note: The smaller ∆A is, the better the above approximation will be. Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 5 4. Suppose that F~ is a smooth vector field and that curl F~ = −x~i + z~k. Estimate the circulation,
around a circle of radius 0.01 in each of the following situations. R C F~ · d~r, (a) Centered at (2, 0, 0), parallel to the yz-plane, and oriented counterclockwise when viewed from the
positive x-axis.
(b) Centered at (0, 0, 3), parallel to the xy-plane, and oriented counterclockwise when viewed from the
positive z-axis.
(c) Centered at (0, 0, 3), contained in the yz-plane, and oriented counterclockwise when viewed from the
positive x-direction.
Note: For each of the following calculations, we will use Formula (∗∗) from the previous
exercise. (a) We have z
Z F~ · d~r ≈ ~
curl F~ (2, 0, 0) · ∆A = −2~i · π(0.01)2 ~i = −2π(0.01)2 ≈ −0.000628. C ∆A
x C
y (b) We have
Z F~ · d~r ≈ ~
curl F~ (0, 0, 3) · ∆A = 3 ~k · π(0.01)2 ~k = 3π(0.01)2 ≈ 0.000942. C C y x (c) We have
Z z
∆A z
F~ · d~r ≈ ~
curl F~ (0, 0, 3) · ∆A = 3 ~k · π(0.01)2 ~j = 0. C ∆A C x 6 Developed by Jerry Morris y 5. Use the “curl test” to determine whether each of the following vector fields is conservative. If it is
conservative, find a function f such that F~ = ∇f.
(a) F~ = yexy ~i + xexy ~j + (2z + y) ~k
(b) F~ = 2xyz ~i + x2 z ~j + (x2 y + 1) ~k
(a) Calculating, we have
curl F~ = = (1 − 0)~i + (0 − 0) ~j + ∂F3
∂F2
−
∂y
∂z ~i + = ~i. ∂F3
∂F1
−
∂z
∂x ~j + ∂F1
∂F2
−
∂x
∂y ~k
(xyexy + exy ) − (xyexy + exy ) ~k Since curl F~ 6= ~0, we see that F~ is not conservative.
Remark: The above calculation could have been shortened by noticing that the ~i
component of curl F~ was nonzero, and concluding that curl F~ 6= ~0 without having to
calculate the ~j or the ~k component of curl F~ . (b) Calculating, we have
curl F~ = = (x2 − x2 )~i + (2xy − 2xy) ~j + (2xz − 2xz) ~k ∂F3
∂F2
−
∂y
∂z ~i + ∂F3
∂F1
−
∂z
∂x ~j + ∂F1
∂F2
−
∂x
∂y ~k = ~0. Therefore, since curl F~ = ~0, the vector field F~ is conservative. We will now find a
potential function, f, for F~ :
fx (x, y, z) = 2xyz =⇒ f (x, y, z) = x2 yz + P (y, z) fy (x, y, z) = x2 z =⇒ f (x, y, z) = x2 yz + Q(x, z) fz (x, y, z) = x2 y + 1 =⇒ f (x, y, z) = x2 yz + z + R(x, y) Since each of the above three representations for f must agree, we see that
f (x, y, z) = x2 yz + z
is our potential function. This can be checked easily by verifying that F~ = ∇f. Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 7 Section 20.2 – Stokes’ Theorem
Preliminary Example. Let S be the surface shown below, oriented upward, and let C be the curve running
along the boundary of S, oriented counterclockwise. Note that S has been divided into M small “patches”. Si = ith surface patch (oriented upward) Ci
~i
∆A =
= curve along boundary of Si
area vector of Si ∆Ai
Si
C We will now calculate the sum of the circulation
of F~ around each of the M patches. Note that the
orientation of each curve Ci has been chosen to
correspond to the orientation of the surface, S,
according to the right-hand rule. We therefore have S
(1) M Z
X
i=1 F~ · d~r Z = Ci F~ · d~r ←− because of "cancellation" along the "inside"
edges (labeled green in the above diagram) C However, by the result of Exercise 3 in the previous section, we also have (2) M Z
X
i=1 F~ · d~r Ci sample point on Si
↓
M
X
~
~i
(curl F )(Pi ) · ∆A
≈ ←− by Equation (∗∗) on page 5 i=1 Therefore, by combining Equations (1) and (2) above, we see that
Z F~ · d~r ≈ C M
X ~i ,
(curl F~ )(Pi ) · ∆A i=1 and this approximation becomes more accurate as the number of patches, M, increases.
take the limit as M → ∞, we obtain Stokes’ Theorem: If we Stokes’ Theorem. Let S be a smooth oriented surface with piecewise smooth, oriented boundary C,
and let F~ be a smooth vector field on an open region containing S and C. Then
Z
Z
~
~
curl F~ · dA
F · d~r =
C S The orientation of C is determined from the orientation of S according to the right hand rule. Notes.
1. The integral
2. The integral R C R S F~ · d~r is a line ~ is a
curl F~ · dA flux integral around the
integral over "boundary" of a surface S
a surface S . . 3. Stokes’ Theorem is mainly useful to replace a tedious line integral with an easier flux integral, or vice versa. 8 Developed by Jerry Morris The Boundary of a Surface
Definition. The boundary of a surface, S, is the curve (or curves) C running around the edge of S (like
the hem on clothing). Once an orientation for S has been chosen, the corresponding orientation for C is
determined by the right-hand rule.
Examples. For each of the following surfaces, draw in or describe both possible orientations. Then, for each
surface orientation, draw in the induced orientation of the boundary curve(s).
1. Cylinder (without top and bottom) n
C n C "outward" orientation "inward" orientation 2. Hemisphere (with open bottom) n n
rC
"upward" orientation r C
"downward" orientation 3. Punctured Yield Sign C out of page Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 C into page 9 Exercises and Examples
1. Let F~ (x, y, z) = x2 z ~i + y 2 exz ~j + z 2 ey ~k, and let S be the portion of the paraboloid z = 9 − x2 − y 2 that lies
R
~
above the xy-plane, oriented upward. Calculate S curl F~ · dA.
Since the boundary, C, of the surface S appears
to be a circle in the xy-plane, which is easy
to parameterize, we will use Stokes’ Theorem as
follows:
Z
Z
~ =
curl F~ · dA
F~ · d~r
S z
S C Note that C is oriented counterclockwise,
as viewed from the positive z-axis, so
that its orientation corresponds with the
upward orientation of S. We must now find and
parameterize C: 3 x y Ck 3 Calculation of Boundary Curve C:
Since the boundary curve consists of the intersection of S with the plane z = 0, we
calculate as follows:
z
z = 9 − x2 − y 2
= 0 =⇒ 9 − x2 − y 2 = 0 =⇒ x2 + y 2 = 9 ←− circle, radius 3 Next, we parameterize C as follows:
x =
y =
z =
Calculation of Integral:
Z 3 cos t 3 sin t 0 ~r(t) = 3 cos t~i + 3 sin t ~j + 0 ~k, 0 ≤ t ≤ 2π Continuing the integral calculation from above, we have ~
curl F~ · dA = S Z F~ · d~r ←− by Stokes’ Theorem C = Z t=2π (x2 z ~i + y 2 exz ~j + z 2 ey ~k) · ~r ′ (t) dt t=0 = Z 2π (0~i + (3 sin t)2 e0 ~j + 0 ~k) · (−3 sin t~i + 3 cos t ~j) dt 0 = Z 2π 27 sin2 t cos t dt 0 = 10 2π 9 sin3 t 0 = 9(sin 2π)3 − 9(sin 0)3 = 0. Developed by Jerry Morris 2. Let F~ (x, y, z) = −2y~i + 2x~j. Use Stokes’ Theorem to find R C F~ · d~r, where C is a circle... (a) parallel to the yz-plane, of radius a, centered at a point on the x-axis, and oriented counterclockwise as
viewed from the positive x-axis.
S = disk forming interior of C, oriented in
the positive x-direction z Since C forms the boundary of S, we have Z F~ · d~r C 4~k
~n dA
Z z }| {
z}|{
~
=
curl F~ · dA C n S = Z 4~k · ~i dA xt y S S = Z 0 dA S = 0. (b) parallel to the xy-plane, of radius a, centered at a point on the z-axis, and oriented counterclockwise as
viewed from above.
S = disk forming interior of C, oriented in
the positive z-direction tz Since C forms the boundary of S, we have n
C Z C F~ · d~r ~n dA
4~k
Z z }| {
z}|{
~
~
curl F · dA
=
S = Z 4~k · ~k dA x S t y S = 4 Z dA
S = 4 (πa2 )
= 4πa2 . Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 11 C2 3. Let F~ (x, y, z) = −y ~i + x ~j + zex+y ~k, and let E be the solid region
that lies above the plane z = 0, below the plane z = 1, and inside
the half of the cylinder x2 + y 2 = 1 for which x ≤ 0. Let S be the
surface of E, EXCLUDING THE SEMICIRCULAR TOP PANEL
that lies in the plane z = 1 (see diagram to the right). Calculate
R
~ where S is given the outward orientation.
curl F~ · dA,
S C1 z
x ÖÓ
n
y R
~ would be
Since S is comprised of three separate surfaces, it is likely that S curl F~ · dA
unpleasant to evaluate directly as a flux integral. We will therefore attempt to use
Stokes’ Theorem. Note that the boundary curve (highlighted above) can be broken into two
pieces: a line segment C1 and a semicircle C2 , oriented as indicated in the above diagram.
We therefore have the following:
Z
Z
Z
~ =
F~ · d~r
F~ · d~r +
curl F~ · dA
S Calculation of F~
d~r R C1 F~ · d~r: Along C1 , we have the following: 0
↓
= −y ~i + x ~j + zex+y ~k
←− parallel to ~j Calculation of R C2 F~ · d~r:
x =
y =
z = C2 C1 Since F~ · d~r = 0 everywhere along
C1 , we have
Z
F~ · d~r = 0.
C1 Along C2 , we have the following: cos t − sin t 1 3π
π
≤t≤
2
2 ~r(t) = cos t~i − sin t ~j + ~k, Therefore, continuing with the integral calculation from above, we have
Z
Z
Z
~ =
F~ · d~r
F~ · d~r +
curl F~ · dA
S C2 C1 = Z F~ · d~r ←− C2 = Z t= 3π
2 Z t= 3π
2 Z 3π
2 t= π
2 = since R C1 F~ · d~r = 0 (−y ~i + x ~j + zex+y ~k) · (− sin t~i − cos t ~j) dt (y sin t − x cos t) dt t= π
2 = = Z (− sin2 t − cos2 t) dt π
2
3π
2 (−1) dt
π
2 = −π. 12 Developed by Jerry Morris 4. Let F~ (x, y, z) = (2y + x2 ez )~i + (4x + ey ) ~j + cos(x2 yz) ~k, and let S be the portion of the sphere
x2 + y 2 + z 2 = 4 for which z ≥ 0, oriented upward.
(a) Describe and draw a flat surface T that has the same boundary curve as S. Also state the proper
orientation for T so that it corresponds to the orientation on the boundary curve induced by S.
R
~ must equal
(b) Assuming the orientations for S and T specified above, explain why S curl F~ · dA
R
~
~
curl F · dA.
T
R
~
curl F~ · dA.
(c) Evaluate
S (a) Let T be the portion of the plane z = 0 that lies inside the
sphere x2 + y 2 + z 2 = 4 (see diagram to the right). Then T must
be oriented upward to match the orientation of the boundary
curve, C.
t S T
C (b) Because the surfaces S and T both have C as their boundary curve, we have the
following:
Z
Z
~
curl F~ · dA
←−
Stokes’ Theorem applied to S
F~ · d~r =
C Z S F~ · d~r = C Z ~
curl F~ · dA ←− Z ~ =
curl F~ · dA Z Stokes’ Theorem applied to T T Therefore, we must have ~
curl F~ · dA. T S R
~ and use part (b).
(c) Since T is a simpler surface than S, we will evaluate T curl F~ · dA
We begin by making some observations about our integral. Note that the normal vector
to T is ~n = ~k, so only the ~k component of curl F~ will be relevant to our calculations.
Therefore, we will not explicitly calculate the ~i and ~j components of curl F~ ; we will
simply write "P " and "Q" in their places:
~n
curl F~ = ~k ←− unit normal to T
= P ~i + Q ~j + (4 − 2) ~k Calculating, we have
Z
~
curl F~ · dA = Z ~
curl F~ · dA ) ←− curl F~ · ~n dA = 2 dA by part (b) T S = Z curl F~ · ~n dA T = Z 2 dA ←− by the above calculations T = 2 (Surface Area of T )
= 2(4π)
= 8π. Note: This strategy could also have been applied to the previous exercise. Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 13 5. Let F~ = F1 ~i + F2 ~j + F3 ~k be a smooth vector field, and let S be a piecewise smooth surface that completely
encloses a solid region in R3 ; in other words, S is a closed surface. Let S have the outward orientation, and
suppose that S has been divided into M small patches Si , each having boundary curve Ci and area vector
~ i , where the orientations of Si and Ci are chosen to correspond to the outward orientation of S (see
∆A
diagram below).
(a) Explain why M Z
X
i=1 F~ · d~r must equal zero. Ci Since S is totally enclosed, it has no boundary
curve (contrary to the surface in the proof of
Stokes’ Theorem). Therefore, in calculating the
above sum, every patch edge is internal and is
traversed in both directions one time, resulting
in cancellation. Thus, this sum must equal zero. Ci ∆ Ai Si (b) Let Pi be a point chosen on the surface Si for each value of i. Use the fact that
Z
~
~
F~ · d~r
(curl F )(Pi ) · ∆Ai ≈
Ci and your answer to part (a) to calculate the value of
We have
Z ~ ≈
curl F~ · dA S R S M
X ~
curl F~ · dA. ~i
(curl F~ )(Pi ) · ∆A i=1 ≈ M Z
X
i=1 = F~ · d~r Ci 0. In the limit as M → ∞, we therefore obtain
Z
~ = 0.
curl F~ · dA
S 14 Developed by Jerry Morris (c) Calculate div (curl F~ ).
First, recall that
curl F~ = ∂F2
∂F3
−
∂y
∂z ~i + ∂F3
∂F1
−
∂z
∂x ~j + ∂F1
∂F2
−
∂x
∂y ~k. Therefore, we have
div (curl F~ ) =
=
= ∂
∂x ∂F3
∂F2
−
∂y
∂z
(F3 )yx − (F2 )zx +
(F1 )zy − (F1 )yz + = 0 + 0 + 0 = 0. (d) Use the Divergence Theorem to calculate ←− R S
∂ ∂F2
∂F1
∂F3
∂F1
+
−
−
∂z
∂x
∂z ∂x
∂y
(F1 )zy − (F3 )xy + (F2 )xz − (F1 )yz
(F2 )xz − (F2 )zx + (F3 )yx − (F3 )xy
∂
∂y + since mixed partial der...

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