Thursday, December 3 ∗∗ Stokes’ Theorem

1. Let S be the portion of the cylinder of radius 2 about the x-axis where −1 ≤ x ≤ 1.

(a) Draw a picture of S and compute its area without doing any integrals. Hint: How could

you make this cylinder out of paper?

SOLUTION: We have Area(S) = 8π since we can cut S along a line in the x y-plane andflatten it out into a 2×4π rectangle.

(b) Find a parameterization r(u, v) of S.

SOLUTION:

Let u = x and v = angle about the x −axis, so that

r (u, v) = (u,2cos v,2sin v)

with domain D = {−1 ≤ u ≤ 1 and 0 ≤ v ≤ 2π}.(c) Does the normal vector field associated to your parameterization point into or out of S?

First, try to determine this without doing any calculations, and then check your answer by

evaluating ru × rv .SOLUTION:

By the right hand rule, the vector points inwards. Check:

ru × rv =

∣∣∣∣∣∣∣i j k1 0 0

0 −2sin v 2cos v

∣∣∣∣∣∣∣= 〈0,−2cos v,−2sin v〉At u = 0, v = 0 we have ru × rv = 〈0,−2,0〉 which points inwards.

(d) If necessary, change your parameterization so that the normal vector field points inwards.

SOLUTION:

If your vector points outwards, interchanging the role of u and v will reverse its direction.

(e) Now consider the vector field F = 〈− z, xz, −x y〉. Compute curlF.SOLUTION:

curlF = 〈−2x, y −1, z〉

(f) Check that curlF is the sum of G = 〈−2x, −1, 0〉 and H = 〈0, y, z〉.SOLUTION:

G+H = 〈−2x,−1,0〉+〈0, y, z〉 = 〈−2x, y −1, z〉 = curlF(g) Use geometric arguments to determine whether the flux of G is positive, zero, or negative.

Remember that we have oriented S so that the normals point inwards. Do the same for Hand curlF.

SOLUTION:

Since every normal vector to S has 0 x-component, the flux of G is the same as the fluxof 〈0,−1,0〉, which by symmetry must be equal to 0. For H, at each point of S H pointsoutward from the surface, so the flux is negative (since the normals point inwards). In

fact, H =−2n.(h) Using your parametrization, directly compute the flux of curlF.

SOLUTION:

∫ ∫S

curlF ·nd A =∫ 1−1

∫ 2π0

〈−2u,2cos v −1,2sin v〉 · (ru × rv )d vdu

=∫ 1−1

∫ 2π0

4(−cos2 v+cos v+sin2 v)d vdu = 4∫ 1−1

∫ 2π0

−1+cos vd vdu = 4∫ 1−1

−2πdu =−16π

(i) Check your answer in (h) using Stokes’ Theorem. Note here that ∂S has two boundary

components, and make sure that your orient them correctly.

SOLUTION:

The boundary components should be oriented like so:

Parametrize C1 by 〈1,2cos t ,2sin t〉 for 0 ≤ t ≤ 2π and C2 by 〈−1,2cos t ,−2sin t〉 for 0 ≤ t ≤2π. Now∫

C1F·dr =

∫ 2π0

〈−2sin t ,2sin t ,−2cos t〉·〈0,−2sin t ,2cos t〉d t =∫ 2π

0−4(sin t 2t+cos2 t )d t =−8π.

Similarly,∫

C2F ·dr =−8π. Thus ∫∂S F ·dr =−8π+ (−8π) =−16π= ∫ ∫S(curlF) ·nd A as guar-

anteed by Stokes’ theorem.

(j) Check your answer in (h) a second time by using what you learned in (g) to compute the

flux of G and H.

SOLUTION:

We know curlF = G+H and the flux of G through S is 0. Since H =−2n, H ·n =−2. Hence∫ ∫S

(curlF) ·nd A =∫ ∫

SH ·nd A =

∫ ∫S−2d A =−2Area(S) =−2 ·8π=−16π

2. Consider the surface S shown below, which is oriented using the outward pointing normal.

S E C T I O N 17.3 Divergence Theorem 981

26. The curl of a vector field F at the origin is v0 = ⟨3, 1, 4⟩. Estimatethe circulation around the small parallelogram spanned by the vectorsA =

〈0, 12 ,

12〉

and B =〈0, 0, 13

〉.

27. You know two things about a vector field F:(i) F has a vector potential A (but A is unknown).

(ii) The circulation of A around the unit circle (oriented counterclock-wise) is 25.Determine the flux of F through the surface S in Figure 22, orientedwith an upward-pointing normal.

S

y

x

z

1

Unit circle

FIGURE 22 Surface S whose boundary is the unit circle.

28. Suppose that F has a vector potential and that F(x, y, 0) = k. Findthe flux of F through the surface S in Figure 22, oriented with anupward-pointing normal.

29. Prove that curl(f a) = ∇f × a, where f is a differentiable func-tion and a is a constant vector.

30. Show that curl(F) = 0 if F is radial, meaning that F =f (ρ) ⟨x, y, z⟩ for some function f (ρ), where ρ =

√x2 + y2 + z2.

Hint: It is enough to show that one component of curl(F) is zero, be-cause it will then follow for the other two components by symmetry.

31. Prove the following Product Rule:

curl(f F) = f curl(F) + ∇f × F

32. Assume that f and g have continuous partial derivatives of order 2.Prove that

∮

∂Sf ∇(g) · dr =

∫∫

S∇(f ) × ∇(g) · dS

33. Verify that B = curl(A) for r > R in the setting of Example 4.

34. Explain carefully why Green’s Theorem is a special caseof Stokes’ Theorem.

Further Insights and Challenges35. In this exercise, we use the notation of the proof of Theorem 1 andprove

∮

CF3(x, y, z)k · dr =

∫∫

Scurl(F3(x, y, z)k) · dS 12

In particular, S is the graph of z = f (x, y) over a domain D, and C isthe boundary of S with parametrization (x(t), y(t), f (x(t), y(t))).(a) Use the Chain Rule to show that

F3(x, y, z)k · dr = F3(x(t), y(t), f (x(t), y(t))(fx(x(t), y(t))x

′(t) + fy(x(t), y(t))y′(t))

dt

and verify that∮

CF3(x, y, z)k · dr =

∮

C0

〈F3(x, y, z)fx(x, y), F3(x, y, z)fy(x, y)

〉· dr

where C0 has parametrization (x(t), y(t)).

(b) Apply Green’s Theorem to the line integral over C0 and show thatthe result is equal to the right-hand side of Eq. (12).

36. Let F be a continuously differentiable vector field in R3, Q a point,and S a plane containing Q with unit normal vector e. Let Cr be a circleof radius r centered at Q in S, and let Sr be the disk enclosed by Cr .Assume Sr is oriented with unit normal vector e.(a) Let m(r) and M(r) be the minimum and maximum values ofcurl(F(P )) · e for P ∈ Sr . Prove that

m(r) ≤ 1πr2

∫∫

Srcurl(F) · dS ≤ M(r)

(b) Prove that

curl(F(Q)) · e = limr→0

1

πr2

∫

CrF · dr

This proves that curl(F(Q)) · e is the circulation per unit area in theplane S .

17.3 Divergence TheoremWe have studied several “Fundamental Theorems.” Each of these is a relation of the type:

Integral of a derivativeon an oriented domain

= Integral over the orientedboundary of the domain

Here are the examples we have seen so far:

• In single-variable calculus, the Fundamental Theorem of Calculus (FTC) relates theintegral of f ′(x) over an interval [a, b] to the “integral” of f (x) over the boundaryof [a, b] consisting of two points a and b:

∫ b

af ′(x) dx

︸ ︷︷ ︸Integral of derivative over [a, b]

= f (b) − f (a)︸ ︷︷ ︸“Integral” over the boundary of [a, b]

(a) Suppose F is a vector field on R3 which is equal to curlG for some unknown vector field G.Suppose the line integral of G around the unit circle (oriented counter-clockwise) in thex y-plane is 25. Determine the flux of F through S.

SOLUTION:

By Stokes:∫ ∫

S F ·dS =∫∂S G ·dr = 25.

(b) Suppose H is a vector field on R3 which is equal to curlB for some unknown vector field B.If H(x, y,0) = k, find the flux of H through the surface S.SOLUTION:

Let D be the unit disc with upwards normal n = k = 〈0,0,1〉 Then S and D have the sameoriented boundary, the counter-clockwise oriented unit circle C . Hence by Stokes:∫ ∫

SH ·dr =

∫C

B ·dr =∫ ∫

DH ·dR =

∫ ∫D

k ·kd A = Area(D) =π.