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Homework
Math 223-04 (Salomone) Due Wednesday, April 25
Name:
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P #1.
Suppose that F (x, y, z) = 〈P(x, y, z), Q(x, y, z), R(x, y, z)〉 is a vector field and h(x, y, z) is a scalar function. (a) Prove that ∇ · (∇ × F ) = 0, that is, div(curl F ) = 0.
Solution: Since curl( F ) = 〈Ry − Qz, Pz − Rx, Qx − Py〉, we have
div(curl( F )) =
∂x (Ry^ −^ Qz)^ +^
∂y (Pz^ −^ Rx)^ +^
∂z (Qx^ −^ Py) = Ryx − Qzx + Pzy − Rxy + Qxz − Pyz = [Ryx − Rxy] + [Qzx − Qxz] + [Pzy − Pyz] = 0 ,
by Clairaut’s theorem of equality of mixed partials.
(b) Prove that ∇ × (∇h) = 0 , that is, curl(grad h) = 0.
Solution: Since ∇h = 〈hx, hy, hz〉 we have
curl(grad h) =
∂y hz^ −^
∂z hy,
∂z hx^ −^
∂x hz,
∂x hy^ −^
∂y hx
= 〈hzy − hyz, hxz − hzx, hyx − hxy〉 = 〈 0 , 0 , 0 〉,
by Clairaut’s equality of mixed partials.
(c) What is curl(div F )?
Solution: this is an undefined quantity. Since div( F ) is a scalar function and not a vector field, its curl (direction and magnitude of greatest circulation density) is a nonsensical notion.
P #2. (MC §20.3, #8, 9, 15)
(a) Compute the curl of the following vector fields.
i. F (x, y, z) =
x^2 , y^3 , z^4
Answer: 〈 0 , 0 , 0 〉 (All relevant partial derivatives are zero.)
ii. F (x, y, z) =
ex, cos y, ez^2
Answer: 〈 0 , 0 , 0 〉 (All relevant partial derivatives are zero.)
(b) What ”theorem” about the curl of vector fields can you determine from the answers to part (a)? Carefully state and prove your result.
Possible answer: Given a vector field F = 〈P, Q, R〉, if P = P(x) depends only on x, Q = Q(y) depends only on y, and R = R(z) depends only on z, then curl( F ) = 0.
Proof: Under these hypotheses,
curl( F ) =
Ry − Qz, Pz − Rx, Qx − Py
P #4.
Suppose that F is a vector field in R^3 , and F = ∇ × G. Let S 1 and S 2 be oriented surfaces which have a common oriented boundary, C. (a) Use Stokes’ theorem to explain why
s S 1 F^ ·^ n^ dA^ =^
s S 2 F^ ·^ n^ dA.
Solution: According to Stokes’ theorem, x
S 1
F · n dA =
x
S 1
(∇∇ × G ) · n dA
C
G · d r
=
x
S 2
(∇ × G ) · n dA
x
S 2
F · n dA.
(b) Use the divergence theorem to explain why
s S 1 F^ ·^ n^ dA^ =^
s S 2 F^ ·^ n^ dA. Hint: Consider the closed surface between S 1 and S 2. How is the outward flux from this closed surface related to these integrals? Pay close attention to orientations, and use a result from question 1 of this homework.
Solution: Together, S 1 and S 2 form a closed surface. If we call the region they enclose W, then by the divergence theorem, {
S 1 −S 2
F · n dA =
y
W
∇ · F dV.
Since the divergence theorem requires one to compute outward flux, we must orient S 1 and S 2 oppositely. If we suppose that outward flux requires us to orient S 2 oppositely, then the flux on the left hand side is equal to x
S 1
F · n dA −
x
S 2
F · n dA =
y
W
∇ · F dV.
On the other hand, the integrand on the right is ∇ × F = ∇ · (∇ × G ) = 0 by the result of question 4a. Thus x
S 1
F · n dA −
x
S 2
F · n dA = 0 ,
giving the desired result.
