cos t − t sin t, sin t + t cos t, 2t

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1. (12 points) Compute Z

C

√z ds where C is parametrized by g(t) =¡

t cos t, t sin t, t²¢

, 0 ≤ t ≤ 2π.

Solution: Since g⁰(t) = ¡

cos t − t sin t, sin t + t cos t, 2t¢ , ds = |g⁰(t)| dt

= p

cos²t − 2t sin t cos t + t² sin²t + sin²t + 2t sin t cos t + t² cos²t + 4t²dt

= √

1 + 5t²dt,

so Z

C

√z ds = Z _2π

0

t√

1 + 5t²dt = 1 15

£¡1 + 20π²¢_3/2

− 1¤ .

2. (12 points) Find the positively oriented simple closed curve C that maximizes the line integral Z

C

£¡x²y + 1 3y³¢

dx +¡

x − xy²¢ dy¤

.

Solution: If S is the region inside C, by the Green’s theorem, we have Z

C

£¡x²y + 1 3y³¢

dx +¡

x − xy²¢ dy¤

= Z Z

S

¡1 − x²− 2y²¢ dx dy.

The integrand is positive inside the ellipse x²+ 2y² = 1 and negative outside, so the integral is maximized by taking C to be the ellipse x²+ 2y² = 1 and S to be the region inside.

3. (a) (5 points) Let S be a regular region in the xy−plane. Show that the area of S =

Z

∂S

x dy = − Z

∂S

y dx.

Solution: By using the Green’s theorem, we have Z

∂S

x dy = Z Z

S

dx dy = Area of S and

− Z

∂S

y dx = Z Z

S

dx dy = Area of S.

(b) (5 points) Let R be a regular region in R³ with piecewise smooth boundary. Show that the volume of R =

Z Z

∂R

F · n dA where F (x, y, z) = (x, −y, z).

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Solution: By using the divergence theorem, we have Z Z

∂R

F · n dA = Z Z Z

R

div F dV = Z Z Z

R

(1 − 1 + 1) dV = the volume of R.

4. (16 points) Let

G(φ, θ) =¡

(2 + cos φ) cos θ , (2 + cos φ) sin θ , sin φ¢

with 0 ≤ φ , θ ≤ 2π

be a parametrization of the torus T obtained by revolving the circle (x − 2)² + z² = 1 in the xz−plane about z axis. Find the area of T.

Solution: Since

∂G

∂φ = ¡

− sin φ cos θ , − sin φ sin θ , cos φ¢

∂G

∂θ = ¡

− (2 + cos φ) sin θ , (2 + cos φ) cos θ , 0¢ , so

∂G

∂φ × ∂G

∂θ =¡

− (2 + cos φ) cos φ cos θ , (2 + cos φ) cos φ sin θ , −(2 + cos φ) sin φ¢ . Hence,

¯¯

¯¯∂G

∂φ × ∂G

∂θ

¯¯

2

= (2 + cos φ)²(cos²φ cos²θ + cos²φ sin²θ + sin²φ) = (2 + cos φ)², and

the area of T = Z _2π

0

Z _2π

0

(2 + cos φ) dφ dθ = 8π².

5. Let a > 0 and S denote the sphere x² + y²+ z² = a². For each (x, y, z) ∈ R³\ {(0, 0, 0)}, let F be a vector field defined by F (x, y, z) = 1

(x²+ y² + z²)^1/2 (x , y , z) = (x²+ y²+ z²)⁻¹² (x , y , z).

(a) (6 points) Find div F.

Solution:

div F = 3(x²+ y²+ z²)^−1/2− 1 2

2(x²+ y²+ z²)

(x²+ y²+ z²)^3/2 = 2(x²+ y²+ z²)^−1/2

(b) (4 points) Find an unit normal vector field n of S.

Solution: n = 1 a

¡x, y, z¢ .

(c) (6 points) Use the divergence theorem to evaluate the surface integral Z Z

S

F · n dA.

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Solution: For 0 < r < a, let B_r(0) denote the ball x²+ y²+ z² < r². Since F is C¹ on the region Ωr = Ba(0) \ Br(0), the divergence theorem implies that

Z Z

∂Ωr

F · n dA = Z Z Z

Ωr

div F dV = Z Z Z

Ωr

2(x²+ y²+ z²)^−1/2dV Now

r→0lim⁺ Z Z

∂Ωr

F · n dA = Z Z

S

F · n dA − lim

r→0⁺

Z Z

∂Br(0)

F · n dA

and

r→0lim⁺

¯¯

¯¯ Z Z

∂Br(0)

F · n dA

¯¯

¯¯ ≤ lim_r→0+

Z Z

∂Br(0)

kF k dA = lim

r→0⁺

Z Z

∂Br(0)

dA = lim

r→0⁺ 4πr² = 0.

Also,

0 ≤ lim

r→0⁺

Z Z Z

Br(0)

2(x²+ y²+ z²)^−1/2dV

= lim

r→0⁺

Z _2π

0

Z _π

0

Z _r

0

2ρ⁻¹ρ² sin φ dρ dφ dθ

= lim

r→0⁺ 2π¡ ρ²|^r₀¢ ¡

− cos φ|^π₀¢

= lim

r→0⁺ 4πr²

= 0.

Hence,

Z Z

S

F · n dA = lim

r→0⁺

Z Z

∂Ωr

F · n dA

= lim

r→0⁺

Z Z Z

Ωr

2(x²+ y²+ z²)^−1/2dV

= Z Z Z

Ba(0)

2(x²+ y²+ z²)^−1/2dV

= Z _2π

0

Z _π

0

Z _a

0

2ρ⁻¹ρ² sin φ dρ dφ dθ

= 2π¡ ρ²|^a₀¢ ¡

− cos φ|^π₀¢

= 4πa²

(d) (4 points) Evaluate the surface integral Z Z

S

F · n dA directly without using the divergence theorem.

Solution:

Z Z

S

F · n dA = Z Z

S

x²+ y²+ z²

a(x²+ y²+ z²)^1/2 dA = Z Z

S

a² a² dA =

Z Z

S

dA = Area of S = 4πa².

6. (a) (6 points) Let S denote the sphere x²+ y²+ z² = a² with unit outward normal n. Let F be

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a C¹ vector field on S. Use the Stokes’s theorem to show that Z Z

S

(curl F ) · n dA = 0.

[Hint: If S⁺ = S ∩{z ≥ 0} and S⁻ = S ∩{z ≤ 0}, then S⁺and S⁻satisfy that S = S⁺∪S⁻, and they share the same boundary ∂S⁺ = ∂S⁻ of opposite orientations.]

Solution: By using the Stokes’s theorem, we have Z Z

S⁺

(curl F ) · n dA = I

∂S⁺

F · dx = − I

∂S⁻

F · dx = − Z Z

S⁻

(curl F ) · n dA.

Hence, Z Z

S

(curl F )·n dA = Z Z

S⁺∪S⁻

(curl F )·n dA = Z Z

S⁺

(curl F )·n dA+

Z Z

S⁻

(curl F )·n dA = 0.

(b) (6 points) Let S be a closed surface (i.e. a surface with no boundary) in R³ with unit outward normal n, and let F be a C¹ vector field on S. Use the Stokes’s theorem to show

that Z Z

S

(curl F ) · n dA = 0.

Solution: Let S₁, S₂ ⊂ S be two connected domains in S such that S₁∪S₂ = S, S₁∩S₂ =

∂S₁ = ∂S₂ is a simple closed C¹ curve in S. Then, by using the Stokes’s theorem, one can

show that Z Z

S1

(curl F ) · n dA = − Z Z

S2

(curl F ) · n dA, and, hence,

Z Z

S

(curl F ) · n dA = Z Z

S1

(curl F ) · n dA + Z Z

S2

(curl F ) · n dA = 0.

Alternatively, by using the Stokes’s theorem “directly”, we get Z Z

S

(curl F ) · n dA = I

∂S

F · dx = Z

∅

F · dx = 0.

7. Let F (x, y, z) = (x−z, x+y, y +z), and let S denote the upper half of the sphere x²+y²+z² = a² with unit outward normal n, R be the circular domain x²+ y² ≤ a² in the xy−plane, i.e. z = 0, and ~k = (0, 0, 1).

(a) (6 points) Find curl F.

Solution:

curl F =

¯¯

i j k

∂

∂x

∂

∂y

∂

∂z x − z x + y y + z

¯¯

= (1, −1, 1).

(b) (6 points) Show that

Z Z

(curl F ) · ~k dA,

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Hint: S ∪ R is a closed surface in R³.

Solution: Since S ∪ R is a closed surface in R³, Z Z

S∪R

(curl F ) · n dA = 0, and, hence,

Z Z

S

(curl F ) · n dA = − Z Z

R

(curl F ) · n dA = − Z Z

R

(curl F ) · (−~k) dA = Z Z

R

(curl F ) ·~k dA

(c) (6 points) Evaluate Z Z

S

(curl F ) · n dA.

Solution:

Z Z

S

R

(curl F ) · ~k dA = Z Z

R

(1, −1, 1) · (0, 0, 1) dA = πa².