standard of evalutaion

1062®201-04
06-10í+ãTŒU–

1. (20 points) Evaluate the integrals.

(a) (10 points) ∫

1 0 ∫

2√ y

√y

e^x3dx dy +∫

4 1 ∫

2

√y

e^x3dx dy.

(b) (10 points) ∬

R

cos (y − 2x

y + x ) dA, where R is the trapezoidal region with vertices (1,0),(2,0),(0,2) and (0,1).

Solution:

(a) Let D1, D2 be the region such that

∫ ∫

D1

e^x3dA =∫

1 0 ∫

2√ y

√y

e^x3dxdy ,

∫ ∫

D2

e^x3dA =∫

4 1 ∫

2e^x3

√y

dxdy

. Therefore, ∫

4 1 ∫

2e^x3

√y

dxdy +∫

1 0 ∫

2√ y

√y

e^x3dxdy = ∫ ∫

D1∪D2

e^x3dA = ∫

2 0 ∫

x²

x2 4

e^x3dydx =

∫

2 0

3

4x²e^x3dx = 1

4(e⁸−1).

standard of evalutaion

Simple calculation error 8pt

Right integration range after changing the order of x and y 5pt

Right integration range after changing the order of x and y but wrong calculation 4pt (b) Let u = x + y, v = y − 2x, then we have x = 1

3(u − v), y = 1

3(v + 2u). (1 point)

∂(x, y)

∂(u, v) = RR RR RR RR RR RR RR R

1 3

−1 2 3 3

1 3

RR RR RR RR RR RR RR R

= 1

3 (2 points)

Boundary: v = u, v = −2u, u = 1 and u = 2. (2 points)

∬R

cos(y − 2x

x + y )dA =∬

D

cos(v u) ∣

∂(x, y)

∂(u, v) ∣dA (2 points)

= ∬D

cos(v u)

1 3dA

=1 3∫

2 1 ∫

u

−2u

cos(v u)dvdu

= 1 3∫

2 1

(u sin(v

u) ∣^v=u_v=−2u)du

= 1 3∫

2 1

u(sin 1 + sin 2)du

=1

2(sin 1 + sin 2) (3 points)

2. (16 points) Evaluate the integrals.

(a) (8 points) ∫

2 0 ∫

1 0 ∫

1 y

e^−z2dz dy dx.

(b) (8 points) ∭

E

x²dV , where E is the solid that lies in the first octant within the cylinder x²+y²= 1 and below the cone z² =4x²+4y².

Solution:

(a) ∫

2 0 ∫

1 0 ∫

1 y

e^−z2dzdydx

= ∫

2 0 ∫

1 0 ∫

0

z e^−z2dydzdx (3 pt)

= ∫

2 0 ∫

1 0

ze^−z2dzdx, let u = −z²du = −2zdz (3pt)

= ∫

2 0 ∫

−1 0

e^ududx

= ∫

2 0

− 1 2e⁻¹+

1

2dx (2pt)

=1 − e⁻¹

(b) (Method 1) Use cylindrical coordinates

∭E

x²dV =

4pts

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ

∫

π/2

0 ∫

1 0 ∫

2r 0

(r cos θ)²r dz dr dθ =

1pt

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ

∫

π/2

0 ∫

1 0

2r⁴cos²θ dr dθ = 3pts

ªπ 10 or

∭E

x²dV =∫

π/2

0 ∫

2 0 ∫

1 z/2

(r cos θ)²r dr dz dθ = ⋯ = π 10 (Method 2) Use spherical coordinates

∭E

x²dV =

4pts

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ

∫

π/2

0 ∫

π/2 tan^{−1 1}₂ ∫

1 sin φ

0

(r sin φ cos θ)²r²sin φ dr dφ dθ =

1pt

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5∫

π/2

0 ∫

π/2 tan^{−1 1}₂

cos²θ sin²φdφ dθ

= 1 5∫

π/2 0

cos²θ dθ∫

π/2 tan^{−1 1}₂

csc²φ dφ = 1 5⋅

π

4 ⋅2 = π

10 (3pts), where∫ csc²φ dφ = − cot φ + C

(Method 3) By symmetry and use cylindrical coordinates

∭E

x²dV = 1

2∭

E

x²+y²dV = 1 2∫

π/2

0 ∫

1 0 ∫

2r 0

r²⋅r dz dr dθ = π 10

Remark 8‹/¤f

θÄ ë∫ ^2π

0 ∫ ^π/4

0 —úTH2π

5 π + 2

20 —6E—Ôgè—0π

40—5

zÄ ë∫ ^4r

2

0 —0π

6ëJacobian r—4zÄ ë∫ ²

0 —0π

8—3

3. (10 points) Let E be the tetrahedron bounded by the planes x + y + z = 3, x = 2z, y = 0, and z = 0 which is completely occupied by a solid with the density function ρ(x, y, z) = y. Find the total mass of the solid.

x

y

z

3 3

E

Solution:

The total mass of the solid is given by M ∶=∭

E

y dV . (3 points)

The region E can be described as

E = {(x, y, z) ∈ R³∣ (z, x) ∈ D , 0 ≤ y ≤ 3 − z − x } , where D is the triangular region in the (z, x)-plane given by

D ∶= {(z, x) ∈ R²∣0 ≤ z ≤ 1 , 2z ≤ x ≤ 3 − z } .

(correct description of the region (reflected in the regions of integration): 4 points) [Method 1] By a direct computation, the total mass is found to be

M =∭

E

y dV =∫

1 0 ∫

3−z 2z ∫

3−z−x 0

y dy dx dz

= ∫

1 0 ∫

3−z 2z

[ y²

2]

y=3−z−x

y=0

dx dz = 1 2∫

1 0 ∫

3−z 2z

(3 − z − x)² dx dz

= 1 2∫

1 0

[−

(3 − z − x)³

3 ]

x=3−z

x=2z

dz = 1 6∫

1 0

(3 − 3z)³ dz

= 3³ 6 [−

(1 − z)⁴

4 ]

1

0

= 3³ 24 = 9

8 . [Method 2] Consider the change of variables

u = 3z , v = x + z , w = x + y + z , which transforms E into a region in the uvw-space given by

E ≅ {(u, v, w) ∈ R³∣0 ≤ u ≤ v ≤ w ≤ 3 }

= {(u, v, w) ∈ R³∣0 ≤ w ≤ 3 , 0 ≤ v ≤ w , 0 ≤ u ≤ v } . The required Jacobian can be calculated from

∂(u, v, w)

∂(x, y, z) = RR RR RR RR RR RR RR

0 0 3 1 0 1 1 1 1 RR RR RR RR RR RR RR

=3 ⇒

∂(x, y, z)

∂(u, v, w) = 1 3 .

Therefore, the total mass is found to be

M =∭

E

y dV =∫

3 0 ∫

w

0 ∫

v 0

(w − v)1

3 du dv dw

= 1 3∫

3 0 ∫

w 0

(wv − v²) dv dw

= 1 3∫

3 0

(w³ 2 −w³

3 ) dw = 1 3 ⋅ 6∫

3

0 w³ dw

= 3⁴ 3 ⋅ 6 ⋅ 4 =

9 8.

(calculation + answer: 3 points)

4. (12 points) Evaluate the line integral ∫

C

sin πx dx + (e^y2 +x²)dy along the following choices of the curve C.

x y

(a) (4 points) C = C0 is the line segment from (−1, 0) to (0, 0).

(b) (8 points) C = C₁∪C₂, where C₁ is the polar curve r = 2 sin θ, 0 ≤ θ ≤ π

2 and C₂ is the cardioid r = 1 + sin θ, π

2 ≤θ ≤ π.

Solution:

(a) Describe C₀ as r(t) = (t − 1, 0), 0 ≤ t ≤ 1 (ëúC0„Ãx—1). Then

∫C0

sin(πx) dx + (e^y2 +x²)dy

= ∫

1 0

sin(π(t − 1)) ⋅ 1 + (1 + (t − 1)²) ⋅0 dt ((ÃxãÛÚM—1)

= ∫

1 0

sin(π(t − 1)) dt = −1

π cos(π(t − 1))∣

1 0

=

−2

π (—úcºTH—2).

(b) If D is the region bounded by C₀, C₁, and C₂, then by Green’s theorem, we have

∫C0∪C1∪C2

sin(πx) dx + (e^y2 +x²) dy =∬

D

2x dA (( Green’s theorem—3)

= ∫

π/2

0 ∫

2 sin θ 0

2 ⋅ r cos θ ⋅ r drdθ +∫

π π/2∫

1+sin θ 0

2 ⋅ r cos θ ⋅ r drdθ

((ÃxãÛbM—3’Œhcº„qŸLf)

= 4

3(sin θ)⁴∣

π/2 0

+ 1

6(1 + sin θ)⁴∣

π π/2

= 4 3+

−5 2 =

−7

6 (Ÿ—úbM—2).

Therefore,

∫C1∪C2

sin(πx) dx + (e^y2+x²) dy = −7 6 +

2 π.

5. (16 points) Let F(x, y) = P(x, y) i + Q(x, y) j, where P(x, y) = xy

√x²+ y², Q(x, y) = √x²+ 2y² x²+ y². (a) (3 points) Compute ∂P

∂y and ∂Q

∂x. Is F conservative on the right half plane D= {(x, y)∣x > 0}? Justify your answer.

(b) (4 points) Compute ∫_CF ⋅ dr, where C is any curve in the right half plane D from (1, 1) to (2, 2).

(c) (4 points) Compute∮_CF ⋅ dr, where C is a positively oriented circle centered at(0, 0) with radius r > 0.

(d) (4 points) Compute ∮_CF ⋅ dr, where C is any positively oriented simple closed curve, C ⊂ R²/{(0, 0)}.

(Hint: You need to discuss whether C encloses(0, 0) or not.) (e) (1 point) Is F conservative on R²/{(0, 0)}? Justify your answer.

Solution:

(a)

∂P

∂y = x

√

x²+y²−¹₂xy^√2y

x²+y²

x²+y² =

x

√

x²+y²− ^xy

2

√ x²+y²

x²+y² =

x³+xy²−xy² (x²+y²)

3 2

=

x³ (x²+y²)

3 2

(1%)

∂Q

∂x = 2x

√

x²+y²− (x²+2y²)¹₂^√2x

x²+y²

x²+y² = 2x³+2xy²−x³−2xy² (x²+y²)

3 2

= x³

(x²+y²)

3 2

(1%)

Since ∂P

∂y =

∂Q

∂x on D = {(x, y)∣x > 0} and D is simply-connected, by Green’s theorem ⃗F is conservative on D. (1%)

(b)

Let ⃗r(t) = ⟨t, t⟩, 1 ≤ t ≤ 2. (1%)

∫

(2,2) (1,1)

F ⋅ d⃗⃗ r = ∫

2 1

⟨ t²

√

2t, 3t²

√

2t⟩ ⋅ ⟨1, 1⟩dt (2%)

=

√

2t²∣²₁=3

√

2 (1%) (c)

r(t) = ⟨r cos t, r sin t⟩, 0 ≤ t ≤ 2π (1%)⃗

∮C

F ⋅ d⃗⃗ r = ∫

2π 0

⟨r²cos t sin t

r ,r²(1 + sin²t)

r ⟩ ⋅ ⟨−r sin t, r cos t⟩dt (2%)

= ∫

2π 0

(−r²cos t sin²t + r²(1 + sin²t) cos t)dt =∫

2π

0 r²cos tdt = 0 (1%) (d)

Case 1 (2%): For any simple closed curve C in R²/{(0, 0)} that enclosed (0, 0). Let D be the region bounded between C and C_r, and ⃗F is defined on D and ∂P

∂y,∂Q

∂x are continuous on D. By Green’s theorem,

∬D

(

∂Q

∂x −

∂P

∂y)dA = 0 =∮

C

F ⋅ d⃗⃗ r −∮

Cr

F ⋅ d⃗⃗ r for some r small enough such that C_r is inside C. Therefore,

∮C

F ⋅ d⃗⃗ r =∮

Cr

F ⋅ d⃗⃗ r = 0 by (c)

where Cr is a circle with x²+y² =r².

Case 2 (2%): For any simple closed curve C in R²/{(0, 0)} that does not enclosed (0, 0). Let D be the region bounded by C. By Green’s theorem,

∬D

(

∂Q

∂x −

∂P

∂y)dA =∬

D

0dA =∮

C

F ⋅ d⃗⃗ r = 0 (e) (1%)

Yes, since ∮

C

F ⋅ d⃗⃗ r = 0 for any simple close curve C in R²/{(0, 0)}, we get ⃗F is conservative in R²/{(0, 0)}, which is connected.

6. (10 points) Find the area of the part of the surface x² +y² +z² = 1 that lies within the cylinder x²+y²+x = 0 and above z = 0.

Solution:

Solution I. The equation of the cylinder is (x +1 2)

2

+y² = 1

4. Set D = {(x, y) ∶ (x +1 2)

2

+y²= 1 4}. Area =∬

D

√

1 + z_x²+z²_ydA We compute z_x= −

x

√

1 − x²−y² and z_y= −

y

√

1 − x²−y² and use polar coordinates to obtain Area =∬

D

1

√

1 − x²−y²dA =∫

3π 2 π 2

∫

−cos θ 0

1

√

1 − r² r dr dθ =∫

3π 2 π 2

[−

√ 1 − r²]

−cos θ

0 dθ

= ∫

3π 2 π 2

1 − ∣ sin θ∣ dθ = 2 ⋅∫

π

π 2

1 − sin θ dθ = π − 2.

By symmetry, one may consider D = {(x, y) ∶ (x − 1 2)

2

+y² =1

4} and compute likewise.

Solution II. The surface is parametrized by

r(u, v) = (sin u cos v, sin u sin v, cos u), 0 ≤ u ≤ π 2, π

2 +u ≤ v ≤ 3π 2 −u.

We compute ∣r_u×r_v∣ =sin u and integrate by parts to obtain Area =∫

π 2

0 ∫

3π 2 −u

π 2+u

sin u dv du =∫

π 2

0

(π − 2u) sin u du = [−(π − 2u) cos u − 2 sin u]^π/2₀ =π − 2.

Grading scheme

3 points for the region, 5 points for the integrand, 1 point for calculation and 1 point for the correct answer.

(01-02í) Suppose that f(x, y, z) is a scalar function with continuous second partial derivatives. Fix a point P0 = (x0, y0, z0). Consider spheres Sρ centered at P0 with radius ρ > 0.

(a) (2 points) Parametrize S_ρwith spherical coordinates r(ϕ, θ) = (x₀+ρ sin ϕ cos θ, y₀+ρ sin ϕ sin θ, z₀+ ρ cos ϕ), 0 ≤ ϕ ≤ π, and 0 ≤ θ ≤ 2π. Write down the double integral in ϕ and θ that represents the average value of f on S_ρ.

(b) (4 points) Let function A(ρ) be the average value of f on S_ρ, for ρ > 0. Evaluate A^′(ρ) in terms of ∬

Sρ

∇f ⋅ dS.

(c) (4 points) If ∇²f = f_xx+f_yy+f_zz is always positive, show that A(ρ) is increasing. If ∇²f (x, y, z) = 0 for all (x, y, z), compute A(ρ).

Solution:

1. { r_φ= (ρ cos φ cos θ, ρ cos φ sin θ, −ρ sin φ) rθ = (−ρ sin φ sin θ, ρ sin φ cos θ, 0)

⇒r_φ×r_θ=ρ²sin φ(sin φ cos θ, sin φ sin θ, cos φ) (1 point)

A(ρ) = 1

Area(S_ρ)∬

Sρ

f (x, y, z)dS = 1 4πρ²∫

2π

0 ∫

π 0

f (r(φ, θ)) ∣r_ρ×r_θ∣dφdθ

∵r(φ, θ) is Spherical coordinate, ∴ ∣r_φ×r_θ∣ =ρ²sin φ

⇒A(ρ) = 1 4π∫

2π

0 ∫

π 0

f (r(φ, θ)) sin φ dφdθ (1 point)

2. A^′(ρ) = d dρ(

1 4π∫

2π

0 ∫

π 0

f (r(φ, θ)) sin φ dφdθ) = 1 4π∫

2π

0 ∫

π 0

d

dρf (r(φ, θ)) sin φ dφdθ

= 1 4π∫

2π

0 ∫

π 0

(∇f (r(φ, θ)) ⋅ d

dρr(φ, θ)) sin φ dφdθ (2 points)

∵ d

dρr(φ, θ) = (sin φ cos θ, sin φ sin θ, cos φ), (1 point)

∴A^′(ρ) = 1 4π∫

2π

0 ∫

π 0

∇f (r(φ, θ)) ⋅r_φ×r_θ

ρ² dφdθ = 1 4πρ²∬

Sρ

∇f ⋅ dS (1 point) 3. By Divergence Theorem,

A^′(ρ) = 1 4πρ² ∬

Sρ

∇f ⋅ dS = 1 4πρ² ∭

Bρ

div(∇f )dV (1 point)

= ∭Bρ

∇ ⋅ (∇f ) dV =∭

Bρ

∇²f dV > 0, where Bρ is the ball centered at P0 with radius ρ.

That is, B_ρ= {(x, y, z) ∈ R³∶x²+y²+z² ≤ρ}

⇒A(ρ) is increasing, if ∇²f is always positive. (1 point) If ∇²f = 0 for all (x, y, z), then A^′(ρ) =∭

Bρ

∇²f dV = 0, that is A(ρ) is a constant. (1 point)

⇒A(ρ) = A(0) = 1 4π∫

2π

0 ∫

π 0

f (x₀, y₀, z₀)sin φ dφdθ = f(x₀, y₀, z₀) (1 point)

7. (14 points) Let F = (x − y)i + (y − z)j + (z − x)k be a vector field on R³. (a) (2 points) Compute curl F on R³.

(b) (6 points) Let S1 be a parametric surface given by r(r, θ) = r cos θi + 2r sin θj + (9 − r²)k for r ∈ [0, 3] and θ ∈ [0, 2π], which comes with the standard orientation given by the normal vector r_r×r_θ. Find the flux of curl F across S₁.

(c) (6 points) Let S₂ be a surface defined by the equation x² 9 +

y²

36−z²=1 for z ∈ [0, 1] and endowed with the orientation given by the downward normal vector. Find the flux of curl F across S2.

x

y

z

S1

x

y

z

S2

n n

Solution:

(a)

curl ⃗F = RR RR RR RR RR RR RR RR RR

⃗i ⃗j k⃗

∂

∂x

∂

∂y

∂

∂z x − y y − z z − x

RR RR RR RR RR RR RR RR RR

= (1, 1, 1)

div F = 3 (b) Method 1: Direct Calculation

r(r, θ) = (r cos θ, 2r sin θ, 9 − r⃗ ²) r⃗_r= (cos θ, 2 sin θ, −2r) r⃗_θ= (−r sin θ, 2r cos θ, 0) r⃗_r× ⃗r_θ= (4r²cos θ, 2r²sin θ, 2r)

∴ ∬S1

curl ⃗F ⋅ d ⃗S =∬

S1

curl ⃗F ⋅ ( ⃗r_r× ⃗r_θ)dS

= ∫

2π

0 ∫

3 0

4r²cos θ + 2r²sin θ + 2rdrdθ

=18π Method 2: Stokes’ Theorem

At z = 0, r = 3, ⃗r = (3 cos θ, 6 sin θ, 0),

C = {(x, y, z)∣x = 3 cos θ, y = 6 sin θ, z = 0, where θ ∈ [0, 2π]}

∴ ∬S1

curl ⃗F ⋅ d ⃗S =∮

C

F ⋅ d⃗⃗ r

= ∫

2π 0

27 sin θ cos θ + 18 sin²θdθ 18π

Method 3: Stokes’ Theorem Let S₃= {(x, y, z)∣x²

9 + y²

36 ≤1, z = 0} where ⃗n = ⃗k

∴ ∬S1

curl ⃗F ⋅ d ⃗S =∮

C

F ⋅ d⃗⃗ r =∬

S3

curl ⃗F ⋅ d ⃗S

= ∬S1

curl ⃗F ⋅ ⃗kdS

= ∬S3

1dS

=18π (Area of ellipse) (c) Solution 1. (Direct Compute)

S₂ ∶=r(z, θ) = ⟨3

√

1 + z²cos θ, 6

√

1 + z²sin θ, z⟩,

0 ≤ z ≤ 1, 0 ≤ θ ≤ 2π (1 point)

r_z = ⟨ 3z

√

1 + z²cos θ, 6z

√

1 + z²sin θ, 1⟩

r_θ = ⟨−3

√

1 + z²sin θ, 6

√

1 + z²cos θ, 0⟩ (1 point)

r_z×r_θ= ⟨−6

√

1 + z²cos θ, −3

√

1 + z²sin θ, 18z⟩ (1 point)

Take the negative-z direction, thus

∬S2

curl F ⋅ dS =∬

D

curl F ⋅ [−(r_z×r_θ)]dA (1 point)

= ∫

2π

0 ∫

1 0

6

√

1 + z²cos θ + 3

√

1 + z²sin θ − 18z dz dθ

= ∫

2π

0 ∫

1 0

−18z dz dθ

= −18π. (2 points)

Solution 2. (Use Stokes’ theorem once)

Let C₁ be the boundary of the oval x² 9 +

y²

36 =2, z = 1 and C₂ be the boundary of the oval x²

9 + y²

36 =1, z = 0.

By Stokes’ theorem, we know that

∬S2

curl F ⋅ dS =∮

C1

F ⋅ dr +∮

C2

F ⋅ dr

where C₁ is negative oriented, C₂ is positive oriented. (2 points) Thus, we have

r₁ = ⟨3√

2 cos θ, 6√

2 sin θ, 1⟩, 0 ≤ θ ≤ −2π

r₂ = ⟨3 cos θ, 6 sin θ, 0⟩, 0 ≤ θ ≤ 2π. (2 points)

∬S2

curl F ⋅ dS

= ∫

−2π 0

⟨3

√

2 cos θ − 6

√

2 sin θ, 6

√

2 sin θ − 1, 1 − 3

√

2 cos θ⟩

⋅⟨−3

√

2 sin θ, 6

√

2 cos θ, 0⟩ dθ + ∫

2π 0

⟨3 cos θ − 6 sin θ, 6 sin θ, 3 cos θ⟩ ⋅ ⟨−3 sin θ, 6 cos θ, 0⟩ dθ

= −36π + 18π = −18π. (2 points)

Solution 3. (Use Stokes’ theorem twice)

Let C₁ be the boundary of the oval D₁= { x²

9 + y²

36=2, z = 1} and C₂ be the boundary of the oval D₂ = {

x² 9 +

y²

36 =1, z = 0}. Then (2 points)

∬S2

curl F ⋅ dS =∮

C1

F ⋅ dr +∮

C2

F ⋅ dr =∬

D1

curl F ⋅ dS +∬

D2

curl F ⋅ dS

where D₁ is oriented downward, D₂ is oriented upward. (2 points)

∬D1

curl F ⋅ dS

= ∬D1

⟨1, 1, 1⟩ ⋅ ⟨0, 0, −1⟩ dA

= − ∬

D1

dA

= −A(D₁).

Where A(D) is the area of the region D.

Similarly,

∬D2

curl F ⋅ dS

=A(D₂).

∬S2

curl F ⋅ dS = −A(D₁) +A(D₂)

= −π ⋅ 3

√ 2 ⋅ 6

√

2 + π ⋅ 3 ⋅ 6

= −18π (2 points)

Note:

1. If you do wrong on orientation, the most score you get is 3.

(01-02í) Let F = (x − y)i + (y − z)j + (z − x)k be a vector field on R³.

(b) (6 points) Let S1 be the part of paraboloid z = x²+ (y − 1)² thea is below the plane z = 5 − 2y with downward orientation. Find the flux of F across S1, ∬

S1

F ⋅ dS.

Solution:

(Method I) Let E is the solid bounded by the paraboloid z = x²+ (y − 1)² and the plane z = 5 − 2y.

Then by Divergence Theorem, the flux of F across the boundary of E, (2 points)

∬S1

F ⋅ dS +∬

S^′

F ⋅ dS =∭

E

divF dV ⇒∬

S1

F ⋅ dS =∭

E

3dV −∬

S^′

F ⋅ dS

∵z = x²+ (y − 1)²=5 − 2y ⇒ x²+y²=4,

∴the projection of the intersection of the paraboloid and the plane to xy− plane is a circle centered at (0, 0) with radius 2.

∭EdV =∫

2π

0 ∫

2 0 ∫

5−2r sin θ

r²−2r sin θ+1rdzdrdθ (By Cylindrical coordinate)

= ∫

2π

0 ∫

2 0

(5 − 2r sin θ) − (r²−2r sin θ + 1)rdzdrdθ = 2π ⋅∫

2 0

r(4 − r²)dr = 8π (2 points)

∬S^′F ⋅ dS =∬

x²+y²≤4F ⋅ (0, 2, 1)dA =∬

x²+y²≤42y − (5 − 2y) − xdA

= ∬x²+y²≤4

4y − x − 5dA = −5 ⋅ 2²π = −20π

⇒ ∬S1

F ⋅ dS =∭

E

3dV −∬

S^′

F ⋅ dS = 3 ⋅ 8π + 20π = 44π (2 points)

(b) (Method II) S₁ ∶r(u, v) = (u, v, u²+ (v − 1)²) ⇒r_u×r_v= (−2u, −2(v − 1), 1)

∬S1

F ⋅ dS =∬

u²+v²≤4F ⋅ (−ru×rv)dudv (2 points)

= ∬u²+v²≤42u(u − v) + 2(v − 1)(v − u²− (v − 1)²) − (u²+ (v − 1)²−u)dudv

= ∬u²+v²≤42u²+2v²+2u²−2(v − 1)³−u²− (v − 1)²dudv (By Symmetirc)

= ∬u²+v²≤43u²+2v²−2(v³−3v²+3v − 1) − (v²−2v + 1)dudv

= ∬u²+v²≤4

3u²+7v²+1 dudv =∫

2π

0 ∫

2 0

(3r²cos²θ + 7r²sin²θ) r drdθ + 4π (3 points)

= ∫

2π

0 ∫

2 0

5r³ drdθ + 4π (∵∫

2π 0

sin²θdθ =∫

2π 0

cos²θdθ)

= 2π ⋅5 4r⁴∣

2

0

+4π = 44π (1 points)

8. (12 points) Let S be the boundary surface of the union of the balls x²+y²+z²≤1 and x²+y²+(z−1)²≤1.

x x

y y

z z

(a) (5 points) Use spherical coordinates to parametrize S.

(b) (7 points) Find ∬

S

F ⋅ dS where F = i + j + z²k and S is given the outward orientation.

Solution:

(a) (sol 1.)

upper surface: < 2 sin φ cos φ cos θ, 2 sin φ cos φ sin θ, 2 cos²φ >, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π 3 lower surface: < sin φ cos θ, sin φ sin θ, cos φ >, 0 ≤ θ ≤ 2π,π

3 ≤φ ≤ π (sol 2.)

upper surface: < 2 sin φ cos φ cos θ, 2 sin φ cos φ sin θ, cos φ + 1 >, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ 2π 3 lower surface: < sin φ cos θ, sin φ sin θ, cos φ >, 0 ≤ θ ≤ 2π,π

3 ≤φ ≤ π

score: the correct answer of upper surface gets 3 points, lower surface gets 2 points. Only write the correct Spherical coordinate, doesn’t write the range of θ, φ gets 1 point separately. The correct range of θ on both surface gets 1 point. The correct range of φ on both surface gets 1 point separately.

(b) (sol 1.)

Use divergence theorem: ∫ ∫ F dS =∫ ∫ ∫ divF dV upper surface: ∫

2π

0 ∫

π 3

0 ∫

2 cos φ 0

2ρ cos φρ²sin φdρdφdθ = 21π 8 lower surface: ∫

2π

0 ∫

π

π 3

∫

1 0

2ρ cos φρ²sin φdρdφdθ = −3π 8 total: 9π

4 (sol 2.)

upper surface: J = RR RR RR RR RR RR RR

i j k

2 cos(2φ) cos θ 2 cos(2φ) sin θ −2 sin(2φ)

−sin(2φ) sin θ sin(2φ) cos θ 0

RR RR RR RR RR RR RR

= < −2 sin²(2φ) cos θ, 2 sin²(2φ) sin θ, sin(4φ) >

2π ^π₃

<1, 1, 4 cos⁴φ >< −2 sin²(2φ) cos θ, 2 sin²(2φ) sin θ, sin(4φ) > dφdθ

= 87π 32

lower surface: J = RR RR RR RR RR RR RR

i j k

cos φ cos θ cos φ sin θ −sin φ

−sin φ sin θ sin φ cos θ 0 RR RR RR RR RR RR RR

= < sin²φ cos θ, sin²φ sin θ, sin φ cos θ >

∫

2π

0 ∫

π

π 3

<1, 1, cos²φ >< sin²φ cos θ, sin²φ sin θ, sin φ cos θ > dφdθ

=

−15π 32 total: 9π

4

score: Knowing to use divergence theorm gets 2 points, other method 1point. Use the upper answer of Spherical coordinate(ρ²sin φ) gets 2 points. The correct interval of integral gets 1 point separately, however you do write the correct value of integral.