Sphere ρ = 2 is equivalent to x2+y2+z2= 4

(1)

1012微微微甲甲甲01-04班班班期期期末末末考考考解解解答答答和和和評評評分分分標標標準準準 1. (10%) Evaluate

ˆ ₂

0

ˆ ₈

x³

x⁵

px⁶+ y²dydx.

Solution:

Interchange the order of the iterated integral, we have ˆ ₂

0

ˆ ₈

x³

x⁵

px⁶+ y²dydx = ˆ ₈

0

ˆ ^√³_y

0

x⁵

px⁶+ y² dxdy (3%)

= ˆ ₈

0

1 62p

x⁶+ y²

√3

y

0

dy (3%)

= 1 3

ˆ ₈

0

(

√

2 − 1)y dy (2%)

= 1 3(√

2 − 1)1 2y²

8

0

(1%)

= 32 3 (√

2 − 1) (1%)

(2)

2. (15%) Find the volume of the solid common to the balls ρ ≤ 2√

2 cos φ and ρ ≤ 2.

Solution:

Method 1.

Sphere ρ = 2 is equivalent to x²+y²+z²= 4; sphere ρ = 2√

2 cos φ is equivalent to x²+y²+(z −√ 2)² = 2. These two spheres intersect at φ = π/4.

Therefore,

Volume = ˆ _2π

0

dθ ˆ ^π

4

0

dφ ˆ ₂

0

ρ²sin φ dφ I

+ ˆ _2π

0

dθ ˆ ^π

2

π 4

dφ

ˆ ₂^√_{2 cos φ}

0

ρ²sin φ dφ II

= 2π(− cos φ)

π 4

0

· (ρ³ 3 )

2 0+ 2π

ˆ ^π

2

π 4

sin φ ·1 3(2√

2 cos φ)³dφ

= 2π

3 (8 − 4√

2) +32√ 2π 3

ˆ ^π

2

π 4

sin φ cos³φ dφ

= 2π

3 (8 − 4√

2) +2π 3

= 16

3 π − 2√ 2π.

Note:

θ : 0 ∼ 2π, (2 points) , Jacobi factor ρ²sin φ, (5 points) .

For part I,, φ : 0 ∼ π/4, (2 points) , ρ : 0 ∼ 2, (1 point) , answer 2π

3 (8 − 4√

2), (1 point)

For part II, φ : π/4 ∼ π/2, (2 points) , ρ : 0 ∼ 2√

2 cos φ, (1 point) , answer 2π

3 , (1 point) Method 2.

Volume = ˆ _2π

0

dθ ˆ ^√₂

0

dr

ˆ ^√_4−r²

√2−√ 2−r²

r dz

= 2π ˆ ^√₂

0

r(p

4 − r²+p

2 − r²−√ 2) dr

= 16 3 π − 2

√ 2π.

Page 2 of 8

(3)

Note:

θ : 0 ∼ 2π, (2 points) , z :√ 2 −p

2 − r²∼p

4 − r², (3 points) , r : 0 ∼√

2, (3 points) ,

Jacobi factor r, (5 points) , answer 2π 3 (8 − 3

√

2), (2 points) . Method 3.

Volume = ˆ ₂

√ 2

π(4 − z²) dz I

+ 4 3π(√

2)³·1 2 II

= 16 3 π − 2

√ 2π.

(4)

3. (15%) Evaluate the integral

¨

Ω

sin(3x²− 2xy + 3y²)dxdy, where Ω is the ellipse 3x²− 2xy + 3y² ≤ 2. You may try the change of variables x = u + kv, y = u − kv for some constant k.

Solution:

Follow the hint, set x = u + kv, y = u − kv. Put in equation of the ellipse: (3 pts) 3x²− 2xy + 3y² = 4u²+ 8k²v² ≤ 2

We can choose k = 1

√

2, then the equation becomes 4u²+ 4v² ≤ 2. Calculate the Jacobian (value of J 2 pts , absolute value 2 pts):

|J (u, v)| = |∂(x, y)

∂(u, v)| = |

1 k

1 −k

| = | − 2k| =√ 2

Then ¨

Ω

sin(3x²− 2xy + 3²) dxdy

=

¨

u²+v²≤¹₂

sin 4(u²+ v²)√

2 dudv (integrand 1 pt, domain 2 pts)

=√ 2

ˆ _2π

0

ˆ √¹ 2

0

sin 4r²rdrdθ (Jacobian of polar coordinates 3 pts)

=2√ 2π

ˆ √¹ 2

0

1

8sin 4r²d(4r²)

=

√2π

4 (1 − cos 2) (2 pts. If make a slight mistake, get 1 pt)

Scoring to steps of this problem:

1. Change of variable in u, v: get 2 pt if the relation is correct.

2. Jacobian of your variable: get 2 pts for the value and 2 pts for absolute value.

3. Write down the correct integrand for your new variables: get 1 pt.

4. Your integral domain is correct: get 2 pt.

5. Change u, v into polar coordinates. If the Jacobian is correct: get 3 pts.

6. Your result fits the correct answer: get 2pts, and get 1 pt if you just make a slight mistake.

Page 4 of 8

(5)

4. (15%) For y > 0, let

F(x, y, z) = (e^−xln y − z)i + 2yz − e^−x/y j + (y²− x)k and G(x, y, z) = e^−xln y i + 2yz − e^−x/y j − x k.

(a) Show that the vector function F is a gradient on {(x, y, z)| y > 0} by finding an f such that ∇f = F.

(b) Evaluate the line integral ˆ

C

G(r) · dr, where C is the curve given by r(u) = (1 + u²)i + e^uj + (1 + u)k, u ∈ [0, 1].

Solution:

(a) f = −e^−xln y + y²z − xz + c, where c ∈ R.( 5 points) (b) G = F + zi − y²k, and r(0) = (1, 1, 1), r(1) = (2, e, 2).

Then ˆ

C

G(r) · dr = ˆ

C

F (r) · dr + ˆ

C

zdx − y²dz

= f (2, e, 2) − f (1, 1, 1) + ˆ ₁

0

[(1 + u)2u − e^2u]du(4points)

= −e⁻²+ 2e²− 4 + (2

3u³+ u²−1

2e^2u)|¹₀(4 points)

= 3

2e²− e⁻²−11

6 (2points)

(6)

5. (15%) Let C be a piecewise-smooth Jordan curve that does not pass through the origin.

Evaluate ffi

C

−y⁵

(x²+ y²)³dx + xy⁴

(x²+ y²)³dy for the following two cases, where C is traversed in the counter- clockwise direction.

(a) C does not enclose the origin.

(b) C does enclose the origin.

Solution:

(a)

Let Ω be the region enclosed by C. Since Ω does not enclose the origin, the functions P (x, y) = −y⁵

(x²+ y²)³ and

Q(x, y) = xy⁴ (x²+ y²)³ are well-defined and differentiable in Ω. We have

∂Q

∂x(x, y) = ∂P

∂y(x, y) = y⁶− 5x²y⁴ (x²+ y²)⁴ in Ω. Therefore, by Green’s theorem,

˛

C

P dx + Qdy = ˆ ˆ

Ω

∂Q

∂y − ∂P

∂x

dxdy = 0.

Grading Policy:

Application of Green’s theorem: 2%

Correct calculation of partial derivatives: 2%

Correct answer 3%

(b)

Let C_r be the curve θ 7→ (r cos θ, r sin θ), θ ∈ [0, 2π], where r is small enough such that C_r lies in the interior of the region bounded by C. Let Ω be the region bounded by C and Cr. By Green’s theorem,

we have ˛

C

P dx + Qdy −

˛

Cr

P dx + Qdy = ˆ ˆ

Ω

∂Q

∂y −∂P

∂x

dxdy,

where P (x, y) and Q(x, y) are defined as in (a). Since Ω does not contain the origin, the right hand side of the above equation is 0. Thus

˛

C

P dx + Qdy =

˛

Cr

P dx + Qdy

= ˆ _2π

0

−r⁵sin⁵θ

r⁶ (−r sin θ) + r⁵cos θ sin⁴θ

r⁶ (r cos θ)

dθ

= ˆ _2π

0

sin⁴θdθ

= 3π 4 . Grading Policy:

Valid application of Green’s theorem: 3%

Correctly transforming the line integral to the ordinary integral: 2%

Correct answer: 3%

Page 6 of 8

(7)

6. (15%) Let S be the triangular region with vertices (0, 0, 0), (a, 0, 0), and (a, a, a), a > 0, with upward unit normal n, and C be the positively oriented boundary of S. Let

F = y − z cos(x²) i + 2x − sin(z²) j + 3z − tan(y²) k.

(a) Find a parametrization of S and find the upward unit normal n. (Hint. Consider the projection of S to xy-plane.)

(b) Evaluate ∇ × F.

(c) Evaluate

˛

C

F · dr.

Solution:

(a) i) S : {(x, y, y)| 0 ≤ y ≤ x ≤ a } (2 points) ii) n = 1

√2(0, −1, 1) (2 points)

(b)

i j k

∂

∂x

∂

∂y

∂

∂z F_x F_y F_z

(2 points)

=2z cos z²− 2y sec²y² i − cos x² j + k (2 points) (c)

˛

c

F·dr =

¨

S

∇ × F · n dσ = ˆ _a

0

ˆ _x

0

√1

2(1 + cos x²) · |r_x× r_y|dydx (4 points)

= ˆ _a

0

x(1 + cos x²)dx = x²+ sin x² 2 |^a₀ = 1

2(a²+ sin a²) (3 points)

(8)

7. (15%) Let S₁ be the surface {(x, y, z)| z = x²+ y², z ≤ y}, S₂be the surface {(x, y, z)| z = y, x²+ y² ≤ z}, and V(x, y, z) = −yi + xj + zk.

(a) Compute directly the downward flux of V across S₁.

(b) Use the divergence theorem to compute the upward flux of V across S₂.

Solution:

7.(a)

S1: f1(x, y) = (x, y, x²+ y²), x²+ y² ≤ y

∂f₁

∂x =(1, 0, 2x)

∂f1

∂y =(0, 1, 2y)

∂f1

∂x ×∂f1

∂y =(−2x, −2y, 1)

d area =(2x, 2y, −1) dx dy (since the direction is downward) (1 pt) V =(−y, x, z)

Let x = r cos θ, y = r sin θ. Then r²≤ r sin θ ⇒ 0 ≤ r ≤ sin θ, 0 ≤ θ ≤ π.

¨

S1

V · d area =

¨

S1

−2xy + 2xy − (x²+ y²) dx dy (1 pt)

= ˆ _π

θ=0

ˆ _{sin θ}

r=0

−r²r dr dθ (3 pts)

= −1 4

ˆ _π

0

sin⁴θ dθ = −1 4

3 4 1 2π

= −3π

32 (2 pts) (b)

Let x = r cos θ, y = r sin θ.

Ω : x²+ y²≤ z ≤ y ⇒

(r²≤ z ≤ y r²≤ r sin θ ⇒







r² ≤ z ≤ y 0 ≤ r ≤ sin θ 0 ≤ θ ≤ π By divergent theorem,

˚

Ω

∇ · V d volumn =

¨

∂Ω=S1+S2

V · d area =

¨

S1

V · d area +

¨

S2

V · d area (1 pt).

˚

Ω

∇ · V d volumn = ˆ _π

θ=0

ˆ _{sin θ}

r=0

ˆ _{r sin θ}

z=r²

1 r dz dr dθ (3 pts)

= ˆ _π

θ=0

ˆ _{sin θ}

r=0

(r sin θ − r²) r dr dθ

= ˆ _π

θ=0

sin θr³ 3

sin θ 0 −r⁴

4

sin θ 0 dθ

= ˆ _π

0

1

3sin⁴θ − 1

4sin⁴θ dθ

= 1 12

3 4 1 2π

= π

32 (2 pts)

⇒

¨

S2

V · d area = π

32 − −3π 32 = π

8 (2 pts)

Page 8 of 8