1. (15%) (a) (7%) Sketch the region of integration and evaluate the integral Z 2 1

1. (15%) (a) (7%) Sketch the region of integration and evaluate the integral Z 2

1

Z x

√x

sin y

y dy dx + Z 4

2

Z 2

√x

sin y

y dy dx.

(b) (8%) Express and compute the moment of inertia Iz of the solid hemisphere x²+ y²+ z² = 4, z ≥ 0, in spherical coordinates. Take density function δ(x, y, z) = 1.

(a)

(2 points)

Z 2 1

Z x

√x

sin y

y dydx + Z 4

2

Z 2

√x

sin y

y dydx = Z 2

1

Z y²

√y

sin y

y dxdy (4 points)

= sin 2 − cos 2 − sin 1. (1 points)

(b)

Z 2π 0

Z ^π₂

0

Z 2 0

(ρ sin φ)²ρ²sin φdρdφdθ (7 points)

= 128

15 π (1 points)

2. (10%) Z Z

R

(x + y)e^x2−y2dA, where R is the rectangle enclosed by the lines x − y = 0, x − y = 2, x + y = 0, and x + y = 3.

Let u = x − y, v = x + y, then x = ^u+v2 , y = ^u−v₂ (2 points)

⇒ ∂(x, y)

∂(u, v) =      

1 2

1 2 1 2 −¹₂

= −1

2 (2 points) Z Z

R

(x + y)e^x2−y2dA = Z 3

0

Z 2 0

ue^uv    −1

2    

dvdu (3 points)

= 1 2

Z 3 0

(e^2v − 1)du = 1

4(e⁶ − 7) (3 points) 沒用 Jacobian (+0 2)

用 ^∂(x,y)_∂(u,v) 當 Jacobian (+0 4) 算 Jacobian 可是積分中沒用 (+0 6) 積分中 Jacobian 沒絕對值 (+0 8)

3. (10%) Find the circulation of the field F = −x²y i + xy²j + z²k around the curve C in which the circular cylinder x² + y² = 2x meets the cone z = 2px²+ y², counterclockwise as viewed from above.

SOL1

Let g(x, y, z) = z − 2px²+ y²(or z² − 4(x²− y²)), (1 point) then ∇g = (√^−2x

x²+y²,√^−2y

x²+y², 1) ( or (−8x, −8y, 2z) ) (1 point) and ∇ × F = (x²+ y²)k (1 point)

I

CF · dt = Z Z

S∇ × F · ~ndσ (by Stoke’s thm.)

= Z Z

{(x−1)²+y²≤1}

(x² + y²)k · ∇g

|∇g|

|∇g|

|∇g · k|dxdy (2 points)

= Z Z

{(x−1)²+y²≤1}

(x² + y²)dxdy = Z ^π₂

−^π2

Z 2 cos θ 0

r³drdθ (3 points)

= Z ^π₂

−^π

(2 cos θ)⁴

4 dθ = 3π

2 (2 points)

SOL2

x²+ y² = 2x ⇒ (x − 1)²+ y² = 1 ⇒ x = cos θ + 1, y = sin θ ⇒ z = 2px²+ y² = 2p2(cos θ + 1)

Let C(θ) = (cos θ + 1, sin θ, 2p2(cos θ + 1)) (3 points 看參數化對不對) Z

C(θ)F · dT = Z 2π

0

(cos θ + 1, sin θ, 2p2(cos θ + 1)) · (− sin θ, cos θ, −2 sin θ

p2(cos θ + 1))dθ (3points)

= Z 2π

0

(cos θ + 1)²sin²θ + (cos θ + 1) cos θ sin²θ − 8p2(cos θ) + 1 sin θ dθ

= 3

2π (4 points)

4. (10%) Evaluate the line integral Z

C

(ye^xy + 3x²y)dx + (xe^xy + x³ + 1)dy, where C is the circle x²+ y² = 1 connecting the points A = (1, 0) and B = (0, 1). Orientation from B to A.

Let F (x, y) = (M(x, y), N(x, y)) = (ye^xy + 3x²y, xe^xy+ x³+ 1)

∂M

∂y = e^xy + xye^xy+ 3x²,∂N

∂x = e^xy + xye^xy+ 3x² ( 3 points )

∂M

∂y = ∂N

∂x ⇒ F (x, y) is conservative .

∴there exists a function f : R² → Rsuch that∇f = F. (1 point)

∂f

∂x = ye^xy+ 3x²y ⇒ f(x, y) = e^xy + x³y + g(y) (2 points)

∂f

∂y = xe^xy + x³+ g^′(y) = xe^xy + x³ + 1 ⇒ g^′(y) = 1 ⇒ g(y) = y ( 2 points )

∴f (x, y) = e^xy+ x³y + y 原式 =

Z A

B ∇f · dr = f(A) − f(B) = f(1, 0) − f(0, 1) = (1 + 0 + 0) − (1 + 0 + 1) = −1

5. (25%) A piece of surface S is cut out from z = xy + 1 by x² + y² = 1, and given a vector field F(x, y, z) = (xz + y, y − 3x, x²)

(a) (5%) Compute surface area of S.

(b) Suppose ∂S is the boundary of S and is oriented counterclockwise when viewed from above.

(i) (5%) Compute the line integral I

∂SF · dr directly.

(ii) (5%) Compute the line integral I

∂SF · dr by any theorem.

(c) (5%) Surface z = xy + 1, x²+ y² = 1 and xy plane enclose a solid region V , find the total flux of F across V .

(d) (5%) (i) Find the total flux of curl F across V .

(ii) Find the total flux of curl G across V , for G is any other given smooth vector field in R³.

φ(x, y, z) = (x, y, xy + 1)

(a) φx = (1, 0, y) φx× φ^y = (−y, −x, 1) φ^y = (0, 1, x) A =R R

D|φx× φy|dxdy =R R

R

√y²+x²+1

1 dA (2 points)

=R2π 0

R1 0

√r²+ 1rdrdθ = 2π(1 + r^3/2)|¹0 = ^2π₃ (2√

2 − 1) (3 pionts) (b) (1)

r(θ) = (cos θ, sin θ, cos θ sin θ + 1)

F (θ) = (cos²θ sin θ + cos θ + sin θ, sin θ − 3 cos θ, cos²θ) (1point) r^′(θ) = (− sin θ, cos θ, cos²θ − sin²θ)(1point)

I

Fdr = Z 2π

0 − sin²θ cos²θ − sin θ cos θ − sin²θ + sin θ cos θ − 3 cos²θ + cos⁴θ − cos²θ sin²θ dθ

= Z 2π

0 −1

2sin²2θ − 1 − cos 2θ

2 − 31 + cos 2θ

2 +(1 + cos 2θ)²

4 dθ

= −15

4 π (3 points) (2)

∇F = (0, −x, −4) (1 point)

I

Fdr = Z Z

S∇ × F · ~ndσ = Z Z

D

(x²− 4)dxdy (1 point)

= Z 2π

0

Z 1 0

(r²cos²θ − 4)rdrdθ

= Z 2π

0

 1

4cos²θ − 2

 dθ

= 1

π − 2 · 2π = −15

π (3 points)

(c)

Z Z

∂V F · ~ndA =

Z Z Z

V

divFdσ

=

Z Z Z

V

(z + 1)dσ = Z 2π

0

Z 1 0

Z r²cos θ sin θ+1 0

(z + 1)dzrdrdθ

= 73

48π (3 points) (d)

(1)R R

∂V ∇ × F · ~ndS =H

φF · dr = 0 (2 points) (2)R R

∂V ∇ × G · ~n =R R R

V div(∇ × G)dS = 0 (3 points)

6. (15%) Find the outward flux of the vector field V(x, y, z) = (_r^x3 + 2y + 3z)i + (_r^y3 + x + 3z)j + (_r^z3 + x + 2y)k across the boundary of the ellipsoid region D = 11x² + 12y²+ 13z² ≤ 14 where r =px²+ y²+ z². (No points without complete computation and explanation.)

V is not differentiable in D since v = 0 at (x, y, z) = (0, 0, 0) so we take the ball B = {(x, y, z)|x²+ y²+ z² < 1} in D and consider the region Ω = D − B with ∂Ω = ∂D + ∂B

Since V is differentiable in Ω, we can apply divergence theorem. (4 points)

divV = ∂

∂x(x

r³ + 2y + 3z) + ∂

∂y(y

r³ + 3z) ∂

∂z(z

r³ + x + 2y)

= 0 (3 points)

⇒

Z Z

∂ΩV · ~ndσ =

Z Z Z

Ω

divV dV = 0

⇒

Z Z

∂DV · ~ndσ = − Z Z

∂BV · ~ndσ (2 points)

~n = (−x, −y, −z) at ∂B , −V · ~n = 1 + 3xy + 4xz + 5yz (1 point)

Let ∂B⁺ = {(x, y, z) ∈ ∂B|z > 0} and ∂B⁺ = {(x, y, z) ∈ ∂B|z < 0} C = {(x, y)|x²+ y² ≤ 1}

Then −R R

∂BV · ~ndσ = −R R

∂B⁺V · ~ndσ + −R R

∂B⁻V · ~ndσ (2 points)

Let r⁺(x, y) = (x, y,p1 − x²− y²) and r⁻(x, y) = (x, y, −p1 − x²− y²), Then |rx⁺ × r⁺y| =

|rx⁻× ry⁻| = √ ¹

1−x²−y²

− Z Z

∂BV · ~ndσ = − Z Z

∂B⁺V · ~ndσ + − Z Z

∂B⁻V · ~ndσ = Z Z

C

2 1 + 3xy p1 − x²− y²dA

= Z 2π

0

Z 1 0

21 + 3r²cos θ sin θ

√1 − r² rdrdθ

= 4π (3 points)

7. (15%) Find T (unit tangent vector), N (principal unit vector), B (unit binormal vector), κ (curvature) and τ (torsion) for the space curve r(t) = (e^tsin 2t)i + (e^tcos 2t)j + 2e^tk at t = π

2. We calculate r^′(t), r^′′(t), r^′′′(t) directly:

r^′(t) = (e^tsin 2t + 2e^tcos 2t, −2e^tsin 2t + e^tcos 2t, 2e^t), r^′(^π₂) = e^π2(−2, −1, 2) r^′′(t) = (−3e^tsin 2t + 2e^tcos 2t, −4e^tsin 2t − 3e^tcos 2t, 2e^t), r^′′(^π₂) = e^π2(−4, 3, 2) r^′′′(t) = (−11e^tsin 2t − 2e^tcos 2t, 2e^tsin 2t − 11e^tcos 2t, 2e^t), r^′′′(^π₂) = e^π2(2, 11, 2) and |r^′(t)| = 3e^t

So, we have

T|^t=π2 = r^′(t)

|r^′(t)|

   t=^π₂

= (e^tsin 2t + 2e^tcos 2t, −2e^tsin 2t + e^tcos 2t, 2e^t) 3e^t

   t=^π₂

=



−2 3, −1

3,2 3



Since dT

dt = 2 cos 2t − 4 sin 2t

3 ,−4 cos 2t − 2 sin 2t

3 , 0



we get the unit normal vector

N|t=^π₂ =

dT dt

|^dT_dt|     t=^π₂

= (−²₃,⁴₃, 0) q(−²₃)²+ (⁴₃)²+ 0²

=



− 1

√5, 2

√5, 0



By definition, we have B|^t=π2 = (T × N)|^t=π2 = −4/2√

5, −2/3√

5, −5/3√ 5
Next, at t = ^π₂

κ =    

dT ds

=    

dT dt

dt ds    

=    

dT dt

1

|r^′(t)| = 2√ 5 9e^π2 Finally, at t = ^π₂

τ = (r^′ × r^′′) ∧ r^′′′

|r^′ × r^′′|² =         

−2 −1 2

−4 3 2

2 11 2         

180e^π2 = − 4 9e^π2