微甲

992微甲^08-12班期末考解答和評分標準

1. (10%) Find the total electric charge over the region

R ={(x, y) : − 1 ≤ x + y ≤ 1 and − 1 ≤ x − y ≤ 1}

with charge density (per unit area) ρ(x, y) =|x| + |y|. (Hint: Use symmetry.) Sol:

By symmetry,

ZZ

R

ρ(x, y) dA = 4· Z ₁

0

Z _1−x

0

(x + y) dydx. (6 pts) Compute the integral in the right hand side,

Z ₁

0

Z _1−x

0

(x + y) dydx = Z ₁

0

(yx + 1

2y²)¯¯¯^1−x

0

dx = Z ₁

0

(1 2− 1

2x²) dx = 1

3. (4 pts) Therefore,

ZZ

R

ρ(x, y) dA = 4 3.

2. (10%) Evaluate the integral Z 2

−2

Z ^√4−x² 0

e^−(x2+y2)dydx.

Sol:

Let x = r cos θ, y = r sin θ. (6 pts) Then

Z 2

−2

Z ^√4−x² 0

e^−(x2+y2)dydx = Z 2

0

Z π 0

e^−r2r dθdr (2 pts)

= π· Z 2

0

e^−r2r dr

= π·³−1

2 e^−r2´¯¯¯²

0

= π

2(1− e⁻⁴). (2 pts) 3. (10%) Find

ZZZ

E

xyz dV , where

E ={(x, y, z) : x ≥ 0, y ≥ 0, z ≥ 0 and 36x²+ 16y²+ 9z² ≤ 144}.

Sol:

Let x = 1

6r sin φ cos θ, y = 1

4r sin φ sin θ, z = 1

3r cos φ, where 0≤ φ ≤ π

2, 0≤ θ ≤ π 2

then we have r²sin φ

72 =k ∂(x, y, z)

∂(r, φ, θ) k (4%)

⇒

ZZZ

E

xyz dV = Z ^π

2

0

Z ^π

2

0

Z 12 0

1

72r³(sin φ)²cos φ sin θ cos θ 1

72r²sin φ drdφdθ (3%)

= 1 72²

Z 12 0

r⁵dr Z ^π

2

0

(sin φ)³cos φ dφ Z ^π

2

0

sin θ cos θ dθ

= 1

72² ×12⁶ 6 ×1

4 × 1

2 = 12 (3%) 4. (10%) Evaluate

ZZ

R

sin

³2y− x 2y + x

´

dA, where R is the region enclosed by 2y + x = 1, 2y + x = 2, 2y− x = 0 and 2y + 5x = 0.

Sol:

Let v = 2y + x, u = 2y− x. Hence x = (v− u)

2 , y = (u + v) 4 .

Then the range 2y + x = 1, 2y + x = 2 imply v = 1,v = 2, 2y− x = 0, 2y + 5x = 0 imply u = 0, u = 3

2v.

So the integral become Z Z

sin (2y− x

2y + x) dxdy = Z 2

1

Z ³

2v

0

sin (u v)1

4dudv (a)

the 1

4 is the Jacobian ∂(x, y)

∂(u, v) from changing of coordinate.

Z ₂

1

Z ³

2v

0

sin (u v)1

4dudv =1 4

Z ₂

1

−v cosu v

¯¯¯

3 2v

0

dv (1)

=1 4

Z 2 1

−v cos3

2+ v dv (2)

=1 4

v²

2(1− cos3 2)¯¯¯²

1

(3)

=3

8(1− cos3

2) (4)

Correction rule:

(1)write down the complete integral(a) without answer you got 5 points. Missing some part will cost 1 to 2 points.

(2)write down the integral (1) you got 2 points.

(3)Complete answer cost the remained 3 point.

5. (10%) Let F = cos yi + (z²− x sin y)j + 2(y + 1)zk. Find a scalar function f(x, y, z) such that

∇f = F and then evaluate Z

C

F· dr, where C is the line segment form (1, 0, 0) to (2, 2π, 1).

Sol:

∂f

∂x = cos(y), ∂f

∂y = z²− x sin(y), ∂f

∂z = 2(y + 1).

We have

f (x, y, z) = x cos(y) + yz²+ z². (6pts) Z

C

f ¦ dr = f(2, 2π, 1) − f(1, 0, 0) = 2 + 2π. (4pts)

6. (8%) Find the line integral Z

C

6y²dx + 4x³dy, where C is the arc of the parabola y = 1− x² from (1, 0) to (0, 1) and then to (−1, 0).

Sol:

(Solution 1)

Let x = t, y = 1− t², t : 1→ −1.

Z

C

6y²dx + 4x³dy = Z

C

6(1− t²)²dt + 4t³(−2t)dt....(4pts)

The answer is −16

5 . (4pts) (Solution 2)

Let C₁ := {y = 0|x : 1 → −1}. Let D be the region bounded by C and C1. Using Green’s theorem,

Z

C

6y²dx + 4x³dy = Z

D

(12x²− 12y)dA + Z

C1

6y²dx + 4x³dy. (2pts) Z

D

(12x²− 12y)dA = −16

5 . (4pts) Z

C1

6y²dx + 4x³dy = 0. (2pts)

7. (10%) Evaluate the line integral I

C

(x²− y) dx + (1 + y²) dy, where C is the loop of the four leaved rose r = cos 2θ, −π

4 ≤ θ ≤ π

4, oriented counterclockwise.

Sol:

Let D be the region enclosed by the curve C. By Green’s Theorem,

I

C

(x²− y)dx + (1 + y²) dy = ZZ

D

1 dA (4pts)

= Z ^π

4

−π 4

Z cos 2θ 0

r drdθ (4pts)

= 1 2

Z ^π

4

−π 4

cos²2θ dθ

= 1 2

Z ^π

4

−π 4

1

2(1 + cos 4θ) dθ

= π

8 (2pts)

8. (10%) Find the surface integral ZZ

S

F· dS, where F = x²yi + xy²j + 2xyzk and

S = {(x, y, z) : x²+ y²+ z² = 2, x≥ 0, y ≥ 0, z ≥ 1} with the normal pointing upwards.

Sol:

S₁ ={(x, y, z)|x²+ z² ≤ 2, x ≥ 0, y = 0, z ≥ 1}

S2 ={(x, y, z)|y²+ z² ≤ 2, x = 0, y ≥ 0, z ≥ 1}

S₃ ={(x, y, z)|x²+ y² ≤ 1, z = 1}

E is the area enclosed by S, S₁, S₂, S₃ Z

E

divFdV = Z

S

F· dS + Z

S1

F· dS1+ Z

S2

F· dS2+ Z

S3

F· dS3 (2 pts) On S₁,∵ y = 0, then F = 0 ⇒

Z

S1

F· dS1 = 0 (1 pt) On S2,∵ x = 0, then F = 0 ⇒

Z

S2

F· dS2 = 0 (1 pt) Z

S3

F· dS3 = Z

S3

−2xy dS3 =− Z ^π

2

0

Z ₁

0

2r³sin θ cos θ drdθ =−1

4 (2 pts) Z

E

divF dV = Z

E

6xy dV = Z ^π

2

0

Z 1 0

Z ^√2−r² 1

6r³sin θ cos θ dzdrdθ

= 3 Z ^π

2

0

sin 2θ dθ Z 1

0

r³(√

2− r²− 1) dr

∵ Z ^π

2

0

sin 2θ dθ = 1 and Z ₁

0

r³√

2− r²dr = 1 2

Z ₁

0

u√

2− u du = 8√ 2− 7 15 then

Z

E

divF dV = 8√ 2− 7

5 −3

4 (4 pts)

⇒ Z

S

F· dS = 8√ 2− 7

5 − 1

2

9. (10%) (a) Find curl v, where v = −y³i + x³j + e^z2k and evaluate ZZ

S

curl v· dS, where S is the portion of the surface of z = x³+ y³− 3xy within the cylinder x² + y² = a² with upward normal. (b) Use Stokes’ Theorem to evaluate

Z

C

v· dr, where C is the boundary of S oriented counterclockwise when viewed from above.

Sol:

(a) curl v =∇ × v = (0, 0, 3x²+ 3y²) (2 pts) g(x, y) = (x, y, x³+ y³− 3xy)







g_x = (1, 0, 3x²− 3y) g_y = (0, 1, 3y²− 3x)

g_x× gy = (−3x²+ 3y, 3x− 3y², 1) (2 pts) ZZ

S

curl v· dS = ZZ

D

curl v· (gx× gy) dA (D :{(x, y)|x²+ y² ≤ a²}) (2pts)

= ZZ

D

3(x²+ y²) dxdy

= 3 Z _2π

0

Z _a

0

r²· r drdθ

= 6π·1 4r⁴¯¯¯^a

0

= 3

2πa⁴ (2pts) (b)

Z

C

v· dr = ZZ

S

curl v· dS = 3

2πa⁴ (2 pts)

10. (12%) Let S₁ be the upper semi-sphere x²+ y²+ z² = 1 with z≥ 0, S2 the unit disk x²+ y² ≤ 1 on xy-plane and V the region enclosed by S₁∪ S2. Endow S₁∪ S2 with outward normal. Let F = xz²i + (yx²+ e^z)j + (y²z + cos(x²+ y²))k. (a) Find divF and evaluate

ZZZ

V

divF dV .

(b) Use Divergence Theorem to evaluate ZZ

S1∪S2

F· dS and then find ZZ

S1

F· dS.

Sol:

(a) i.

divF = z²+ x²+ y².

ii. ZZZ

V

divFdV = Z ₁

0

Z _2π

0

Z ^π

2

0

ρ⁴sin φdφdθdρ = 2 5π.

(b) i. By the divergence theorem, we have ZZ

S1∪S2

F· dS = ZZZ

V

divFdV = 2 5π.

ii.

ZZ

S1

F· dS = ZZ

S1∪S²

F· dS − ZZ

S2

F· dS

= 2 5π−

ZZ

S2

F· (0, 0, −1)dS

= 2

5π− (−

Z ₁

0

Z _2π

0

r cos r²dθdr)

= (2

5+ sin 1)π.

Grading policy:

1. divF. (2 pts) 2.

ZZZ

V

divFdV . (4 pts)

3. Apply the divergence theorem (2 pts). Here you will get full points even if making a mistake in calculation of

ZZZ

V

divFdV .

4.

ZZ

S1

F· dS (4 pts) or ZZ

S2

F· dS (2 pts. Note the normal vector is (0, 0, −1)). That is, you could receive partial credit even without the result of (a).