982微積分甲08-13班期末考解答與評分標準
1. (8%) Evaluate Z 4
0
Z 2
y 2
ex2dx dy.
Sol:
Z 4 0
Z 2
y 2
ex2dxdy
= Z Z
D
ex2dA
= Z 2
0
Z 2x
0
ex2dydx (4 pts)
= Z 2
0
2xex2dx (2 pts)
= ex2¯¯¯2
0
= e4− 1 (2 pts)
2. (10%) Find the area of the surface z = 2
3(x32 + y32), 0≤ x, y ≤ 1.
Sol:
Parametrization:
x = µ
u, v,2 3
³
u32 + v32
´¶
and u, v ∈ [0, 1]
xu =
³ 1, 0, u12
´
xv =
³ 0, 1, v12
´
(3 pts) (xu× xv) =
³−u12,−v12, 1
´
|xu× xv| =√
1 + u + v
Area of surface = Z 1
0
Z 1
0
√1 + u + v du dv (4 pts)
= 2 3
Z 1 0
h
(1 + u + v)32 i1
0
dv
= 2 3
Z 1 0
(2 + v)32 − (1 + v)32 dv
= 4 15
h
(2 + v)52 − (1 + v)52i1 0
= 4 15
h
352 − 252 − 252 + 1 i
(3 pts)
Calculation errors: −1 pt.
Minor mistakes: −1 pt.
3. (10%) Evaluate Z 2
−2
Z √4−x2
−√ 4−x2
Z 4−x2−y2 0
(x2+ y2) dz dy dx.
Sol:
Z 2
−2
Z √4−x2
−√ 4−x2
Z 4−x2−y2
0
(x2+ y2) dzdydx
= Z 2π
0
Z 2 0
Z 4−r2 0
r2· r dzdrdθ (x = r cos θ, y = r sin θ) (5 pts)
= 32π
3 (5 pts) 4. (10%) Evaluate
ZZ
x2+xy+y2≤1
e−(x2+xy+y2)dx dy.
Sol:
ZZ
x2+xy+y2≤1
e−(x2+xy+y2)dxdy
= ZZ
(x+y/2)2+(√
3y/2)2≤1
e−((x+y/2)2+(√3y/2)2)dxdy
= ZZ
u2+v2≤1
eu2+v2|
¯¯¯¯
¯¯¯
1 −1/√ 3 0 2/√
3
¯¯¯¯
¯¯¯|dudv (4 pts)
= 2
√3 Z 2π
0
Z 1 0
e−r2rdrdθ (前面做對,到此累計得 7 pts)
=2π(1− e−1)
√3 (前面做對,到此累計得 10 pts)
5. (16%) Let F(x, y, z) = y2i + (2xy + e3z)j + 3ye3zk
(a) Find the potential function of F.
(b) Compute the line integral Z
C
F· dr, where C : r(t) = h4t, 3 cos t, 3 sin ti, 0 ≤ t ≤ π 2. Sol:
(a) F (x, y, z) = (2xy, 2xy + e3z, 3ye3z).
Assume f is the potential function (Real value function such that ∇f = F ), then fx = 2xy imply f (x, y, z) = xy2+ h(y, z).
Then take derivative with y variable on both side we got fy = 2xy + hy(y, z).
By definition of F we also have, hy(y, z) = e3z ⇒ h(y, z) = ye3z+ g(z).
Since fz = 3ye3z, this imply fz = hz = 3ye3z+ gz(z) = 3ye3z, i.e, gz = 0.
So f = xy2+ ye3z+ C for some constant.
(b)
Z
C
F dr = Z
C
∇fdr = f(r(π/2)) − f(r(0))
= f (2π, 0, 3)− f(0, 3, 0) = 2π × 02+ 3× 0 × e0− (0 × 32+ 3e0) =−3 評分標準:
(a) If the answer is right then you got 8 point.
If not, you can gain 2 at each step while you are trying to solve h and g.
(b) Z
C
F dr = Z
C
∇fdr = f(r(π/2)) − f(r(0)) This step cost 4 point.
= f (2π, 0, 3)− f(0, 3, 0) This step cost 2 point.
= 2π× 02+ 3× 0 × e0− (0 × 32+ 3e0) =−3. Answer cost 2 point.
6. (16%) Let P = y3
(x2+ y2)2, Q = − xy2
(x2+ y2)2. Let C be the counterclockwise ellipse x2
9 + y2
16 = 1, and D be the region inside C but outside the unit circle x2+ y2 = 1.
(a) Evaluate ZZ
D
³∂Q
∂x − ∂P
∂y
´ dA.
(b) Evaluate Z
C
P dx + Q dy.
Sol:
(a) ∂Q
∂x = 3x2y2− y4
(x2+ y2)3 (2 pts), ∂P
∂y = 3x2y2 − y4
(x2+ y2)3 (2 pts) so ∂Q
∂x − ∂P
∂y = 0 ZZ
D
(∂Q
∂x − ∂P
∂y)dA = 0 (2 pts)
(b) Let E be the counterclockwise unite circle. Then ∂D = C∪ E By Green theorem,
Z Z
D
(∂Q
∂x −∂P
∂y)dA = Z
C
P dx + Qdy + Z
−E
P dx + Qdy So
Z
C
P dx + Qdy = Z
E
P dx + Qdy (4 pts) To compute
Z
E
P dx + Qdy, parametrize E ={(cos θ, sin θ)|0 ≤ θ ≤ 2π}}
Then Z
C
P dx + Qdy = Z
E
P dx + Qdy = Z 2π
0
(sin θ)3d cos θ− cos θ(sin θ)2d sin θ (2 pts)
= Z 2π
0
[−(sin θ)4− (cos θ)2(sin θ)2]dθ
= Z 2π
0
[−(sin θ)2]dθ = Z 2π
0
h−1− cos 2θ 2
i
dθ (2 pts)
=−θ
2 +cos 2θ 4
¯¯¯2π
0
=−π (2 pts)
7. (10%) Evaluate ZZ
S
x2
p1 + x2+ y2 dS, where S is the helicoid with equation r(u, v) = u cos vi + u sin vj + vk, 0≤ u ≤ 1, 0 ≤ v ≤ 2π.
Sol:
γu = (cos v , sin v , 0 ) γv = (−u sin v , u cos v , 1 ) γu× γv = (sin v ,− cos v , u) (3 pts)
| γu× γv | =√
u2+ 1 (2pts)
We have
ZZ
S
x2
p1 + x2+ y2dS = Z 2π
0
Z 1 0
u2cos2v
√1 + u2
√1 + u2du dv (2 pts)
= Z 2π
0
Z 1 0
u2cos2v du dv
= π
3 (3 pts) 8. (10%) Compute the integral
ZZ
S
curlF·dS, where F(x, y, z) = (ez2+ y)i + (4z−y)j+(8x sin y)k and S is the part of z = 4− x2− y2 above the xy-plane with orientation given by the upward unit normal vector.
Sol:
Solution 1: By Stokes’ Theorem
ZZ
S
curlF· dS = I
∂S
F· dr (2 pts)
The boundary of S is the circle r(t) = 2 cos ti + 2 sin tj + 0k. (2 pts) And
r0(t) = (−2 sin ti + 2 cos tj + 0k) (1 pt)
= I
∂S
F(r(t))· r0(t) dt (2 pts)
= Z 2π
0
((e02 + 2 sin t)i + (4× 0 − 2 sin t)j
+ (8 cos t sin(sin t))k)· (−2 sin ti + 2 cos tj + 0k) dt (1 pt)
= Z 2π
0
−2 sin t − 4 sin2t− 4 sin t cos t dt (1 pt)
= Z 2π
0
−2 sin t − 2 + 2 cos 2t − 2 sin 2t dt
= (2 cos t− 2 + 2 sin 2t + 2 cos 2t)¯¯¯2π
0
=−4π (1 pt)
Solution 2: curlF = (8x cos y− 4)i + (2zez2 − 8 sin y)j − k (2 pts) The surface of S is S(x, y) = (x, y, 4− x2− y2) (1 pt)
Sx = (1, 0,−2x), Sy = (0, 1,−2y), Sx× Sy = (−2x, 2y, 1) (2 pts) ZZ
S
curlF· dS = ZZ
D
(8x cos y− 4)i + (2zez2 − 8 sin y)j − k · (−2x, 2y, 1) dA (2 pts)
Because of the symmetric − ZZ
S
dA =−4π (3 pts)
Solution 3: curlF = (8x cos y− 4)i + (2zez2 − 8 sin y)j − k (2 pts)
Let D be the disk 0 = 4− x2− y2 where the orientation given (0, 0,−1) ZZ
S
curlF· dS = ZZZ
V
div(curlF)dV − ZZ
D
curlF· dA (4 pts)
= 0− ZZ
D
dA =−4π (3 pts)
(1 pt for knowing ZZZ
V
div(curlF)dV = 0)
Solution 4: By Stokes’ Theorem
ZZ
S
curlF· dS = I
∂S
F· dr (2 pts)
curlF = (8x cos y− 4)i + (2zez2 − 8 sin y)j − k (2 pts)
Let D be the disk 0 = 4− x2− y2 where the orientation given (0, 0, 1) By Stokes’ Theorem
I
∂S
F· dr = ZZ
D
curlF· dA (3 pts)
=− ZZ
D
dA =−4π (3 pts)
9. (10%) Evaluate ZZ
S
F·dS, where F(x, y, z) = 2y2j−zk and S is the surface of the solid enclosed by y = x2+ z2 and y = 1 with outward normal vector.
Sol:
method 1: Let F = P i + Qj + Rj = 0i + 2y2j− zk, then divF = ∂P
∂x + ∂Q
∂y +∂R
∂z = 4y− 1 (2 pts) By the divergence theorem, we have
ZZ
S
F· dS = ZZZ
R
divF dV (3 pts)
where R is the region bounded by S. Hence ZZ
S
F· dS = Z 2π
0
Z 1 0
Z 1 r2
(4y− 1)r dydrdθ (3 pts)
= 2π Z 1
0
r− 2r5+ r3dr = 5π
6 (2 pts) method 2:
ZZ
S
F· dS = ZZ
S
F· n dS
= ZZ
S1
F· n dS + ZZ
S2
F· n dS (2 pts)
where n is the outer normal of S, S1 = S ∩ {y = x2+ z2}, and S2 = S ∩ {y = 1}.
ZZ
S1
F· n dS = ZZ
A
(0, 2(x2+ z2)2,−z) · (2x, −1, 2z) dA (3 pts)
= Z 2π
0
Z 1 0
(−2r4− 2r2sin2θ)r drdθ
=−7π
6 (3 pts) ZZ
S2
F· n dS = ZZ
A
(0, 2,−z) · (0, 1, 0) dA
= Z 2π
0
Z 1 0
2rdrdθ = 2π (2 pts)
