1012微
微甲01-04班
班期 末 考 解 答 和 評 分 標 準
ˆ
2ˆ 8
1. (10%) Evaluate
0
x3
x5
p
dydx.
x6 + y 2
Solution:
Interchange the order of the iterated integral, we have
ˆ
0
2ˆ 8
x3
ˆ
x5
p
dydx =
x6 + y 2
8ˆ
0
ˆ
8
=
0
1
3
ˆ
√
3
y
x5
p
dxdy (3%)
x6 + y 2
0
√
3y
1 p 6
2
2 x + y dy (3%)
6
0
8
√
( 2 − 1)y dy (2%)
1 2 8
1 √
= ( 2 − 1) y (1%)
3
2 0
√
32
= ( 2 − 1) (1%)
3
=
0
Page 1 of 8
√
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 +y 2 +z 2 = 4; sphere ρ = 2 2 cos φ is equivalent to x2 +y 2 +(z − 2)2 =
2. These two spheres intersect at φ = π/4.
Therefore,
ˆ
ˆ
2π
π
4
ˆ
ˆ
2
2π
ˆ
ˆ
π
2
√
2 2 cos φ
ρ2 sin φ dφ +
dθ
dφ
π
0
0
I
4
ˆ π
π ρ3 2
2
1 √
4
= 2π(− cos φ) · ( ) + 2π
sin φ · (2 2 cos φ)3 dφ
π
3 0
3
0
√ ˆ π 4
√
32 2π 2
2π
(8 − 4 2) +
sin φ cos3 φ dφ
=
π
3
3
4
√
2π
2π
=
(8 − 4 2) +
3
3
√
16
= π − 2 2π.
3
Volume =
dθ
0
dφ
0
0
ρ2 sin φ dφ
II
Note:
θ : 0 ∼ 2π, (2 points) , Jacobi factor ρ2 sin φ, (5 points) .
For part I,, φ : 0 ∼ π/4, (2 points) , ρ : 0 ∼ 2, (1 point) , answer
√
2π
(8 − 4 2), (1 point)
3
√
2π
For part II, φ : π/4 ∼ π/2, (2 points) , ρ : 0 ∼ 2 2 cos φ, (1 point) , answer
, (1 point)
3
Method 2.
ˆ
ˆ
2π
Volume =
√
dθ
0
ˆ
= 2π
√
ˆ
2
dr
0
2
r(
√
√
4−r2
√
2− 2−r2
r dz
p
p
√
4 − r2 + 2 − r2 − 2) dr
0
√
16
= π − 2 2π.
3
Page 2 of 8
Note:
p
p
√
√
θ : 0 ∼ 2π, (2 points) , z : 2 − 2 − r2 ∼ 4 − r2 , (3 points) , r : 0 ∼ 2, (3 points) ,
Jacobi factor r, (5 points) , answer
√
2π
(8 − 3 2), (2 points) .
3
Method 3.
ˆ
4 √ 3 1
2
π(4
−
z
)
dz
+
π( 2) ·
√
3
2 II
2
I
√
16
= π − 2 2π.
3
2
Volume =
Page 3 of 8
¨
sin(3x2 − 2xy + 3y 2 )dxdy, where Ω is the ellipse 3x2 − 2xy + 3y 2 ≤ 2. You
3. (15%) Evaluate the integral
Ω
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 + 3y 2 = 4u2 + 8k 2 v 2 ≤ 2
1
We can choose k = √ , then the equation becomes 4u2 + 4v 2 ≤ 2. Calculate the Jacobian (value of
2
J 2 pts , absolute value 2 pts):
√
1 k ∂(x, y)
| = | − 2k| = 2
|J(u, v)| = |
| = |
1 −k ∂(u, v)
Then
¨
sin(3x2 − 2xy + 32 ) dxdy
Ω
¨
√
sin 4(u2 + v 2 ) 2 dudv
=
(integrand 1 pt, domain 2 pts)
u2 +v 2 ≤ 21
√ ˆ
= 2
0
√
2π ˆ
ˆ
1
√
2
sin 4r2 rdrdθ
(Jacobian of polar coordinates 3 pts)
0
√1
2
1
=2 2π
sin 4r2 d(4r2 )
8
0
√
2π
=
(1 − cos 2) (2 pts. If make a slight mistake, get 1 pt)
4
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
4. (15%) For y > 0, let
F(x, y, z) = (e−x ln y − z)i + 2yz − e−x /y j + (y 2 − x)k and
G(x, y, z) = e−x ln 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.
ˆ
(b) Evaluate the line integral
G(r) · dr, where C is the curve given by r(u) = (1 + u2 )i + eu j + (1 + u)k,
C
u ∈ [0, 1].
Solution:
(a) f = −e−x ln y + y 2 z − xz + c, where c ∈ R.( 5 points)
(b) G = F + zi − y 2 k, and r(0) = (1, 1, 1), r(1) = (2, e, 2).
Then
ˆ
ˆ
ˆ
zdx − y 2 dz
F (r) · dr +
C
ˆ 1
[(1 + u)2u − e2u ]du(4points)
= f (2, e, 2) − f (1, 1, 1) +
G(r) · dr =
C
C
0
2
1
= −e + 2e − 4 + ( u3 + u2 − e2u )|10 (4 points)
3
2
3 2
11
−2
= e − e − (2points)
2
6
−2
2
Page 5 of 8
5. (15%) Letffi C be a piecewise-smooth Jordan curve that does not pass through the origin.
−y 5
xy 4
Evaluate
dx
+
dy for the following two cases, where C is traversed in the counter2
2 3
(x2 + y 2 )3
C (x + y )
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) =
−y 5
(x2 + y 2 )3
Q(x, y) =
xy 4
(x2 + y 2 )3
and
are well-defined and differentiable in Ω. We have
∂P
y 6 − 5x2 y 4
∂Q
(x, y) =
(x, y) = 2
∂x
∂y
(x + y 2 )4
in Ω. Therefore, by Green’s theorem,
˛
ˆ ˆ ∂Q ∂P
P dx + Qdy =
−
dxdy = 0.
∂y
∂x
C
Ω
Grading Policy:
Application of Green’s theorem: 2%
Correct calculation of partial derivatives: 2%
Correct answer 3%
(b)
Let Cr be the curve θ 7→ (r cos θ, r sin θ), θ ∈ [0, 2π], where r is small enough such that Cr 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
˛
˛
ˆ ˆ ∂Q ∂P
P dx + Qdy −
P dx + Qdy =
−
dxdy,
∂y
∂x
C
Cr
Ω
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
˛
˛
P dx + Qdy =
P dx + Qdy
C
r ˆC2π
−r5 sin5 θ
r5 cos θ sin4 θ
=
(−r sin θ) +
(r cos θ) dθ
r6
r6
ˆ0 2π
sin4 θdθ
=
0
=
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
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(z 2 ) j + 3z − tan(y 2 ) 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.
˛
F · dr.
(c) Evaluate
C
Solution:
(a)
i) S : {(x, y, y)| 0 ≤ y ≤ x ≤ a } (2 points)
1
ii) n = √ (0, −1, 1) (2 points)
2
i
j
k ∂
∂
∂ (b) (2 points)
∂x ∂y ∂z Fx Fy Fz = 2z cos z 2 − 2y sec2 y 2 i − cos x2 j + k (2 points)
˛
¨
ˆ aˆ x
1
√ (1 + cos x2 ) · |rx × ry |dydx (4 points)
(c)
F·dr =
∇ × F · n dσ =
2
c
S
0
0
ˆ a
2
2
x + sin x a 1 2
x(1 + cos x2 )dx =
=
|0 = (a + sin a2 ) (3 points)
2
2
0
Page 7 of 8
7. (15%) Let S1 be the surface {(x, y, z)| z = x2 +y 2 , z ≤ y}, S2 be the surface {(x, y, z)| z = y, x2 +y 2 ≤ 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 + y 2 ), x2 + y 2 ≤ y
∂f1
=(1, 0, 2x)
∂x
∂f1
=(0, 1, 2y)
∂y
∂f1 ∂f1
×
=(−2x, −2y, 1)
∂x
∂y
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.
¨
¨
V · d area =
−2xy + 2xy − (x2 + y 2 ) dx dy
S1
ˆ
S1
π
ˆ
(1 pt)
sin θ
−r2 r dr dθ
=
(3 pts)
ˆ
−1 3 1
−1
sin4 θ dθ =
π
4 0
4 42
−3π
=
(2 pts)
32
θ=0
r=0
π
=
(b)
Let x = r cos θ, y = r sin θ.
 2

r ≤ z ≤ y
⇒ 0 ≤ r ≤ sin θ


0≤θ≤π
(
r2 ≤ z ≤ y
Ω : x2 + y 2 ≤ z ≤ y ⇒
r2 ≤ r sin θ
By divergent theorem,
˚
¨
∇ · V d volumn =
Ω
¨
¨
ˆ
π
ˆ
sin θ
ˆ
r sin θ
∇ · V d volumn =
Ω
S2
S1
∂Ω=S1 +S2
˚
V · d area
V · d area +
V · d area =
1 r dz dr dθ
z=r2
θ=0 r=0
ˆ π ˆ sin θ
(r sin θ − r2 ) r dr dθ
=
θ=0
ˆ π
r=0
r3 sin θ r4 sin θ
− dθ
3 0
4 0
θ=0
ˆ π
1
1
=
sin4 θ − sin4 θ dθ
3
4
0
1 31
π
=
π =
(2 pts)
12 4 2
32
=
sin θ
¨
⇒
V · d area =
S2
π
−3π π
−
=
32
32
8
Page 8 of 8
(2 pts)
(3 pts)
(1 pt).