1012微微微甲甲甲07-11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%) Find the volume of the solid bounded below by the cone z2= 4(x2+ y2) and above by the ellipsoid 4(x2+ y2) + z2 = 8.
Solution:
Method 1 Use cylindrical coordinates:
Note that the region is bounded below by the cone and above by the ellipsoid, so it only consist of the part above the xy-plane!
V =
¨
x2+y2≤1
(p
8 − 4(x2+ y2) −p
4(x2+ y2)) dxdy
= ˆ 2π
0
ˆ 1
0
2(p
2 − r2− 2r) rdrdθ
=8π 3 (
√ 2 − 1) Scoring to parts of this method:
1. Integral domain x2+ y2 ≤ 1: 2 pts 2. Upper bound of z , i.e. p
8 − 4(x2+ y2): get 2 pts 3. Lower bound of z , i.e. p
4(x2+ y2): get 2 pts 4. Jacobian of polar coordinates: get 2 pts
5. Your result fits the correct answer: get 2 pts, if you make a slight mistake, get 1 pt.
Method 2 Use spherical coordinates:
Let (x, y, z) = (√
2r sin φ cos θ,√
2r sin φ sin θ, 2√
2r cos φ). The Jacobian is
|J (u, v)| = |∂(x, y, z)
∂(r, θ, φ)| = |
√
2 sin φ cos θ √
2r sin φ sin θ 2√ 2 cos φ
−√
2r sin φ sin θ
√
2r sin φ cos θ 0
√
2r cos φ cos θ √
2r cos φ sin θ −2√ 2r sin φ
|
= 4√
2r2sin φ Then the corresponding domain is
0 ≤ r ≤ 1 0 ≤ φ ≤ π 0 ≤ θ ≤ 2π4
V = ˆ π
4
0
ˆ 2π
0
ˆ 1
0
4
√
2r2sin φdrdφdθ Scoring to parts of this method:
1. Integral domain for r: 2 pts 2. Integral domain for φ: get 2 pts
3. Jacobian of spherical coordinate: get 4 pts, and if you miss the multiple before r2sin φ, get 2 pts.
4. Your result fits the correct answer: get 2 pts, if you make a slight mistake, get 1 pt.
Page 1 of 10
2. (12%) Evaluate
Ω
e−4x2−9y2dxdy, where Ω is the region satisfying 2x ≤ 3y and x ≥ 0.
Solution:
Let
x = r 2cos θ y = r
3sin θ (3%) then
J = r
6 (3%)
Therefore ˆ ˆ
Ω
e−4x2−9y2dxdy
=
ˆ π/2
π/4
ˆ ∞
0
e−r2|J |drdθ (3%)
= π
48 (3%)
Page 2 of 10
3. (12%) Evaluate the surface integral
¨
S
(x2+ y2)zdσ, where S is the part of the plane z = 4 + x + y that lies inside the cylinder x2+ y2 = 4.
Solution:
To parametrize this surface: r = (x, y, 4 + x + y), where x2+ y2 ≤ 4.
So dσ = s
∂z
∂x
2
+ ∂z
∂y
2
+ 1 · dxdy =√ 3dxdy
¨
S
(x2+ y2)zdσ
=
¨
x2+y2≤4
(x2+ y2)(4 + x + y)√ 3dxdy
= ˆ 2π
0
ˆ 2
0
(r2)(4 + r cos θ + r sin θ)√ 3rdrdθ
= ˆ 2π
0
ˆ 2
0
(r2)(4)√
3rdrdθ (by symmetry)
= ˆ 2π
0
ˆ 2
0
4
√
3r3drdθ
= ˆ 2π
0
16√ 3dθ
= 32
√ 3π
評分標準如下:
將曲面參數化: 2分
算出曲面的Jacobian: 2分
將面積分換成可處理的積分式，並且寫出完整的積分區域: 4分
計算積分: 4分 其餘錯誤酌量扣分
將此積分視為三重積分者一律不給分!!!!
Page 3 of 10
ˆ
C
(2x sin(πy) − ez) dx + (πx2cos(πy) − 3ez) dy − xezdz
along the curve C = n
(x, y, z)|z = lnp
1 + x2, y = x, 0 ≤ x ≤ 1 o
.
Solution:
Since F = (2x sin(πy) − ez)i + (πx2cos(πy) − 3ez)j − xezk has F + 3ezj = ∇(x2sin(πy) − xez) (4 points), so
ˆ
C
F · dr = −√ 2 −
ˆ 1
0
3ezdy (3 points)
= −√ 2 − 3
ˆ 1
0
p1 + x2dx (3 points)
= −(5√
2 + 3 ln(√
2 + 1))/2 (2 points)
Page 4 of 10
5. (10%) Evaluate ffi
r=1−cos θ
(x2y + y)dx − (xy2− x)dy with the curve oriented counterclockwise.
Solution:
By Green’s theorem,
˛
r=1−cos θ
(x2y + y)dx − (xy2− x)dy =
¨
Ω
−(x2+ y2)dA
= ˆ 2π
0
ˆ 1−cos θ
0
−r2· rdrdθ
= −1 4
ˆ 2π
0
(1 − cos θ)4dθ
= −35 16π.
Green’s thm : 3pts , region : 3pts ,Jacobian : 2pts , computation : 2pts.
Page 5 of 10
6. (12%) Let V = (2x − y)i + (2y + z)j + x2y2z2k and let S be the upper half of the ellipsoid 4 +
9 +
25 = 1.
Find the flux of curlV in the direction of the upper unit normal n (pointing away from the origin.).
Solution:
Solution 1
By using Stokes’ Theorem,
¨
S(O × V) · n ds =
˛
C
V(r)dr (4 points)
C : r(θ) = 2 cos θi + 3 sin θj , θ ∈ [0, 2π]
˛
C
V(r)dr = ˆ 2π
0
(10 sin θ cos θ + 6 sin2θ) dθ (4 points)
= [5
2(− cos 2θ) + 3(θ −1
2sin 2θ)] |2π0
= 6π. (4 points)
Solution 2
O × V = (2yx2z2− 1)i + (−2xy2z2)j + k. (3points)
∵ O · (O × V)= 0 (2points)
By using divergence theorem,
∴
¨
(O × V)·n ds =
˚
O · (O × V)dv = 0 (3points) In the suface of the buttom, n = −k. (1points)
We can find that the solution is 0 − [−(2 · 3π)] = 6π. (3points) ps. Using Stokes’ Theorem twice is permitted.
ps2. If students observe the unit normal of the bottom surface, and they only calculate the k-component of curl V. They do NOT pay for without calculating other components.
Page 6 of 10
7. (12%) Evaluate the flux of
V(x, y, z) = (z2x + y2z)i + 1
3y3+ z tan x
j + (x2z + 2y2+ 1)k
across S: the upper half sphere x2+ y2+ z2 = 1, z ≥ 0 with normal pointing away from the origin.
Solution:
S :
(x2+ y2+ z2 = 1
0 ≤ z upward.
S1 :
(z = 0
x2+ y2≤ 1 downward.
Ω :
(x2+ y2+ z2 ≤ 1 0 ≤ z
Let
x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ
.
Then
Ω :
(ρ2 ≤ 1
0 ≤ ρ cos φ ⇒
0 ≤ ρ ≤ 1 0 ≤ φ ≤ π/2 0 ≤ θ ≤ 2π
S :
(ρ2 = 1
0 ≤ ρ cos φ ⇒
ρ = 1 0 ≤ φ ≤ π/2 0 ≤ θ ≤ 2π
.
Let
x = r cos θ y = r sin θ z = z
.
Then
S1 :
(z = 0 r2 ≤ 1 ⇒
z = 0 0 ≤ r ≤ 1 0 ≤ θ ≤ 2π
d area = −k r dr dθ
By divergent theorem,
˚
Ω
∇ · V d volumn =
¨
∂Ω=S+S1
V · d area =
¨
S
V · d area +
¨
S1
V · d area (2 pts).
˚
Ω
∇ · V d volumn =
˚
Ω
z2+ y2+ x2d volumn =
˚
Ω
ρ2· ρ2sin φ dρ dφ dθ
= ˆ 2π
θ=0
ˆ π/2
φ=0
ˆ 1
ρ=0
ρ4sin φ dρ dφ dθ (3 pts)
= ρ5 5
1
0· (− cos φ)
π/2
0 · 2π =2
5π (1 pt)
Page 7 of 10
S1
V · d area =
S1
−(x2z + y2+ 1) r dr dθ
= − ˆ 2π
θ=0
ˆ 1
r=0
(r sin θ)2+ 1 r dr dθ (3 pts)
= −2 · r4 4
1
0· π − π = −3π
2 (1 pt)
⇒
¨
S
V · d area =
˚
Ω
∇ · V d volumn −
¨
S1
V · d area
= 19π
10 (2 pts)
Page 8 of 10
8. (10%) Find stationary points of f = 3xy −x3−y3+2. Determine which are local maximum, local minimum or a saddle point.
Solution:
∇f = (3y − 3x2, 3x − 3y2) = (0, 0) ⇒ y = x2 and x = y2
⇒ (x, y) = (0, 0) and (1, 1) (stationary points) D = fxxfyy− fxy2 = 36xy − 9
D(0, 0) = −9 < 0 ⇒ (0,0) is a saddle point.
D(1, 1) = 27 > 0 and fxx(1, 1) = −6 < 0 ⇒ (1,1) is a local maximum.
2 pts if stationary points are all wrong.
5 pts if D is wrong or you just get one stationary point.
6 − 8 pts if you have some mistakes.
Page 9 of 10
the curve 2x2− 2xy + y2 = 1. (You don’t have to give the locations of these extrema.)
Solution:
by Lagrange multiplier, we have
6x = λ(4x − 2y) and −4y = λ(−2x + 2y) and 2x2− 2xy + y2= 1 if λ = 0 then x = y = 0 it is not satisfies 2x2− 2xy + y2 = 1
so, λ 6= 0 we get 6x
λ − 4x = −2y and −4y
λ − 2y = −2x and 2x2− 2xy + y2 = 1
⇒ (λ6 − 4)
−2 = −2
−4 λ − 2
⇒ λ = 2 or −3
if λ = 2 then x = 2y and y = x 2
from 2x2− 2xy + y2 = 1 we get 8y2− 4y2+ y2= 1 ⇒ y2= 1 5 and 2x2− x2+x2
4 = 1 ⇒ x2 = 4 5 so, 3x2− 2y2 = 2 (maximum) if λ = −3 then 3x = y and x = y
3
from 2x2− 2xy + y2 = 1 we get 2x2− 6x2+ 9x2 = 1 ⇒ x2= 1 5 and 2
9y2−2
3y2+ y2= 1 ⇒ y2= 9 5 so, 3x2− 2y2 = −3 (minimum)
no point if you don’t use Lagrange multiplier.
2 − 3 pts if you don’t get any extreme values.
5 pts if both extreme values are wrong.
7 − 8 pts if one of the extreme values is wrong.
Page 10 of 10
