Name
ID
MATH 253
Exam 2
Sections 201-202
Solutions
∫0 ∫0 ∫0
2
z
/45
12
/12
10
/12
13
/12
11
/12
14
/12
Fall 2006
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
1. Compute
1-9
Total
/105
xz
15x dy dx dz.
a. 4
b. 8
c. 16
Correct Choice
d. 32
e. 64
∫0 ∫0 ∫0
2
z
xz
15x dy dx dz =
2
= z5
2. Compute
z=0
∫0 ∫y
2
4
2
∫0 ∫0
2
z
15xy
xz
dx dz =
y=0
∫ 0 ∫ 0 15x 2 z dx dz = ∫ 0
2
z
2
5x 3 z
z
dz =
x=0
∫ 0 5z 4 dz
2
= 32
y sinx 2  dx dy by interchanging the order of integration.
a. − cos 16
2
cos 16 − 1
2
1 − cos 16
4
cos 16
8
cos 16 − 1
4
b.
c.
d.
e.
Correct Choice
Plot the region.
y
In the original order:
For each y between
1
0 and 2, x runs from y 2 to 4.
In the reversed order:
For each x between
0 and 4, y runs from 0 to
∫0 ∫y
2
4
y sinx 2  dx dy =
2
=
2
∫0 ∫0
4
− cosx 2 
4
x
y sinx 2  dy dx =
4
0
0
x.
0
∫0
4
y2
sinx 2 
2
x
0
dx = 1
2
2
4
x
∫ 0 x sinx 2  dx
4
= 1 − cos 16
4
1
3. Which of the following is the polar plot of r = cos3θ?
y
a.
b.
1
c.
0
2
4
6
x
-1
←Correct Choice
d.
e.
(c) is the rectangular plot of r = cos3θ.
(d) is its polar plot because there are 3 positive loops and
3 negative loops which retrace the positive loops.
4.
Find the area of the region inside the circle r = 4
y
4
outside the polar curve r = 3 + 2 cos2θ with y ≥ 0.
2
The area is given by the integral:
-4
a. A =
∫ π/3 ∫ 3+2 cos2θ r dr dθ
b. A =
∫ π/3 ∫ 4
c. A =
∫ π/3 ∫ 3+2 cos2θ dr dθ
d. A =
∫ π/6 ∫ 3+2 cos2θ dr dθ
e. A =
∫ π/6 ∫ 3+2 cos2θ r dr dθ
5π/3
5π/3
5π/3
5π/6
5π/6
0
2
4
x
4
3+2 cos2θ
dr dθ
4
4
4
r = 3 + 2 cos2θ = 4
A=
-2
Correct Choice
 cos2θ = 1
2
∫∫ dA = ∫ π/6 ∫ 3+2 cos2θ r dr dθ
5π/6
4
 2θ = π
3
 θ = π , 5π
6 6
by symmetry
2
5.
Find the volume of the apple given
in spherical coordinates by ρ = 3ϕ.
The volume is given by the integral:
a. 54π
∫0
b. 54π
∫ 0 ϕ 2 sin ϕ dϕ
c. 27π
∫0
d. 27π
∫ 0 ϕ 2 sin ϕ dϕ
e. 18π
∫ 0 ϕ 3 sin ϕ dϕ
2π
ϕ 2 sin ϕ dϕ
π
2π
ϕ 2 sin ϕ dϕ
π
π
V=
π
Correct Choice
∫∫∫ dV = ∫ 0 ∫ 0 ∫ 0
2π
3ϕ
ρ 2 sin ϕ dρ dϕ dθ = 2π ∫
π
0
ρ3
3
3ϕ
ρ=0
π
sin ϕ dϕ = 18π ∫ ϕ 3 sin ϕ dϕ
0
⃗ = y − z, x + z, y − x + 2z.
6. Find a scalar potential f for the vector field F
Then evaluate f1, 1, 1 − f0, 0, 0:
a. 1
b. 2
Correct Choice
c. 4
d. 5
e. 7
∂xf = y − z
f = xy − xz + gy, z
∂yf = x + z
x + ∂yg = x + z
−x + y + dh = y − x + 2z
dz
∂ z f = y − x + 2z
∂yg = z
dh = 2z
dz
g = yz + hz
f = xy − xz + yz + hz
h = z2
f = xy − xz + yz + z 2
f1, 1, 1 − f0, 0, 0 = 1 − 1 + 1 + 1 − 0 − 0 + 0 + 0 = 2
3
⃗ for any vector field F
⃗.
7. Which vector field cannot be written as ⃗
∇×F
⃗ = xz, yz, z 2 
a. A
⃗
⃗ = z + z + 2z = 4z ≠ 0
∇⋅A
Correct Choice
⃗ = x, y, −2z
b. B
⃗
⃗ = 1+1−2 = 0
∇⋅B
⃗ = xz, yz, −z 2 
c. C
⃗
⃗ = z + z − 2z = 0
∇⋅C
⃗
⃗ = z cos x − z cos x = 0
∇⋅D
⃗ = z sin x, −yz cos x, y sin x
d. D
⃗
⃗ = sin y − sin y = 0
∇⋅E
⃗ = x sin y, cos y, x cos y
e. E
⃗ cannot be A
⃗ =⃗
⃗ = 0, A
⃗.
∇⋅⃗
∇×F
∇×F
Since it is always true that ⃗
⃗, D
⃗, C
⃗ and E
⃗ probably have vector potentials.
On the other hand, B
8. Find the total mass of a plate occupying the region between x = y 2 and x = 4
if the mass density is ρ = x.
a. 64
b.
c.
d.
e.
5
128
5
256
5
32
3
32
9
x = y2 = 4
M=
Correct Choice
y = ±2
∫∫ ρ dA = ∫ −2 ∫ y
2
4
2
x dx dy =
∫ −2
2
x2
2
4
dy = 1
2
2
x=y
5
∫ −2 16 − y  dy = 12 16y − y5
2
2
4
y=−2
= 128
5
9. Find the center of mass of a plate occupying the region between x = y 2 and x = 4
if the mass density is ρ = x.
a.
b.
c.
d.
e.
20 , 0
7
12 , 0
5
512 , 0
7
128 , 0
5
14 , 0
5
M = 128
5
Correct Choice
By symmetry, ȳ = 0.
x-mom = M y =
∫∫ xρ dA = ∫ −2 ∫ y
= 2 128 − 128
7
3
2
= 512
7
4
7
1 ∫ 2 64 − y 6  dy = 1 64y − y
dy
=
7
3 −2
3
x=y 2
My
x-mom
x̄ =
=
= 512 5 = 20
M
M
7 128
7
x 2 dx dy =
2
∫ −2
2
x3
3
4
2
y=−2
4
Work Out: (12 points each. Part credit possible. Show all work.)
10.
Find the dimensions and volume of the largest box
which sits on the xy-plane and whose upper vertices
are on the elliptic paraboloid z = 12 − 2x 2 − 3y 2 .
You do not need to show it is a maximum.
You MUST eliminate the constraint.
Do not use Lagrange multipliers.
Maximize V = 2x2yz = 4xy12 − 2x 2 − 3y 2  = 48xy − 8x 3 y − 12xy 3
V x = 48y − 24x 2 y − 12y 3 = −12y2x 2 + y 2 − 4 = 0
V y = 48x − 8x 3 − 36xy 2 = −4x2x 2 + 9y 2 − 12 = 0
Since x = 0 or y = 0 gives V = 0, we can assume x ≠ 0 and y ≠ 0.
2x 2 + y 2 = 4
2x 2 + 9y 2 = 12
and
8y 2 = 8
Subtract:
 y=1
 x2 = 3
 x= 3
2
2
z = 12 − 2x 2 − 3y 2 = 12 − 2 3 − 31 = 6
2
2x 2 + 1 = 4
Substitute back:
Substitute back:
The dimensions are:
L = 2x =
V = LWH =
The volume is:
W = 2y = 2
6
H=z=6
6 26 = 12 6
11. A pot of water is sitting on a stove. The pot is a cylinder of radius 3 inches and height 4 inches.
If the origin is located at the center of the bottom, then the temperature of the water is
∫∫∫ T dV
T = 102 + x 2 + y 2 − z. Find the average temperature of the water:
T ave =
.
∫∫∫ dV
∫∫∫ dV = ∫ 0 ∫ 0 ∫ 0 r dz dr dθ =
2π
3
4
θ
2π
0
r2
2
3
z
0
4
0
= 2π 9 4 = 36π
2
∫∫∫ T dV = ∫ 0 ∫ 0 ∫ 0 102 + r 2 − zr dz dr dθ = 2π ∫ 0
2π
3
4
3
2
102z + r 2 z − z
2
= πr 2 h
4
r dr
z=0
= 2π ∫ 408 + 4r 2 − 8r dr = 2π ∫ 400r + 4r 3  dr = 2π200r 2 + r 4  3r=0 = 2π1881 = 3762π
3
3
0
0
T ave = 3762π = 209
36π
2
5
12.
Compute
∫∫ x dx dy
over the "diamond"
y
D
6
shaped region bounded by the curves
y = x2
y = 3 + x2
y = 4 − x2
4
y = 6 − x2
2
Use the curvilinear coordinates
u = y + x 2 and v = y − x 2 .
0
0.0
(Half credit for using rectangular coordinates.)
0.5
1.0
1.5
2.0
x
The boundary curves are v = 0, v = 3, u = 4, u = 6
We solve for x and y and compute the Jacobian factor:
u + v = 2y
J=
u − v = 2x 2
∂y
∂u
∂y
∂v
∂x
∂u
∂x
∂v
=
∫∫ x dx dy = ∫ 0 ∫ 4
3
u − v −1/2
2 2
1
2
u − v −1/2
2 2
1
2
−
The integrand is x =
6
D
x=
u−v
y = u+v
2
2
=
u − v −1/2
u − v −1/2
+
4 2
4 2
=
u − v −1/2
2 2
u−v
. So
2
u − v u − v −1/2
du dv =
2
2 2
∫0 ∫4
3
6
1 du dv = 1 32 = 3
4
4
2
13. The sides of a cylinder C of radius 3 and height 4 may be parametrized by
Rh, θ = 3 cos θ, 3 sin θ, h for 0 ≤ θ ≤ 2π and 0 ≤ h ≤ 4.
⃗ for F
⃗ ⋅ dS
⃗ = −yz 2 , xz 2 , z 3  and outward normal.
Compute ∫∫ ⃗
∇×F
C
⃗ = ⃗e h × ⃗e θ , ⃗
⃗ and
HINT: Find ⃗e h , ⃗e θ , N
∇×F
⃗
⃗
∇×F
⃗ h, θ .
R
Rh, θ = 3 cos θ, 3 sin θ, h
⃗e h = 
⃗e θ = 
î
0,
− 3 sin θ,
̂
0,
3 cos θ,
k̂
1 
0 
⃗ = ⃗e h × ⃗e θ = î−3 cos θ − ̂ 3 sin θ + k̂ 0 = −3 cos θ, −3 sin θ, 0
N
⃗ = 3 cos θ, 3 sin θ, 0
Reverse N
6
⃗
⃗ =
∇×F
î
̂
k̂
∂x
∂y
∂z
= î0 − 2xz − ̂ 0 − −2yz + k̂ z 2 − −z 2  = −2xz, −2yz, 2z 2 
−yz 2 xz 2 z 3
⃗
⃗
∇×F
⃗ h, θ
R
= −6h cos θ, −6h sin θ, 2h 2 
⃗
⃗⋅N
⃗ = −18h cos 2 θ − 18h sin 2 θ = −18h
∇×F
∫∫ ⃗∇ × F⃗ ⋅ dS⃗ = ∫ ∫ ⃗∇ × F⃗ ⋅ N⃗ dh dθ = ∫ 0 ∫ 0 −18h dh dθ = −2π9h 2  40 = −288π
2π
4
C
14. The hemispherical surface x 2 + y 2 + z 2 = 9 has surface density ρ = x 2 + y 2 .
⃗ ϕ, θ = 3 sin ϕ cos θ, 3 sin ϕ sin θ, 3 cos ϕ.
The surface may be parametrized by R
Find the mass and center of mass of the surface.
⃗ = ⃗e ϕ × ⃗e θ ,
HINT: Find ⃗e ϕ , ⃗e θ , N
⃗
N
⃗ ϕ, θ .
and ρ R
⃗ ϕ, θ = 3 sin ϕ cos θ, 3 sin ϕ sin θ, 3 cos ϕ
R
î
̂
k̂
⃗e ϕ = 3 cos ϕ cos θ, 3 cos ϕ sin θ, −3 sin ϕ
⃗e θ = −3 sin ϕ sin θ, 3 sin ϕ cos θ,
0 
⃗ = ⃗e ϕ × ⃗e θ = î9 sin 2 ϕ cos θ − ̂ −9 sin 2 ϕ sin θ + k̂ 9 sin ϕ cos ϕ cos 2 θ + 9 sin ϕ cos ϕ sin 2 θ
N
⃗ = 9 sin 2 ϕ cos θ, 9 sin 2 ϕ sin θ, 9 sin ϕ cos ϕ
N
⃗ =
N
2
2
9 sin 2 ϕ cos θ + 9 sin 2 ϕ sin θ + 9 sin ϕ cos ϕ 2 = 9 sin 4 ϕ + sin 2 ϕ cos 2 ϕ = 9 sin ϕ
⃗ ϕ, θ
ρ R
= 3 sin ϕ cos θ 2 + 3 sin ϕ sin θ 2 = 9 sin 2 ϕ
M=
∫∫ ρ dS = ∫∫ ρ
Let
u = cos ϕ
⃗ ϕ, θ
R
⃗ dϕ dθ =
N
π/2
∫0 ∫0
2π
9 sin 2 ϕ9 sin ϕ dϕ dθ = 2π81 ∫
π/2
0
1 − cos 2 ϕ sin ϕ dϕ
du = − sin ϕ dϕ
0
3
M = −162π ∫ 1 − u 2  du = −162π u − u
3
1
0
1
= 162π 1 − 1
3
= 324π = 108π
3
By symmetry x̄ = ȳ = 0
z-mom = M xy =
∫∫ z ρ dS = ∫∫ 3 cos ϕ ρ
= 2π243 ∫
z̄ =
π/2
0
⃗ ϕ, θ
R
cos ϕ sin 3 ϕ dϕ = 486π
⃗ dϕ dθ =
N
sin 4 ϕ
4
π/2
0
π/2
∫0 ∫0
2π
3 cos ϕ9 sin 2 ϕ9 sin ϕ dϕ dθ
= 243π
2
z-mom
= 243π 1 = 9
M
2 108π
8
7