Name
MATH 221
Exam 2
Section 503
Solutions
Fall 2009
1-13
/65
15
/10
14
/15
16
/15
Total
/105
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
1. Compute
2
x
∫ 0 ∫ −x 3y 2 dy dx.
Correct Choice
a. 8
b. 12
c. 16
d. 24
e. 0
2
x
∫ 0 ∫ −x
3y 2 dy dx =
2. Compute
9
∫0 ∫
2
∫0
3
y
y3
x
y= −x
dx =
2
∫0
x 3 − (−x) 3 dx =
π sin(πx 3 ) dx dy.
2
∫0
2x 3 dx =
x4
2
2
x=0
=8
HINT: Reverse the order of integration.
a. 1
b.
c.
d.
e.
y
9
2π
9
1
3
2
3
2π
3
Correct Choice
10
9
∫0 ∫
8
3
y
6
=
4
π sin(πx 3 ) dx dy =
3
∫0
0
x2
∫0 ∫0
[π sin(πx 3 )y] xy=0 dx =
= −1 cos(πx 3 )
3
2
3
2
3
0
π sin(πx 3 ) dy dx
3
∫ 0 πx 2 sin(πx 3 ) dx
= −1 cos(27π) + 1 cos(0) = 2
3
3
3
0 1 2 3
x
1
3.
Find the area inside the circle r = 2 cos θ
but outside the circle r = 1.
a. π − cos π
3
3
2π + cos 2π
3
3
π + sin 2π
3
3
2π − sin π
3
3
2π + sin π
3
3
b.
c.
d.
e.
2 cos θ = 1
A=
4. Compute
cos θ = 1
2

θ = −π, π
3 3

r 2 2 cos θ dθ = 1 π/3 (4 cos 2 θ − 1) dθ
∫ −π/3 ∫ 1
∫ −π/3 2 r=1
2 ∫ −π/3
π/3
π/3
2θ − 1 dθ = 1
(1 + 2 cos 2θ) dθ
∫ −π/3 4 1 + cos
2
2 ∫ −π/3
π/3
θ + sin 2θ
= 1 π + sin 2π − 1 − π + sin −2π
θ=−π/3
2 3
2
3
3
3
π + sin 2π + π + sin 2π = π + sin 2π
3
3
3
3
3
3
∫ ∫ 1 dA =
= 1
2
= 1
2
= 1
2
Correct Choice
2
π/3
∫0 ∫0
2 cos θ
4−x 2
ex
π/3
r dr dθ =
2 +y 2
HINT: Switch to polar coordinates.
dy dx.
a. π e 2
2
π e4
2
π (e 2 − 1)
2
π (e 4 − 1)
2
π (e 4 − 1)
4
b.
c.
d.
e.
Correct Choice
y2
2
∫0 ∫0
1
4−x 2
=
0
0
1
ex
π
2
2 +y 2
dy dx =
1 e r2
2
2
0
π/2
2
∫ 0 ∫ 0 e r r dr dθ
2
= π (e 4 − 1)
4
2
x
2
5. Compute the mass of the solid cone
x 2 + y 2 ≤ z ≤ 4 if the volume density is ρ = z.
a. 4π
b. 8π
c. 16π
d. 32π
Correct Choice
e. 64π
M=
4
z 2 4 dr = 2π 4 r 8 − r 2
z
r
dz
dr
dθ
=
2π
r
∫0 ∫0 ∫r
∫ 0 2 z=r
∫0
2
4
4
4
= 2π 4r 2 − r
= 2π 4 3 − 4
= 2π4 3 1 − 1 = 64π
8 r=0
8
2
∫∫∫ ρ dV =
2π
4
4
x 2 + y 2 ≤ z ≤ 4 if the volume density is ρ = z.
6. Compute the center of mass of the solid cone
a.
b.
c.
d.
e.
dr
0, 0, 5
16
0, 0, 16
Correct Choice
5
0, 0, 1024π
5
0, 0, 5
1024π
0, 0, 8
5
4
z 3 4 dr = 2π 4 r 64 − r 3
2
z
r
dz
dr
dθ
=
2π
r
∫0 ∫0 ∫r
∫ 0 3 z=r
∫0 3 3
4
5
5
= 2π 32r 2 − r
= 2π 2 ⋅ 4 4 − 4
= 2π 4 4 2 − 4 = 1024π
5 r=0
5
5
5
3
3
3
M xy
z̄ =
= 1024π 1 = 16
By symmetry x̄ = ȳ = 0
M
5
5
64π
M xy =
∫∫∫ zρ dV =
2π
4
4
7. Find the average value of the function f =
1
x2 + y2 + z2
spheres x 2 + y 2 + z 2 = 1 and x 2 + y 2 + z 2 = 4.
dr
over the solid region between the two
f dV
HINT: f ave = ∫∫∫
∫∫∫ 1 dV
a. 3
4
b. 5
8
c. 3
7
Correct Choice
d. 4π
e. 8π
∫∫∫ f dV =
π
2π
2
∫0 ∫0 ∫1
1 ρ 2 sin ϕ dρ dθ dϕ = − cos ϕ
ρ2
π
∫∫∫ 1 dV = Volume = 43 π2 3 − 43 π1 3 = 28
3
π
0
(2π)(2 − 1) = 4π
f ave = 4π3 = 3
7
28π
3
2
2
y
∫∫ cos x9 + 4 dx dy over the region inside the ellipse
HINT: Elliptical coordinates are x = 3t cos θ, y = 2t sin θ.
8. Compute
2
x 2 + y = 1.
9
4
a. 6π sin(1) − 6π
Correct Choice
b. 6π sin(1)
c. 6π − 6π cos(1)
d. −6π cos(1)
e. 6π cos(1)
∂(x, y)
∂(t, θ)
J=
3 cos θ
=
2 sin θ
= |6t cos 2 θ + 6t sin 2 θ| = 6t assuming t ≥ 0
−3t sin θ 2t cos θ
2
2
x 2 + y = (3t cos θ) + (2t sin θ) = t 2 < 1
0≤t≤1
9
4
9
4
2
2π 1
y2
1
∫∫ cos x9 + 4 dx dy = ∫ 0 ∫ 0 cos(t 2 ) 6t dt dθ = 2π3 sin(t 2 )| 0 = 6π sin(1)
2
9. Compute
2 ,2
∫ (0,0)
x ds along the parabola y = x 2 parametrized as ⃗r(t) = (t, t 2 ).
a. 13
b.
c.
d.
e.
Correct Choice
6
9
4
27
4
1 (17 3/2 − 1)
12
17 3/2
12
⃗v = (1, 2t)
2 ,2
∫ (0,0)
|v⃗| =
x ds =
10. Compute
∫0
2
1 + 4t 2
x |v⃗| dt =
2 ,2
∫ (0,0)
∫0
x=t
2
t 1 + 4t dt =
2
(1 + 4t 2 ) 3/2
12
2
0
= 13
6
⃗ f ⋅ ds⃗ for f = xy along the parabola y = x 2 parametrized as ⃗r(t) = (t, t 2 ).
∇
a. 2
b. 2 2
Correct Choice
c. 4
d. 4 2
e. 8
⃗v = (1, 2t)
2 ,2
∫ (0,0)
⃗ f = (y, x) = (t 2 , t)
∇
⃗ f ⋅ ds⃗ =
∇
∫0
2
⃗ f ⋅ ⃗v dt =
∇
∫0
⃗ f ⋅ ⃗v = t 2 + 2t 2 = 3t 2
∇
2
3t 2 dt = t 3
2
0
=2 2
4
⃗ = (−yz 2 , xz 2 , z 3 ) counterclockwise around the circle x 2 + y 2 = 4 with
for F
HINT: Parametrize the circle.
11. Compute
z = 4.
∫ F⃗ ⋅ ds⃗
a. 64
b. 128
c. 64π
Correct Choice
d. 128π
e. 256π
⃗r(θ) = (2 cos θ, 2 sin θ, 4)
⃗ (r⃗(θ)) ⋅ ⃗v = 64 sin 2 θ + 64 cos 2 θ = 64
F
12. Compute
⃗ (r⃗(θ)) = (−32 sin θ, 32 cos θ, 64)
F
⃗v = (−2 sin θ, 2 cos θ, 0)
∫∫∫ ∇⃗ ⋅ F⃗ dV
∫ F⃗ ⋅ ds⃗ = ∫ 0
2π
⃗ ⋅ ⃗v dθ =
F
2π
∫0
64 dθ = 128π
⃗ = (xy 2 , yz 2 , zx 2 ) over the solid hemisphere x 2 + y 2 + z 2 ≤ 25 with z ≥ 0.
for F
a. 125 π
3
b. 250 π
3
c. 500 π
3
d. 1250π
Correct Choice
e. 2500π
⃗ = y2 + z2 + x2 = ρ2
⃗ ⋅F
∇
∫∫∫ ∇⃗ ⋅ F⃗ dV =
13.
π/2
2π
5
∫0 ∫0 ∫0
ρ 2 ρ 2 sin ϕ dρ dθ dϕ = − cos ϕ
π/2
0
(2π)
ρ5
5
5
= 2π5 4 = 1250π
0
y
Which of the following vector fields
are plotted at the right?
a. (y 2 , 4x)
c. (4x, y 2 )
2
-4
b. (−y 2 , 4x)
4
-2
2
4
-2
Correct Choice
x
-4
d. (4x, −y 2 )
e. (4xz, 4y)
All of the vectors in the plot have a positive y-component. This only happens for vector field (c).
5
Work Out: (Points indicated. Part credit possible. Show all work.)
14. (15 points) Find the volume of the largest box such the length plus twice the width plus three times
the height is 36.
Maximize V = LWH subject to the constraint g = L + 2W + 3H = 36.
Lagrange Multiplier Method
⃗ V = (WH, LH, LW)
∇
⃗ g = (1, 2, 3)
∇
⃗ V = λ∇
⃗ g or
The Lagrange equations are ∇
WH = λ
LH = 2λ
LH = 2WH


LW = 3WH
LW = 3λ
W= L
2
L
H=
3
where we use H ≠ 0 and W ≠ 0 to give a maximum volume.
Plug into the constraint:
L + 2 L + 3 L = 3L = 36
2
3
L = 12
W=6
H=4
V = 12 ⋅ 6 ⋅ 4 = 288
Eliminate a Variable Method:
L = 36 − 2W − 3H
V = WH(36 − 2W − 3H) = 36WH − 2W 2 H − 3WH 2
V W = 36H − 4WH − 3H 2 = 0
36 − 4W − 3H = 0

V H = 36W − 2W − 6WH = 0
2
36 − 2W − 6H = 0
where we use H ≠ 0 and W ≠ 0 to give a maximum volume.
4W + 3H = 36
2W + 6H = 36

8W + 6H = 72
2W + 6H = 36
6H = 36 − 2W = 36 − 12 = 24

6W = 36

W=6

H=4
L = 36 − 2W − 3H = 36 − 12 − 12 = 12
V = 12 ⋅ 6 ⋅ 4 = 288
⃗ (p, q) = (p, p + q 2 , p − q 2 ) for 0 ≤ p ≤ 3 and 0 ≤ q ≤ 2.
15. (10 points) Find the area of the surface R
⃗e p =
⃗e q =
î
̂
k̂
(1
1
1)
( 0 2q −2q )
⃗ = ⃗e p × ⃗e q = î(−2q − 2q) − ̂ (−2q) + k̂ (2q) = (−4q, 2q, 2q)
N
⃗ = 16q 2 + 4q 2 + 4q 2 = 24 q
N
A=
∫ ∫ 1 dS =
∫∫
⃗ dp dq =
N
2
3
∫0 ∫0
24 q dp dq =
24 p
3
0
q2
2
2
= 6 24 = 12 6
0
6
∫∫ P ∇⃗ × F⃗ ⋅ dS⃗
⃗ = (−yz 2 , xz 2 , z 3 ) over the paraboloid z = x 2 + y 2 with
for F
⃗ (r, θ) = (r cos θ, r sin θ, r 2 ).
z ≤ 4, oriented up and in, and parametrized by R
16. (15 points) Compute
⃗ =
⃗ ×F
∇
î
̂
k̂
∂x
∂y
∂z
−yz 2 xz 2 z 3
⃗
⃗ ×F
∇
⃗e r =
⃗e θ =
= î(0 − 2xz) − ̂ (0 − −2yz) + k̂ (z 2 − −z 2 ) = (−2xz, −2yz, 2z 2 )
⃗ (r, θ)
R
= (−2r 3 cos θ, −2r 3 sin θ, 2r 4 )
î
̂
k̂
( cos θ
sin θ
2r)
(−r sin θ r cos θ 0 )
⃗ = ⃗e r × ⃗e θ = î(0 − 2r 2 cos θ) − ̂ (0 − −2r 2 sin θ) + k̂ (r cos 2 θ − −r sin 2 θ) = (−2r 2 cos θ, −2r 2 sin θ, r)
N
⃗ is correctly oriented up and in.
N
⃗⋅N
⃗ = 4r 5 cos 2 θ + 4r 5 sin 2 θ + 2r 5 = 6r 5
⃗ ×F
∇
z = r2 ≤ 4

r≤2
∫∫ P ∇⃗ × F⃗ ⋅ dS⃗ = ∫ 0 ∫ 0 ∇⃗ × F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 6r 5 dr dθ = 2π[r 6 ] 0 = 128π
2π
2
2π
2
2
7