Name
Sec
MATH 253
Exam 2
Section 501,503
Fall 2009
Solutions
2
/60
15
/15
13
/15
16
/5
14
/10 Total
/105
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
1. Compute
1-12
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
b.
c.
d.
e.
3
3
2π + cos 2π
3
3
π + sin 2π
3
3
2π − sin π
3
3
2π + sin π
3
3
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
b.
c.
d.
e.
2
π e4
2
π (e 2 − 1)
2
π (e 4 − 1)
2
π (e 4 − 1)
4
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
d. 128π
Correct Choice
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 =
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
5
Work Out: (Points indicated. Part credit possible. Show all work.)
13. (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.
14. (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
15. (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
16. (5 points) Mark the Project that you worked on and then answer the questions in 1 or 2 sentences.
Cauchy’s Gradient Method
How did the procedure for finding a maximum differ from the procedure for finding a minimum?
Seeing a Blimp
Approximately how many degrees wide was the blimp? (1 digit accuracy)
The Trash Dumpster
Draw the basic shape of your dumpster. Label your variables. Label your hinges and welds.
Locating an Apartment
What was the geometrical condition satisfied by the location of the apartment?
A diagram may help.
Rectangles and Triangles
Where were the lines in the rectangle which minimized the sum of the areas?
Where were the lines in the triangle which minimized the sum of the areas?
Two diagrams may help.
Exact Gradient Method
What is the relation between the velocity of your curve and the gradient of the function
to find a maximum? What is the relation to find a minimum?
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