Name
1-9
/54
11c
/10
Spring 2009
10
/10
11d
/10
P. Yasskin
11a
/10
11e
/6
11b
/10 Total
/104
Sec
MATH 253
Exam 2
Sections 200,501,502
Solutions
Multiple Choice: (6 points each. No part credit.)
1. Find the volume under z = xy 2 above the rectangle 1 ≤ x ≤ 2 and 0 ≤ y ≤ 2.
a. 2
Correct Choice
b. 4
c. 6
d. 12
e. 16
3
V=
2
2
∫0 ∫1
2
x2
2
xy 2 dx dy =
2. (Non-Honors Only)
x=1
y3
3
Compute
2
4−1
2
=
y=0
2
y
8
3
=4
y
∫ 0 ∫ 0 ∫ x x dz dx dy.
a. 1
2
2
3
3
4
4
5
5
6
b.
c.
d.
e.
2
y
y
∫0 ∫0 ∫x
Correct Choice
x dz dx dy =
=
2
∫0
2
y
∫0 ∫0
y3
y3
−
2
3
xz
y
z=x
dy =
dx dy =
2
∫0
2
y
∫0 ∫0
y3
dy =
6
(xy − x 2 ) dx dy =
y4
24
2
y=0
2
∫0
3
x2y
− x
3
2
y
dy
x=0
= 16 = 2
24
3
1
over the region R in the first octant bounded by y = 9 − x 2 , z = 2 and the
∫∫∫ R z dV
3. Compute
coordinate planes.
Correct Choice
a. 36
b. 54
c. 72
d. 96
e. 108
2
3
9−x 2
∫∫∫ R z dV = ∫ 0 ∫ 0 ∫ 0
4.
z dy dx dz =
2
3
9−x 2
∫ 0 z dz ∫ 0 ∫ 0
1 dy dx =
2
z2
2
3
∫ (9 − x 2 ) dx = 2
0 0
3
9x − x
3
3
0
= 36
Find the mass of the plate bounded by the curves
x = y 2 and x = 4, if the surface mass density is ρ = x.
a.
32
3
b.
64
3
c.
128
3
d.
64
5
e.
128
5
M=
2
y
x
Correct Choice
4
∫ −2 ∫ y
x dx dy =
2
4
x2
2
2
∫ −2
x=y 2
dy =
2
∫ −2
8−
y4
2
dy =
8y −
y5
10
2
y=−2
= 2 16 − 16
5
= 128
5
5. Find the center of mass of the plate bounded by the curves
x = y 2 and x = 4, if the surface mass density is ρ = x.
a. (x̄ , ȳ ) =
12 , 0
7
b. (x̄ , ȳ ) = (2, 0)
c. (x̄ , ȳ ) =
d. (x̄ , ȳ ) =
e. (x̄ , ȳ ) =
My =
x̄ =
2
4
∫ −2 ∫ y
20 , 0
7
24 , 0
7
512 , 0
7
x 2 dx dy =
2
Correct Choice
2
∫ −2
x3
3
My
= 512 ⋅ 5 = 20
M
7
7
128
4
dy = 1
2
3
x=y
y7
∫ −2 (64 − y 6 ) dy = 13 64y − 7
2
2
y=−2
= 2 128 − 128
7
3
= 512
7
ȳ = 0 by symmetry.
2
6.
A styrofoam board is cut in the shape of the
y
right half of the cardioid r = 1 − sin θ.
x
A static electricity charge is put on the board
whose surface charge density is given by ρ e = x.
Find the total charge on the board Q = ∫∫ ρ e dA.
a. 0
b. 2
3
c. 4
Correct Choice
3
d. 8π
3
e. 16π
3
Q = ∫∫ ρ e dA =
π/2
∫ −π/2 ∫ 0
r cos θ r dr dθ =
−(1 − sin θ) 4
12
=
7.
1−sin θ
π/2
−π/2
π/2
1−sin θ
r 3 cos θ
3
∫ −π/2
0
dθ =
π/2
∫ −π/2
(1 − sin θ) 3
cos θ dθ
3
4
= −0 − −2 = 4
12
12
3
Find the volume of the solid above the cone z =
x2 + y2
below the hemisphere x 2 + y 2 + z 2 = 4.
a.
8π
3
b.
8π 2
3
c.
16π
3
d.
8π 2 − 2
3
e.
8π 2 + 2
3
Correct Choice
In cylindrical coordinates, the cone is z = r and the hemisphere is z =
They intersect when z = r = 4 − r 2 , or r 2 = 4 − r 2 , or r = 2 .
V=
∫∫∫ 1 dV =
2π
2
∫0 ∫0 ∫r
4−r 2
r dz dr dθ = 2π ∫
3
(4 − r 2 ) 3/2
= 2π −
− r
3
3
2
0
rz
2
3/2
0
= 2π − 2
3
−
4−r 2
z=r
2
3
dr = 2π ∫
2
0
3
− 2π −
4 − r2 .
r 4 − r 2 − r 2 dr
(4) 3/2
3
= 2π 8 − 4 2
3
In spherical coordinates, the cone is z = r or ρ cos ϕ = ρ sin ϕ or ϕ = π
4
and the hemisphere is ρ = 2.
V=
∫∫∫ 1 dV =
2π
π/4
2
∫0 ∫0 ∫0
ρ 2 sin ϕ dρ dϕ dθ = 2π − cos ϕ
π/4
0
ρ3
3
2
0
= 16π
3
−
2
+1
2
= 8π 2 − 2
3
3
2
4
∫0 ∫y
8. Compute
2
2
a.
1 e 16
4
b.
1 (e 16 − 1)
4
c.
1 (1 − e 16 )
2
d.
1 (e 4 − 1)
4
e.
1 (1 − e 4 )
2
y e x dx dy.
HINT:
Interchange the order of integration.
y
Correct Choice
2
1
0
2
4
∫0 ∫y
0
y e x dx dy =
2
2
2
∫0 ∫0
9. Compute
4
∫0 ∫0
4−x 2
∫0
x
y e x dy dx =
2
4−x 2 −y 2
4
∫0
y2
2
1
2
x
4
2
e x dx = 1
2
y=0
∫ 0 xe x
z cos (x 2 + y 2 + z 2 ) 2 dz dy dx.
3
2
dx =
HINT:
4
x
1 e x2
4
4
x=0
= 1 (e 16 − 1)
4
Convert to spherical coordinates.
a. π sin(4)
b.
c.
d.
e.
2
∫0 ∫0
8
π sin(4)
16
π sin(16)
4
π sin(16)
8
π sin(16)
16
4−x 2
∫0
=
π
2
Correct Choice
4−x 2 −y 2
z cos (x 2 + y 2 + z 2 ) 2 dz dy dx =
sin 2 ϕ
2
π/2
0
2
∫0
ρ 3 cos(ρ 4 ) dρ =
π
2
π/2
π/2
2
∫ 0 ∫ 0 ∫ 0 ρ cos(ϕ) cos(ρ 4 )ρ 2 sin(ϕ) dρ dϕ dθ
1
2
sin(ρ 4 )
4
2
0
= π sin(16)
16
4
Work Out: (Points indicated. Part credit possible. Show all work.)
10.
∫ ∫ y dx dy
(10 points) Compute
over the diamond
y
shaped region bounded by the curves
y = 4x
y= x
y = 1x
y = 4x
4
y
HINT: Let u 2 = xy and v 2 = x . Solve for x and y.
5
4
3
2
1
The boundaries are:
0
u 2 = xy = 1 or u = 1
u 2 = xy = 4 or u = 2
y
v 2 = x = 4 or v = 2
y
v 2 = x = 1 or v = 1
4
2
y
u 2 v 2 = (xy) x
J=
∂(x, y)
∂(u, v)
u 2 = (xy) x
y
v2
= y2
=
1
v
−u
v2
v
u
=
u − − uv
v
v2
= x2
So
= 2u
v
2
2
0
x = uv
=
16 − 2
3
3
2− 1
2
2u 3
3
2
3
4
5
x
y = uv
The integrand is y = uv.
∫ ∫ y dx dy = ∫ ∫ y J du dv = ∫ ∫ uv 2uv du dv = ∫ 1/2 ∫ 1 2u 2 du dv =
1
2
v
u=1
So
2
v=1/2
=7
5
11. Consider the surface, S, given parametrically by
⃗ (p, q) =
R
1 p 2 , q 2 , pq
2
for 0 ≤ p ≤ 3 and 0 ≤ q ≤ 2.
⃗ , and
a. (10 points) Find ⃗
e p , ⃗e q , N
î
̂
⃗ . Simplify
N
⃗
N
by looking for a perfect square.
k̂
⃗e p = (p 0 q)
⃗e q = (0 2q p)
⃗ = ⃗e p × ⃗e q = î(−2q 2 ) − ̂ (p 2 ) + k̂ (2pq) = (−2q 2 , −p 2 , 2pq)
N
⃗ =
N
(−2q 2 ) 2 + (−p 2 ) 2 + (2pq) 2 =
(p 2 + 2q 2 ) 2 = p 2 + 2q 2
4q 4 + p 4 + 4p 2 q 2 =
b. (10 points) Compute the surface area of the surface, S.
HINT:
A=
∫∫
A=
∫ ∫ 1 dS
⃗ dp dq =
N
= 9q + 2q 3
A=
∫∫
=
⃗ dq dp =
N
2
3
∫0 ∫0
2
q=0
3
= (18 + 16) = 34
2
∫0 ∫0
2p 3
+ 16 p
3
3
(p 2 + 2q 2 ) dp dq =
(p 2 + 2q 2 ) dq dp =
3
2
∫0
p3
+ 2pq 2
3
3
2
dq =
∫ 0 (9 + 6q 2 ) dq
dp =
∫0
p=0
OR
3
∫0
p2q +
2q 3
3
2
q=0
3
2p 2 + 16
3
dp
= (18 + 16) = 34
p=0
6
⃗ (p, q) =
R
Recall
1 p 2 , q 2 , pq
2
for 0 ≤ p ≤ 3 and 0 ≤ q ≤ 2.
c. (10 points) Compute the mass of the surface, S, if the surface mass density is ρ(x, y, z) = z.
M=
HINT:
M=
∫ ∫ ρ dS
∫ ∫ ρ N⃗ dp dq =
2
∫0 ∫0
2
81q 2
9q 4
+
8
4
=
⃗ dq dp =
M = ∫∫ρ N
p4
+ 4p 2
2
=
3
pq(p 2 + 2q 2 ) dp dq =
=
q=0
3
2
∫ 0 ∫ 0 pq(p
3
81 + 36
2
2
p=0
∫0
2
3
∫0
3
p4q
+ p2q3
4
= 153
2
+ 2q ) dq dp =
81 + 36
2
=
2
dq =
p=0
81q
+ 9q 3 dq
4
2
∫0
OR
2pq 4
p3q2
+
2
4
2
q=0
dp =
3
∫ 0 (2p 3 + 8p) dp
= 153
2
⃗ = (2x, 2y, z) if the
d. (10 points) Compute the flux through the surface, S, of the vector field F
surface is oriented down and out.
⃗
⃗ ⋅ dS
HINT:
Flux = ∫ ∫ F
⃗ = (−2q 2 , −p 2 , 2pq) is up and in. Reverse it. N
⃗ = (2q 2 , p 2 , −2pq)
N
⃗ = (2x, 2y, z) = (p 2 , 2q 2 , pq)
F
Flux =
∫ ∫ F⃗ ⋅ N⃗ dp dq = ∫ 0 ∫ 0 (2p 2 q 2 + 2p 2 q 2 − 2p 2 q 2 ) dp dq = ∫ 0 ∫ 0 2p 2 q 2 dp dq
=2
2
p3
3
3
0
q3
3
2
0
3
= 2(9) 8
3
2
3
= 48
e. (6 points HONORS ONLY) Find the equation of the plane tangent to the surface,S, at the point
where (p, q) = (2, 1).
⃗ (p, q) =
R
1 p 2 , q 2 , pq
2
⃗ (2, 1) = (2, 1, 2)
P=R
⃗ = (−2q 2 , −p 2 , 2pq)
N
⃗ = (−2, −4, 4)
N
⃗⋅X = N
⃗⋅P
N
−2x − 4y + 4z = −2(2) − 4(1) + 4(2) = −4 − 4 + 8 = 0
x + 2y − 2z = 0
7