Name
ID
MATH 251
Exam 2 Version A
Sections 506
Solutions
1-13
/52
Spring 2013
14
/12
P. Yasskin
15
/28
16
/12
Total
/104
Multiple Choice: (4 points each. No part credit.)
1. Compute
x 2 and the line y
xy dA over the region R between the parabola y
2x.
R
a. 4
b.
c.
d.
e.
3
8
3
32
15
10
3
29
6
Correct Choice
x2
SOLUTION:
2
xy dA
0
R
x
0, 2
y2
x
2
0
x
2
xy dy dx
2
x6
6
1 x4
2
3
2. Compute
2x
2x
0
9 x2
9 x2
2
0
1 16
2
2x
dx
x2
1
2
32
3
2
x 4x 2
x 4 dx
0
1
2
2
4x 3
x 5 dx
0
8
3
x dy dx by converting to polar coordinates.
a. 3
b. 9
c. 18
Correct Choice
d. 27
e. 36
SOLUTION:
3
0
9 x2
9 x2
The region is a semicircle in the right half plane.
/2
3
x dy dx
r cos r dr d
/2
0
sin
/2
/2
r3
3
3
1
1 9
0
18
0
1
3. Find the mass of the triangular plate with vertices
0, 0 ,
and
2, 6
2, 6
if the surface density
2
is
x .
a. 4
b. 6
c. 12
d. 16
Correct Choice
e. 24
SOLUTION:
Edges: y
2
M
3x
2
x 2 dy dx
dA
0
3x and y
3x
2
3x
x2y
3x
0
y
3x
dx
3 x4
2
6x 3 dx
0
4. Find the center of mass of the triangular plate with vertices
b.
c.
d.
e.
2, 6
and
2, 6
if the surface
Correct Choice
SOLUTION:
0 by symmetry.
y
2
3x
x dA
0
My
M
x
0, 0 ,
x .
8 ,0
5
9 ,0
5
5 ,0
8
5 ,0
9
6 ,0
5
My
x
24
0
2
density is
a.
2
192
5 24
2
x 3 dy dx
3x
3x
x3y
3x
0
2
dx
6x 5
5
6x 4 dx
0
2
0
192
5
8
5
5. Which of the following integrals is NOT equivalent to
4
4 z
0
0
y
f x, y, z dx dy dz? The region is
0
shown.
a.
4
4 z
0
d.
e.
f x, y, z dy dx dz
x2
0
4
4 y
2
0
0
y2
b.
c.
4 z
4
y
Correct Choice
f x, y, z dx dz dy
4 y
f x, y, z dz dx dy
0
0
0
2
4 x2
4 z
0
0
x2
2
4
4 y
0
x2
0
SOLUTION:
f x, y, z dy dz dx
f x, y, z dz dy dx
(b) should be
4
4 y
0
0
y
f x, y, z dx dz dy
0
2
6. Find the area of one petal of the 8 petaled daisy r
a.
b.
c.
d.
e.
sin 4 .
2
4
8
Correct Choice
16
32
SOLUTION:
A
0 when 4
r
/4
/4
1 dA
r dr d
0
0
0
0, /4
sin 4
r2
2
/4
1
2
d
0
a.
b.
c.
d.
e.
/4
1
2
d
0
1
cos 8
2
0
d
16
0
x2
7. Find the mass of the solid between the paraboloids z
density is
sin 2 4
/4
sin 8
8
1
4
or
0,
sin 4
y 2 and z
x2
8
y 2 if the volume
z.
Correct Choice
64
32
16
8
4
SOLUTION:
z
2
M
r2
8
2
r2
8
dV
0
2
x2
r
2
0
32r 2
4r 4
2
0
2
rz
2
2
0
8. Compute
8
2
16r 3 dr
64r
2r 2
z r dz dr d
r2
0
r2
8 r2
z r
2
dr
2
r 8
r2
y2
z2
2
r 4 dr
0
64
y 2 z dV over the solid hemisphere 0
x2
2.
a. 0
b. 8
3
c. 16
3
64
d.
9
e. 2
Correct Choice
SOLUTION:
x2
x2
y 2 z dV
y2 z
2
/2
0
2
sin 4
4
2
0
/2
0
sin cos
3
sin 2 cos
2
sin sin
2
6
2
0
2
0
2 1 32
4 3
3
cos
/2
sin d d d
0
6
2
sin 2 cos
sin 3 cos d
dV
2
5
2
sin
d
0
16
3
3
9. Which integral gives the arclength of the ellipse r
2
a.
3 2
2 sin 2 d
3 4
2 cos 2 d
3 2
2 cos 2 d
6 cos , 3 sin , 3 sin
?
Correct Choice
0
2
b.
0
2
c.
0
2
3 sin 2
2 3
d.
d
0
2
54 d
e.
0
SOLUTION:
v
6 sin , 3 cos , 3 cos
2
L
ds
2
3 4 sin 2
|v | d
0
36 sin 2
|v |
2
2 cos 2 d
3 2
0
in a pot of water where the temperature is T
the water as measured by the thermocouple.
fds
HINT: f ave
ds
2 sin 2 d
41
x2
3 cos t, 3 sin t, 4t for 0 t 4 is placed
y 2 z °C. Find the average temperature of
8
4
b. 50
9 cos 2
0
10. A helical thermocouple whose shape is the curve r t
a. 173
9 cos 2
16
c. 1000
160
d. 250
2
40
e. 50
Correct Choice
8
SOLUTION:
v
3 sin t, 3 cos t, 4
4
ds
5 dt
20
T
41
x2
9 sin 2 t
|v |
y2
z
41
9
9 cos 2 t
4t
50
16
5
4t
0
4
T ds
50
4t 5 dt
5 50t
0
T ave
1000
160
20
2
50
2t 2
4
0
1000
160
2
8
4
2xy 2
11. Find a scalar potential, f x, y, z , for F x, y, z
Then compute f 2, 2, 2
2x
2xz, 2x 2 y
3z, x 2
3z 2
3y .
f 1, 1, 1 .
a. 0
b. 1
c. 7
Correct Choice
d. 23
e. 25
SOLUTION:
xf
2xy 2
2x
yf
2x 2 y
3z
2
3y
f x, y, z
x2y2
x2
f 2, 2, 2
f 1, 1, 1
zf
x
2
2xz
3z
x2
12. If f
y2
a.
f
0
b.
f
0
x2
f
x2y2
f
2
z
z3
C
3yz
x z
x2z
3yz
16
4
8
x2z
3
12
h x, z
3yz
8
k x, y
1
0 and
f
2x, 2y, 4z
z
1
3
1
23
0
SOLUTION:
F
1
xz, yz, z 2 , which of the following is false?
e. None of the above. They are all true.
f
g y, z
0
F
d.
x2y2
2z 2 and F
F
c.
f
z
f
2z
13. Compute
2
0
Correct Choice
0 are always true.
F
2
4
F
0
0
0
xz, yz, z 2
G dV for G
over the solid cylinder x 2
y2
25 with 0
z
4.
C
a. 800
Correct Choice
b. 400
c. 200
d. 80
e. 40
SOLUTION:
G
4
2
z
G dV
C
0
0
z
2z
4z
4zr dr d dz
2
5
0
z2
4
0
r2
5
0
2 16 25
800
5
Work Out: (Points indicated. Part credit possible. Show all work.)
12 points
14.
y
x 2 dA over the "diamond"
Compute
D
shaped region bounded by the curves
y
1
x3, y
x3, y
3
x3, y
4
HINT: Define curvilinear coordinates
y
u
x
3
and y
v
x3.
7
so that
u, v
3
x .
x
2 pts
a.
What are the boundaries in terms of u and v?
SOLUTION:
3 pts
b.
u
1, u
3, v
Add the formulas:
Subtract the formulas:
4 pts
d.
2 pts
u
v
v
y
2x 3
x
y
u
y
v
1
3
1
3
x
u
x, y
u, v
1
6
x
v
2/3
1 v u
12
2
v u 2/3
2
1
12
v
u
v
2
v u
2
1/3
.
v
u
2/3
u
2/3
1
2
2
v
2
2/3
u
1
6
2
1
2
1
2
1
2
v
u
2/3
2
Express the integrand in terms of u and v.
x2
SOLUTION:
e.
0
u
x, y
u, v
x, y
u, v
1 pts
2y
Find the Jacobian factor J
SOLUTION:
J
7
Find formulas for x and y in terms of u and v.
SOLUTION:
c.
4, v
v
2/3
u
2
Compute the integral.
SOLUTION:
x 2 dA
D
x 2 J du dv
D
7
3
4
1
v
u
2
2/3
1
6
v
u
2
2/3
du dv
1 2 3
6
1
6
15.
28 points
Consider the elliptical region, E, in the plane z
2
x y
4 oriented upwards.
2
y above the circle
x
2
HINT: This ellipse may be parametrized by R r,
10 pts
a.
r cos , r sin , 2
r cos
Find the normal vector N to the ellipse and its length
r sin
.
N .
N starts hard but simplifies!
Note:
SOLUTION:
î
er
cos
e
cos
r cos
r sin
cos
r
r cos
r sin
k r cos 2
r cos
r
2
0
r sin
0.
r cos cos
r cos
r sin cos
r2
r2
sin
sin
2
N
r2
3r
2
2
A
dS
3 r dr d
E
3 pts
2
N dr d
E
0
2
3
0
2
r2
2
4
3
0
x2
Find the mass of the ellipse if the surface density is
x2
SOLUTION:
2
M
y2
2
dS
E
d.
r sin
Find the surface area of the ellipse.
SOLUTION: 0
c.
N
sin
Oriented correctly because N 3
r, r, r
3 pts
b.
sin
r sin
N
î sin
k
12 pts
0
r2
3 r 3 dr d
2
2
r4
4
3
0
8
3
0
yz, xz, z 2 , compute the surface integral
If F
y2.
F dS
E
î
SOLUTION:
F
x
k
y
î0
z
x
0
y
kz
z
x, y, 2z
yz xz z 2
F
R r,
F N
r cos , r sin , 4
r 2 cos
r 2 sin
2
4r
0
3r 2 sin
2
2
0
0
0
3r 2 cos
0
dr
0
3r 2 cos
4r
4r
3r 2 cos
4r2 dr
4 r2
2
2
Recall: N
2r sin
2r 2 sin
F N dr d
0
2
2r 2 cos
4r
2
F dS
E
2r cos
3r 2 sin
2
0
r, r, r
3r 2 sin
d dr
16
7
16.
12 points
Compute
G dS for G
xz, yz, z 2
over the cylinder x 2
y2
25 for 0
z
4
C
with outward normal.
SOLUTION: R , z
5 cos , 5 sin , z
î
e
k
5 sin
ez
0
5 cos
0
4
G dS
C
G N
2
G N d dz
25z d dz
0
î 5 cos
5 sin
k0
5 cos , 5 sin , 0
Oriented correctly
1
5z cos , 5z sin , z 2
G R ,z
C
N
0
0
25z cos 2
25z sin 2
2
2 25 z
2
4
25z
400
0
8