Name
MATH 251
Exam 2
Sections 515
Solutions
1-9
/54
Fall 2012
10
/20
P. Yasskin
11
/20
12
/6
Multiple Choice: (6 points each. No part credit.)
Total
1. Compute
a. 81
3
3
0
y
/100
4x 2 dx dy.
Correct Choice
b. 72
c. 60
d. 48
e. 32
SOLUTION:
3
0
3
y
4x 2 dx dy
3
0
3
4x
3
3
3
dy
0
x y
2. Which of the following is the polar plot of r
36
4
y3
dy
3
36y
3
108
27
81
0
cos 3 ?
y
a.
y4
3
b.
c.
1
0
2
4
6
x
-1
d.
Correct Choice e.
SOLUTION:
(c) is the rectangular plot of r cos 3 .
(d) is its polar plot because there are 3
positive loops and 3 negative loops which retrace the positive loops with r 1 when
0.
1
3. Find the mass of a triangular plate whose vertices are
0, 0 ,
and
1, 0
1, 3 , if the density is
2x.
a. 1
Correct Choice
b. 2
c. 3
d. 4
e. 5
1
SOLUTION:
M
3x
dA
1
2x dy dx
0
2xy
0
0
3x
y 0 dx
1
6x 2 dx
2x 3
0
1
0
2
4. Find the x-component of the center of mass of a triangular plate whose vertices are
and
1, 3 , if the density is
0, 0 ,
1, 0
2x.
a. 1
4
b. 1
2
c. 3
4
d. 3
2
Correct Choice
e. 3
SOLUTION:
x
5.
My
M
My
3 1
2 2
1
3x
0
0
x dA
2x 2 dy dx
1
2x 2 y
0
3x
dx
y 0
1
0
6x 3 dx
3 x4
2
1
0
3
2
3
4
The surface of an apple is given in spherical coordinates by
3
3 cos
Its volume is given by the integral:
a. V
b. V
c. V
d. V
e. V
2
/2
3 3 cos
1d d d
0
2
0
0
3 3 cos
0
2
0
0
0
2
0
0
3 3 cos
0
2
0
0
1
0
0
0
1d d d
/2
3 3 cos
3
SOLUTION:
2
2
sin d d d
3 cos
2
sin d d d
2
V
Correct Choice
sin d d d
3 3 cos
dV
0
0
2
sin d d d
0
2
6.
Find the area inside the circle r
and outside the limacon r
a. 4
3
2
3
2
3
2
3
2
c. 2
5
3
d.
e. 2
SOLUTION:
x
Correct Choice
Find the angles of intersection: 4 cos
/3
A
1 dA
4 cos
/3
2
1 r dr d
0
/3
y
2 cos .
3
5
3
b.
1
4 cos
16 cos 2
1
1 2 cos
0
4 cos 2
4 cos
4 cos
r 2 1 2 cos
1
/3
d
16 cos
5
5
3
61
cos 2
6 cos 2
4 cos d
5
3 sin 2
4 sin
3 3
2
4 3
2
5
3
So the area element is dA
e.
2 cos
d
1
4 cos d
2 uv du dv
2 uv du dv
2 uv du dv
2 uv du dv
2
2 u2 du dv
v
SOLUTION:
x, y
u, v
3 sin 2
3
4 sin
3
y
x
and v
yx .
dx dy
Correct Choice
uv
v
u2
1
u
5
3
/3
0
3
2
7. Hyperbolic coordinates in quadrant I are given by u
d.
1
3
2
0
0
c.
2
/3
/3
b.
1
2
cos
/3
d
0
a.
2 cos
v
u
v
u
y
v
u
x
v
u
x
2 uv
v
u
y
J
uv
x, y
u, v
2 uv
dA
2 uv du dv
3
8. If F
a.
b.
c.
d.
e.
xe y z , ye x z , ze x y , then
2ze xy
2ze xy
2ze xy
2ze xy
0
2xe yz
2xe yz
2xe yz
2xe yz
2xyze xy
2xyze xy
2xyze xy
2xyze xy
SOLUTION:
9. If f
sin x
SOLUTION:
2xyze yz
2xyze yz
2xyze yz
2xyze yz
Correct Choice
F
y , then
a. 2 sin x y
b. 2 sin x y
c. 2 cos x y
d. 2 cos x y
e. 0
F
0 for any twice differentiable vector field.
f
Correct Choice
f
cos x
y , cos x
y
f
sin x
y
sin x
y
2 sin x
y
4
Work Out: (Points indicated. Part credit possible. Show all work.)
10. (20 points) Compute
x2
9
z
y2
for z
î
F
R r,
x
y
î0
z
r .
x
0
y
k z
z
x, y, 2z
xz z 2
r cos , r sin , 2 9
î
r cos , r sin , 9
k
yz
F
over the cone
5 oriented down and in.
Note: The cone may be parametrized as R r,
SOLUTION:
yz, xz, z 2
F dS for the vector field F
r
r cos , r sin , 2r
18
k
er
cos
sin
1
e
r sin
r cos
0
N î0
r cos
0 r sin
k r cos 2
r sin 2
r cos , r sin , r
r cos , r sin , r
now down and in
Reverse
N
2
2
2
2
F N
r cos
r sin
r 2r 18
3r 2 18r
9 r 5
2
4
0
0
F dS
3r 2
18r dr d
11. (20 points) Compute
9
3x 2
F
2
2
9
0
6
0
2
5
r2
F dV
8
12.
r4
r
6
r3
2
2
64
x2
3y 2
y 2 and above the plane z
x2
4r 2 r dz dr d
y2
4 x2
2
2
4r 3 z
0
8
16
0
32
3
128
1
y2
r2
9
z 5
2
3
4r 2
144
4
160
y2z
over the solid region
5.
5
2
dr
r
x3, y3, x2z
F dV for the vector field F
below the paraboloid z
SOLUTION:
9r 2 40
up and out
2
9
4r 3 4
r2
r
2
r 2 dr
0
128
3
(6 points) At the right is the contour
plot of a function f x, y . If you start
at the dot at
5, 6
and move so that
your velocity is always in the direction
of
f, the gradient of f, roughly
sketch your path on the plot.
NOTE : The numbers on the right are
the values of f on each level curve.
SOLUTION:
The curve starts at 5, 6 goes down and curves to the right towards higher values
of the function f, always perpendicular to each level curve. It should not go up.
5