Name
1-9
/45
MATH 251
Exam 2B
Fall 2015
10
/30
Sections 511/512 (circle one)
Solutions P. Yasskin
11
/15
12
/15
Total
/105
Multiple Choice: (5 points each. No part credit.)
1. Find the volume below the surface z
x 2 and y
between y
y above the region in the xy-plane
3x
2x.
32
a.
b. 32
c. 32
15
d. 92 Correct
15
e. 116
15
x2
Solution: Find the intersections:
2
2x
0
x2
V
3x
y dy dx
3
4
8x
3
x5
10
3x
4
y
2
3xy
0
2
x
2
dx
3 4
6x 2
0, 2
2x 2
3x 3
0
y x2
8
3
8
0
2x
2x
2
2
16
5
x4
2
dx
92
15
x2
2. The temperature on a circular hot plate with radius 2 is T
4.
Find the average temperature.
a. 10
b. 20
c. 4
2
5
5
d. 4
4
e. 5 Correct
Solution: A
x2
T
2
T dA
0
1
A
0
r 2 cos 2
r2
2
0
1
cos 2
2
20
T dA
4
4
4
2
4 r dr d
0
4
T ave
0
2
1 r dr d
2
0
2
r 2 cos
4
2
2
0
8 d
2
r 4 cos 2
4
sin 2
2
2
4r
2
2
d
r 0
2
8
4
16
20
0
5
1
3x 2 below y
3. Find the centroid of the region above y
a.
b.
c.
d.
e.
0, 27
5
0, 36
Correct
5
0, 54
5
0, 1944
5
0, 3888
5
Solution: 3x 2
A
12
1 dA
By symmetry, x
12
3x
x
2
dx
12x
x3
dx
1
2
2
3x
5
2
y dy dx
2
92
5
36 2 5
5 32
25 9
2
y2
2
9
5
12
3x 2
2
2 24
8
32
2
12 2
2
9x 4 dx
1 12 2 x
2
5
9x
5
2
2
36 2 5
5
36
5
9x 2
4. Find all critical points of the function f x, y
4y 2
A
2, 3
B
2, 3
C
2, 3
D
2, 3
E
3, 2
F
3, 2
G
3, 2
H
3, 2
Note 432
2
2
12
y dA
Ay
A
1 dy dx
2
0.
2
12 2 2
4
2
3x 2
2
Ay
x2
12
2
y
12.
432 . Select from:
xy
2433
a. A, B, C, D
b. E, F, G, H
c. A, D Correct
d. B, C
e. E, H
Solution:
d 9x 2 4y 2 432
18 x 3 y 24
fx
x 3 y 24
xy
dx
x2y
d 9x 2 4y 2 432
8 xy 3 54
fy
xy 3 54
xy
dy
xy 2
Multiply:
x 4 y 4 24 54 2 4 3 4
xy
6
2
x
24
4
x
2
Divide:
y
54
9
3
y2
Multiply results:
xy xy
x2
6 2
4
Must have x 2 4
x
2
3
y
Divide results:
xy x
y2
6 3
9
Must have y 2 9
y
3
2
Looking at x 3 y 24, x and y must be both positive or both negative. So 2, 3 ,
2, 3
2
5.
Select all of the following statements which
are consistent with this countour plot?
A. There is a local maximum at 1, 0 .
B. There is a local minimum at 1, 0 .
C. There is a saddle point at 1, 0 .
D. There is a local maximum at 0, 0 .
E. There is a local minimum at 0, 0 .
F. There is a saddle point at 0, 0 .
a. A,B,D,E
b. C,F
c. A,B,F Correct
d. C,D,E
Solution: A,B but not D,E because there should be circles around a local maximum or minimum.
4
2 xy has a critical point at x, y
2, 1 .
x
y
Use the Second Derivative Test to classify this critical point.
6. The function f
a. Local Minimum Correct
b. Local Maximum
c. Inflection Point
d. Saddle Point
e. Test Fails
4
2
y
fy
x2
y2
4
f yy
f xy 1
y3
1
f yy 2, 1
4
f xy 2, 1
Solution: f x
f xx
f xx
8
x3
2, 1
x
1
D
f xx f yy
2
f xy
4
1
3
Local Minimum
3
Find the mass of the region inside
7.
y
2
the upper half of the cardioid
r
2
1
2 cos
if the surface density is
y.
0
2
4
x
a. 16
b.
c.
d.
e.
3
32
3
4
16
9
32
9
Correct
Solution: The density is
r sin .
y
2 2 cos
M
dA
0
Let u
2
0
2 cos
0
1
6
M
r3
3
0
1
du
2 sin d
du
2
1 u4 0
1 0 64
4 4
6
6
2 2 cos
r sin r dr d
u 3 du
4
1
3
sin d
0
2
2 cos
3
sin d
0
sin d
32
3
8. Find the x-component of the center of mass of the region inside the upper half of the cardioid
r
2
if the surface density is
2 cos
y.
a. 4
b.
c.
d.
e.
5
6
5
8 Correct
5
32
5
256
15
Solution: The density is
r sin and x
y
2 2 cos
Mx
Let u
Mx
x dA
r cos r sin r dr d
0
1
4
2
1
16
2
0
2 cos
4
r cos .
r4 2
4 0
0
2 cos
cos sin d
cos sin d
0
2 cos
0
4
u4 u
du
2 du
1 du
2
2 sin d
1
16
u6
6
5
2u
5
0
4
sin d
256
15
cos
x
Mx
M
1 u 2
2
256 3
15 32
8
5
4
9.
Compute
y
x dA over the “diamond” shaped
40
30
region in the first quadrant bounded by
x2y
16
x2y
81
y
x2
y
20
16x 2
HINTS: Use the coordinates
u
x
y u2v2
v
Find the boundaries and Jacobian.
10
0
0
1
2
3
4
x
a. 19 ln 2
b. 80 ln 2
c. 15 ln 3
d. 65 ln 2 Correct
e. 65 ln 3
u , u2v2
x, y
v
2 2 2
u
2
Boundaries: x y
u 4 16, 81
v u v
2
y
u2v2 v 2
v 4 1, 16
2
x
u
1
2uv 2
x, y
v
Jacobian:
J
u 2u 2 v
u, v
v2
Evaluate the integral:
2 3
u 4u 2 du dv
x dA
x J du dv
u4
v
1 2
Solution: Let
|2u 2
3
2
ln v
2
1
u
2, 3
v
1, 2
2u 2 |
81
4u 2
16 ln 2
65 ln 2
5
Work Out: (Points indicated. Part credit possible. Show all work.)
10.
30 points
Consider the piece of the
paraboloid surface z
12
3x 2
3y 2
above the xy-plane.
Find the mass of the paraboloid if the surface mass density is
z 3x 2 3y 2 .
y
x ,
Find the flux of the electric field E
, 0 down into the paraboloid.
x2 y2 x2 y2
Parametrize the surface as
R r,
r cos , r sin , 12
3r 2
and follow these steps:
a. Find the coordinate tangent vectors:
î
k
er
cos
sin
6r
e
r sin
r cos
0
b. Find the normal vector and check its orientation.
N i 6r 2 cos
j 6r 2 sin
k r cos 2
r sin 2
6 2 cos , 6r 2 sin , r
This is up and out. We need down and in. So reverse the normal.
N
6r 2 cos , 6r 2 sin , r
c. Find the length of the normal vector.
N
36r 4 cos 2
36r 4 sin 2
r2
36r 4
r2
r 36r 2
1
6
d. Evaluate the density
R r,
z
3r 2
12
3x 2
3y 2 on the paraboloid.
3r 2
12
12
3r 2
e. Compute the mass.
Find the limit on r:
2
M
z
2
dS
0
12r 36r 2
0
1 dr d
2
12
0
2 145 3/2
9
2
2
36r 2
3 72
1
3/2
2
0
1
x
f. Evaluate the electric field E
x2
r cos , r sin , 0
r2
r2
E R r,
r
,
y2 x2
y
y2
,0
on the paraboloid.
cos , sin , 0
r
r
g. Compute the flux.
2
2
E dS
2
2
0
0
E N dr d
0
0
2
2
6r dr
0
12
r2
2
cos 6r 2 cos
r
sin 6r 2 sin
r
dr d
2
24
0
7
11.
15 points
Find the point in the first octant on the surface, z x
y
2 2 , closest to the origin.
Solution: Minimize f D 2 x 2 y 2 z 2 subject to the constraint g z x y
Use Lagrange Multipliers
f
2x, 2y, 2z
g
z, z, x y
f
g
2x
z
2y
z
2z
2y
2x
2z
2x
2z
x y,
2x 2 z 2
z
z
z
x y
2x
g zx y
2 x 2x
2 2 x2 2 2
So x 1
y 1
z
2
12.
15 points
4
Draw the region of integration and compute
0
Solution: To reverse the order of integration
plot the region 0
x
4,
x
y
2.
y
2
y3
2 2.
x
y
z
2x
1 dy dx
x
2
1
Include a vertical line to indicate the y limits.
0
Add a horizontal line to indicate the new x limits.
0
1
2
3
Write the new integral and compute it.
2
0
y2
0
y3
2
1 dx dy
0
y3
1 x
y2
0
2
dy
0
y3
1 y 2 dy
2 y3
9
1
3/2 2
0
4
x
2 9 3/2
9
1
52
9
8