Name
1-9
/45
MATH 251
Exam 2A
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
a.
b.
c.
d.
e.
2y above the region in the xy-plane
x
3x.
71
20
71
20
972
5
783 Correct
20
1944
5
x2
Solution: Find the intersections:
3
3x
0
x2
V
3
x
2y dy dx
3
x4
4
12 x
3
xy
x5
5
y2
0
3
3x
x
3
3x
y x2
dx
9x 2
x3
x 4 dx
0
81
4
12 9
0
3x 2
0, 3
243
5
2. The temperature on a hot plate with dimensions 4
783
20
4 and 0
x
y
6 is T
x2y
cos 2 x.
Find the average temperature.
a. 24
b. 768
3
c. 768
3
d. 16
2
e. 16
2
24
1 Correct
2
4
Solution: A
6
1 dy dx
4
4
6
T dA
4
4
x2y
cos 2 x dy dx
0
18x 2
4
T ave
6 8
2 6 43 3
1
T dA
A
48
0
x2y2
2
4
4
cos 2x
dx
2
34
0
768 3
3
768
24
16
48
6
1
6x 3
6
y cos 2 x
3 x
dx
y 0
sin 2x
2
4
4
24
2
1
2
1
2x 2 below y
3. Find the centroid of the region above y
a.
b.
c.
d.
e.
0, 27
5
0, 36
5
0, 54
Correct
5
0, 1944
5
0, 3888
5
Solution: 2x 2
A
x2
18
3
18
1 dA
3
2x
By symmetry, x
1 dy dx
2
3
18 2 3
Ay
A
18
2x
2
dx
x
3
18x
2x
3
3
3
18
y dA
Ay
9
3
2x
5
3
y dy dx
2
43
5
16 3 5
5 72
35 4
3
y2
2
4
5
18
dx
2x 2
2 18 3
3
3
23
3
72
1
2
3
18 2
3
4x 4 dx
1 18 2 x
2
5
4x
5
3
3
16 3 5
5
54
5
4x 2
4. Find all critical points of the function f x, y
9y 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
3
0.
3
y
18.
432 . Select from:
xy
2433
a. A, B, C, D
b. E, F, G, H
c. A, D
d. B, C
e. E, H Correct
Solution:
d 4x 2 9y 2 432
8 x 3 y 54
fx
0
x 3 y 54
xy
dx
x2y
d 4x 2 9y 2 432
18 xy 3 24
fy
0
xy 3 24
xy
dy
xy 2
Multiply:
x 4 y 4 54 24 2 4 3 4
xy
6
2
x
54
9
x
3
Divide:
y
24
4
2
y2
Multiply results:
xy xy
x2
6 3
9
Must have x 2 9
x
3
2
y
Divide results:
xy x
y2
6 2
4
Must have y 2 4
y
2
3
Looking at x 3 y 54, x and y must be both positive or both negative. So 3, 2 ,
3, 2
2
5.
Select all of the following statements which
are consistent with this countour plot?
A. There is a local maximum at 1, 1 .
B. There is a local minimum at 1, 1 .
C. There is a saddle point at 1, 1 .
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. C,D,E
d. A,B,F Correct
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
Solution: f x
f xx
f xx
8
x3
2, 1
4
x2
f yy
1
f yy
y
fy
4
f xy
y3
2, 1
4
2
y2
1
x
f xy 2, 1
1
D
f xx f yy
2
f xy
4
1
3
Local Minimum
3
7.
Find the mass of the 1 of the circle
8
2
2
x y
9
for
0 y x
if the surface density is
y2
1
x.
0
0
1
2
3
x
a. 9
b. 9
2
c. 9 2
9
Correct
2
9
e.
2 2
d.
Solution: In polar coordinates, the density is
/4
M
3
dA
r cos r dr d
0
0
3
r
3
sin
0
r cos .
x
3
/4
9
0
1
2
8. Find the y-component of the center of mass of the 1 of the circle
if the surface density is
a.
b.
c.
d.
e.
9 2
16
9 2
8
81
16
81
32
1
2
8
x.
9
2
x2
y2
9
for
0
y
x
Correct
Solution: In polar coordinates, the density is
/4 3
r4
My
y dA
r sin r cos r dr d
4
0
0
My
2
9
2
81
y
M
16 9
16
x
3
0
r cos and y
/4
sin 2
81
2
4
0
r sin .
1
81
4
16
4
9.
Compute
y
y dA over the “diamond” shaped
region in the first quadrant bounded by
xy 2
xy 2
8
27
y
x
y
15
10
8x
5
HINTS: Use the coordinates
u
x
y uv
v2
Find the boundaries and Jacobian.
0
0
1
2
3
4
5
x
a. 19 ln 2 Correct
b. 80 ln 2
c. 15 ln 3
d. 65 ln 2
e. 65 ln 3
Solution: Let
Boundaries: xy 2
y
x
Jacobian:
u , uv
x, y
v2
u u 2 v 2 u 3 8, 27
v2
2
uv vu
v 3 1, 8
J
1
v2
2u
v3
x, y
u, v
u
v
v
2, 3
1, 2
2 u2
v
u
v2
u
3 u2
v
Evaluate the integral:
2
y dA
y J du dv
1
3
uv 3 u2 du dv
2
v
u3
3
2
ln v
2
1
27
8 ln 2
19 ln 2
5
Work Out: (Points indicated. Part credit possible. Show all work.)
10.
30 points
Consider the piece of the
2x 2
paraboloid surface z
for z
2y 2
18.
Find the mass of the paraboloid if the surface mass density is
Find the flux of the electric field E
Parametrize the surface as
x
x
2
y
2
,
y
x
2
y2
,0
r cos , r sin , 2r 2
R r,
3z
x2
y2
.
down and out of the paraboloid.
and follow these steps:
a. Find the coordinate tangent vectors:
î
k
er
cos
sin
4r
e
r sin
r cos
0
b. Find the normal vector and check its orientation.
N i 4r 2 cos
j 4r 2 sin
k r cos 2
r sin 2
4r 2 cos , 4r 2 sin , r
This is up and in. We need down and out. So reverse the normal.
N
4r 2 cos , 4r 2 sin , r
c. Find the length of the normal vector.
N
16r 4 cos 2
16r 4 sin 2
r2
16r 4
r2
r 16r 2
1
6
3z
d. Evaluate the density
x2
3 2r 2
r2
R r,
on the paraboloid.
y2
6
e. Compute the mass.
Find the limit on r:
2
M
3
dS
4
2r 2
z
0
0
145 3/2
1
18
6 r 16r 2
r
1 dr d
x
f. Evaluate the electric field E
x2
r cos , r sin , 0
r2
r2
E R r,
2
,
y2 x2
3
2
6
16r 2
3 32
y
y2
,0
1
3/2
3
0
on the paraboloid.
cos , sin , 0
r
r
g. Compute the flux.
2
3
E dS
2
3
0
0
E N dr d
0
0
3
2
4r dr
0
8
r2
2
cos 4r 2 cos
r
sin 4r 2 sin
r
dr d
3
36
0
7
11.
15 points
Find the volume of the largest rectangular solid which sits on the xy-plane and has its
upper 4 vertices on the paraboloid z x 2 4y 2 64.
Solution: Minimize V 4xyz subject to the constraint g z x 2 4y 2 64.
Use Lagrange Multipliers
V
4yz, 4xz, 4xy
g
2x, 8y, 1
f
g
4yz
2x
4xz
8y
4xy
4yz
4xz
z
z
z
2
4xy
or
x
and
y
or
x
and
2x
8y
2
2x
8y
z
g z x 2 4y 2 z z
2z 64
z 32
x 2 16
x 4
y2
2
2
V 4xyz 4 4 2 32 1024
12.
15 points
Draw the region of integration and compute
4
2
0
Solution: To reverse the order of integration
plot the region 0
y
4,
y
x
y
2.
Add a vertical line to indicate the new y limits.
1
Write the new integral and compute it.
0
0
0
0
x3
2
1 dy dx
0
x3
1 y
x2
0
1 dx dy
y
3
2
x2
4
z
8
y 2
4
Include a horizontal line to indicate the x limits.
2
x3
y2
2
dx
0
x3
1 x 2 dx
1
2 x3
9
2
x
1
3/2 2
0
2 9 3/2
9
1
52
9
8