Name
1-9
/45
MATH 251
Exam 2C
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.
3y above the region in the xy-plane
x
2x.
32
b. 32
c. 32
15
d. 92
15
e. 116
15
Correct
x2
Solution: Find the intersections:
2
2x
0
x2
V
x
3y dy dx
xy
3
0
3
8x
3
x4
4
3 x5
2 5
y
2
2
2x
8
2
dx
y
8
3
0
x
2x
2
2
6x 2
3 x 4 dx
2
x3
0
x2
4
2x 2
0, 2
16
5
3
116
15
2. The temperature on a triangular hot plate with vertices at 0, 0 , 0, 4 and 2, 0 is T
xy.
Find the average temperature.
a. 1
b.
c.
d.
e.
3
2
3
1
4
3
8
3
Correct
2
Solution: A
4
0
4 2x
T dA
2
xy
2
xy dy dx
0
T ave
2
1 dy dx
0
2
4 2x
0
1 16 x 2
2
2
1
T dA
A
0
3
16 x
3
8 1
3 4
x4
2
0
0
2
2x dx
4 2x
dx
y 0
1 16 2 2
2
2
4x
1
2
x2
2
0
2
x 4
2x
4
2
dx
0
3
16 2
3
24
1
2
2
16x
16x 2
4x 3 dx
0
8
3
2
3
1
5x 2 below y
3. Find the centroid of the region above y
a.
0, 8
b.
0, 10
c.
0, 12
d.
0, 15
e.
0, 16
Correct
Solution: 5x 2
20
1 dA
2
5x
By symmetry, x
20
5x 2 dx
x
2
20x
5x
3
3
2
20
2
800
y
1 dy dx
2
y dA
Ay
A
4
2
5x
2
y dy dx
2
2
y2
2
20
dx
5x 2
40
3
2 40
2
160
3
1
2
2
400
2
25x 4 dx
1 400x
2
5x 5
2
2
160 640
640 3
12
160
4x 2
4. Find all critical points of the function f x, y
9y 2
A
3, 2
B
3, 2
C
3, 2
D
3, 2
E
2, 3
F
2, 3
G
2, 3
H
2, 3
Note 432
2
0.
2
Ay
x2
20
2
A
20.
432 . Select from:
xy
2433
a. A, B, C, D
b. E, F, G, H
c. B, C
d. A, D Correct
e. E, H
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
xy
6
Multiply:
x 4 y 4 54 24 2 4 3 4
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 0, 0 .
B. There is a local minimum at 0, 0 .
C. There is a saddle point at 0, 0 .
D. There is a local maximum at 1, 1 .
E. There is a local minimum at 1, 1 .
F. There is a saddle point at 1, 1 .
a. A,B,D,E
b. C,F
c. C,D,E Correct
d. A,B,F
Solution: D,E but not A,B 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
b. Local Maximum Correct
c. Inflection Point
d. Saddle Point
e. Test Fails
4
x2
Solution: f x
f xx
f xx
8
x3
2, 1
f yy
1
y
2
y2
fy
4
f xy
y3
f yy 2, 1
x
1
4
f xy 2, 1
1
D
f xx f yy
2
f xy
4
1
3
Local Maximum
3
Find the mass of the region in the
7.
upper half of the circle
r
0.2
cos
if the surface density is
a.
b.
c.
d.
e.
y 0.4
1
12
1
9
1
6
1
3
1
2
0.0
y.
0.0
0.5
1.0
x
Correct
Solution: The density is
/2
M
r sin .
/2
dA
1
3
y
cos
r3
3
r sin r dr d
0
4
0
0
/2
cos
4
1 0
3
0
1
4
cos
sin d
0
1
3
/2
cos 3 sin d
0
1
12
8. Find the x-component of the center of mass of the region in the upper half of the circle
r
cos
a.
b.
c.
d.
e.
if the surface density is
1
24
1
4
1
3
1 Correct
2
2
3
Solution: The density is
/2
Mx
y.
y
r sin and x
cos
x dA
r cos r sin r dr d
0
r cos .
r 4 cos cos sin d
4 0
0
Mx
1 12
1
M
24 1
2
/2
0
/2
cos 6
1
1 0 1
1
4
6
4
6
24
0
You can expect this by symmetry.
x
1
4
/2
cos 5 sin d
0
4
9.
Compute
y
x dA over the “diamond” shaped
40
30
region in the first quadrant bounded by
x2y
1
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 Correct
c. 15 ln 3
d. 65 ln 2
e. 65 ln 3
u , u2v2
x, y
v
2 2 2
u
2
Boundaries: x y
u 4 1, 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 1
Solution: Let
u
1, 3
v
1, 2
|2u 2
3
1
ln v
2
1
2u 2 |
81
4u 2
1 ln 2
80 ln 2
5
Work Out: (Points indicated. Part credit possible. Show all work.)
10.
30 points
Consider the piece of the
paraboloid surface z
18
2x 2
2y 2
above the xy-plane.
Find the mass of the paraboloid if the surface mass density is
z 2x 2 2y 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 , 18
2r 2
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 out. We need down and in. 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
d. Evaluate the density
R r,
2r 2
18
2x 2
z
2y 2 on the paraboloid.
2r 2
18
18
2r 2
e. Compute the mass.
Find the limit on r:
2
M
z
3
dS
0
18r 16r 2
0
1 dr d
2
18
0
3 145 3/2
4
3
2
16r 2
3 32
1
3/2
3
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
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
A cardboard box without a lid needs to hold 4000 cm 3 . Find the dimensions of the box
which uses the least cardboard.
Solution: Minimize the surface area A xy
V xyz 4000.
Use Lagrange Multipliers
A
y 2z, x 2z, 2x 2y
V
yz, xz, xy
f
g
y 2z
yz
y 2z
2x 2y
x 2z
1
yz
xz
xy
z
2
2
1
2
and
y
x
y
z
V
12.
xyz
y y
15 points
y3
2
y
2
y3
4000
2xz
2yz subject to the constraint that the volume is
x 2z
2
1
y
z
x y
8000
y
xz
2x 2y
2
2
2
x
y
x
y
and
z
2
xy
20
z
Draw the region of integration and compute
x
3
3
0
y
20
y x3
10
9 dx dy
Solution: To reverse the order of integration
plot the region 0
y
3, y
x
y
3.
3
Include a horizontal line to indicate the x limits.
2
Add a vertical line to indicate the new y limits.
1
Write the new integral and compute it.
0
3
0
x
y x3
3
9 dy dx
0
x3
9
0
1 36 3/2
9
9
3/2
216
27
9
y2
2
x
dx
0
0
1
2
3
0
x3
1
9 x 2 dx
2
3
x
1 x3
9
9
3/2 3
0
21
8