Name
ID
1-10
/50
Fall 2015
11
/30
Version B Solutions P. Yasskin
12
/5
13
/20
Total
/105
MATH 251
Final Exam
Sections 511
Multiple Choice: (5 points each. No part credit.)
1. A triangle has vertices A
3, 2, 4 , B
3, 4, 6 and C
6, 1, 4 . Find the angle at A.
a. 30°
b. 45°
c. 60°
Correct Choice
d. 120°
e. 150°
Solution:
AB
0, 2, 2
AB AC
AB AC
cos
AC
3, 3, 0
6
2 23 2
1
2
120°
2. Find the tangent plane to the graph of z
x2y
y 3 x 3 at x, y
2, 1 .
Where does it cross the z-axis?
a.
72
b.
48
c.
12
Correct Choice
d. 12
e. 48
Solution:
z
f tan x, y
z-intercept
f x, y
x2y
f x x, y
2xy
f y x, y
2
f 2, 1
f tan 0, 0
x
y3x3
3y 3 x 2 f x 2, 1
2 3
3y x
f x 2, 1 x
12
f 2, 1
f y 2, 1
2
16 2
12
16
28
f y 2, 1 y
1
28 1
48
12
16 x
2
28 y
1
1
3. Find the tangent plane to the graph of hyperbolic paraboloid
9x 3 2 4y 1
at the point 2, 2, 3 . Where does it cross the z-axis?
a. 19
2
4z
2
2
1
Correct Choice
2
b. 19
c. 0
d.
19
e.
19
2
Solution:
F
P
18 x
N X
2, 2, 3
3 , 8y
N P
y
9x
1 , 8z
18x
z-intercept at x
F
8y
2
8z
0:
3
2
N
18 2
8z
4y
76
F
1
2
4z
2
18, 8, 8
P
8 2 8 3
76
19
z
8
2
4. Sketch the region of integration for the integral
2
76
16 x 2
2 2
0
ex
2
y2
dy dx in problem (12),
x
then select its value here:
a.
b.
c.
d.
e.
4
4
8
8
16
e 16
1
e8
1
e 16
Correct Choice
1
e8
1
e 16
1
y
Solution: Switch to polar coordinates:
/2
4
/4
0
/2
r2
e r dr d
2
4
/4
1 e 16
2
1
1e
2
r2
3
4
2
0
8
4
e 16
1
1
0
0
1
2
x
2
5. Find the line perpendicular to the curve x 3
a.
11
2t, 1
b.
2
t, 1
11t
c.
2
t, 1
11t
d.
2
11t, 1
t
e.
2
11t, 1
t
Solution:
f
y3
7 at the point 2, 1 .
xy
t
Correct Choice
A vector perpendicular to the curve is the gradient at the point:
3x 2
y, 3y 2
x
v
f
12
2,1
So the perpendicular line is X
1, 3
tv or x, y
P
2
11, 1
2, 1
t 11, 1
2
6. Consider the surface of the cone given in cylindrical coordinates by z
11t, 1
4
t
r above the xy-plane. It
may be parametrized by
R r,
r .
r cos , r sin , 4
Find the z-component of its centroid.
a. 1
3
b. 2
3
c. 4
Correct Choice
3
32 2
d.
3
e. 16 2
Solution:
We first find the normal and its length:
î
N
e
e
î
k
er
cos
sin
1
e
r sin
r cos
0
r sin 2
k r cos 2
r cos
r sin
N
r 2 cos 2
r 2 sin 2
2 r dr d
2
r2
r cos , r sin , r
2r
The surface area is
2
4
A
0
Since z
4
4
r2
2
2
0
16 2
0
r, the z-moment is
2
Az
4
4
4
0
2
r
2 r dr d
2
2
0
r 2 dr
4r
0
r3
3
2 2r 2
4
2
2 32
64 2
3
1
16 2
0
64 2
3
64
3
So the z-component of the centroid is
z
Az
A
4
3
3
7. Death star is basically a spherical shell with a hole cut out of one end, which we will take as centered
at the south pole. In spherical coordinates, it fills the region between 1
the volume.
a. V
21
b. V
63
c. V
84
d. V
105
e. V
126
Solution:
Correct Choice
The volume is
2
2 /3
4
0
0
1
V
1
cos 2
3
2
8.
2 . Find
3
4 and 0
Compute
f ds for f
2
sin d d d
cos 0
4
2
3
1
3
cos
2
3
2 /3
0
3
1
2
4
1
1 21
63
ye x
y
R
3
counterclockwise around
2
the polar curve
1
r
3
cos 6
-4
shown at the right.
-2
-1
2
4
x
-2
Hint: Use a Theorem.
-3
a. 6e 3
b. 3e 6
c. 6e 6
3e 3
d. 3e 3
6e 6
e. 0
Solution:
Correct Choice
By the FTCC,
f ds
fB
fA
0 because B
A no matter what point you start at.
R
4
3
y
9.
Compute
F ds for F
3x 3
2
4y 4
5y, 7x
1
P
counterclockwise around the complete
-3
boundary of the plus sign shown at the right.
-2
-1
1
2
-1
Hint: Use a Theorem.
3
x
-2
-3
a.
40
b.
20
c. 20
d. 40
Correct Choice
e. 240
Solution:
P 3x 3 5y
By Green’s Theorem,
Q
F ds
R
10. Compute
7x
4y 4
xQ
yP
xQ
yP
dx dy
R
7
2 dx dy
5
2
2 area
2 20
40
R
4 x2
F dS over the paraboloid z
y 2 for z
16 oriented down and out for
P
F
y z, x z, z .
Hint: Use a Theorem.
a. 64
b. 32
Correct Choice
c. 16
d. 8
e. 4
Solution:
By Stokes’ Theorem,
F ds oriented clockwise.
F dS
P
P
At the boundary, z 4 x 2 y 2
16. So the boundary is the circle of radius r 2 at height z
parametrized by r
2 cos , 2 sin , 16 . The vector field is F
8 sin , 8 cos , 4 .
The tangent vector is v
2 sin , 2 cos , 0 . Reversed v
2 sin , 2 cos , 0 .
So F v 16 sin 2
16 cos 2
16. And the integral is
2
F ds
P
2
F v dt
0
16,
16 dt
32
0
5
Work Out: (Points indicated. Part credit possible. Show all work.)
11.
(30 points) Verify Gauss’ Theorem
F dV
F dS
V
V
for the vector field F
x 3 z, y 3 z, x 2
between the cone z
x2
y2
y2
2
and the solid
and the plane z
2.
Be careful with orientations. Use the following steps:
Left Hand Side:
a. Compute the divergence:
3x 2 z
F
3y 2 z
3z x 2
y2
b. Express the divergence and the volume element in the appropriate coordinate system:
3zr 2
F
dV
r dr d dz
c. Compute the left hand side:
2
2
z
F dV
0
V
0
2
3
z 5 dz
2 0
Right Hand Side:
2
3zr 2 r dr d dz
2
0
0
3 z6
2 6
2
4
0
26
24
4
3z r
4
z
dz
r 0
16
The boundary surface consists of a cone C and a disk D with appropriate orientations.
d. Parametrize the disk D:
R r,
r cos , r sin , 2
e. Compute the tangent vectors:
er
cos ,
e
r sin ,
sin ,
0
r cos ,
0
f. Compute the normal vector:
N
î0
r sin 2
k r cos 2
0
0, 0, r
This is correctly up.
x 3 z, y 3 z, x 2
g. Evaluate F
F
y2
2
on the disk:
2r 3 cos 3 , 2r 3 sin 3 , r 4
R r,
h. Compute the dot product:
F N
r5
i. Compute the flux through D:
2
2
F dS
D
0
0
r 5 dr d
2
r6
6
2
0
64
3
6
The cone C may be parametrized by R r,
r cos , r sin , r
j. Compute the tangent vectors:
er
cos ,
e
r sin ,
sin ,
1
r cos ,
0
k. Compute the normal vector:
N
î r cos
r sin 2
k r cos 2
r sin
r cos , r sin , r
This is oriented up. Need down. Reverse: N
x 3 z, y 3 z, x 2
l. Evaluate F
F
y2
2
r cos , r sin , r
on the cone:
r 4 cos 3 , r 4 sin 3 , r 4
R r,
m. Compute the dot product:
r 5 sin 4
r 5 cos 4
F N
r5
sin 4
r 5 cos 4
1
n. Compute the flux through C:
Hints: Use these integrals as needed, without proof:
2
2
sin 2 d
0
0
2
2
F dS
0
C
2
cos 2 d
2
sin 4 d
0
r 5 cos 4
sin 4
1 dr d
0
6
r
6
2
0
2
0
0
cos 4 d
cos 4 d
2
3
4
2
sin 2 cos 2 d
0
sin 4 d
0
1
4
2
1d
0
25 3
3
16
2
3 4
4
3
o. Compute the TOTAL right hand side:
F dS
V
F dS
D
F dS
C
64
3
16
3
16
which agrees with (c).
7
y
12.
(5 points) Sketch the region of integration
4
3
for the integral
16 x 2
2 2
0
2
ex
2
y2
dy dx.
x
1
Shade in the region.
0
You computed its value in problem (4).
0
1
2
x
13. (20 points) A cardboard box needs to hold 96 cm 3 . The cardboard for the vertical sides costs 12¢ per
cm 2 while the thicker bottom costs 36¢ per cm 2 . There is no top. What are the length, width, height
and cost of the box which costs the least?
Solution:
C 36LW
The volume constraint is V LWH 96. The cost is
12 2LH 2WH
36LW 24LH 24WH which needs to be minimized.
Lagrange Multipliers:
C
36W 24H, 36L
24H, 24L
24W
V
36W 24H
36L 24H
WH
LH
36LW 24LH
36LW 24WH
LWH
LWH
24L
LW
24LH
LWH
24W
LWH
C
WW 3 W
2
36LW
24LH
96
W3
24WH
24WH
64
36 4 4
W
4
WH, LH, LW
C
V
36LW 24LH 36LW 24WH
24LH 24WH
L W
36LW 24WH 24LH 24WH
36LW 24LH
3W 2H
L
24 4 6
4
H
24 4 6
6
1728¢
8