Name
ID
1-10
/50
Fall 2015
11
/30
Version A 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 its area.
a. 3
b. 2 3
c. 3 3
Correct Choice
d. 6 3
e. 6
Solution:
AB
0, 2, 2
î
AB
AC
AC
3, 3, 0
k
0
2
3
3 0
6î
2
6
6k
2. Find the tangent plane to the graph of z
A
1 AB
2
1 62
2
x3y
y 3 x 2 at x, y
AC
62
62
3 3
1, 2 .
Where does it cross the z-axis?
a.
38
b.
14
c.
10
Correct Choice
d. 10
e. 38
Solution:
z
f tan x, y
z-intercept
f x, y
x3y
y3x2
2
f x x, y
3x y
f y x, y
x3
f 1, 2
f tan 0, 0
2y x f x 1, 2
3y 2 x 2
f x 1, 2 x
10
f 1, 2
3
f y 1, 2
1
22 1
10
22
13
f y 1, 2 y
2
13 2
38
10
22 x
1
13 y
2
1
3. Find the tangent plane to the graph of hyperbolic paraboloid
4x 3 2 4y 1
at the point 4, 2, 1 . Where does it cross the z-axis?
2
9z
2
2
1
11
9
11
3
a.
b.
c. 0
d. 11
Correct Choice
3
e. 11
9
Solution:
F
P
8x
N X
4, 2, 1
3 , 8y
N P
F
1 , 18 z
8x
z-intercept at x
y
4x
8y
3
2
4y
N
18z
32
18z
66
0:
2
F
1
2
9z
2
2
8, 8, 18
P
16
18
66
z
18
66
11
3
1
1
0
x2
4. Sketch the region of integration for the integral
x sin y 2 dy dx in problem (12),
then select its value here:
a. 1 cos 1
4
b. 1 1 cos 1
4
c. 1 cos 1 1
4
d. 1
Correct Choice
cos 1
e. cos 1
1
Solution: Reverse the order of integration:
1
1
0
x2
x sin y
0
Substitute:
1
0
x2
1
y
dy dx
x sin y
0
1
1
2
x 2 sin y 2
2
u
y2
x sin y 2 dy dx
1 cos y 2
4
1
0
y
1
0
dx dy
y
sin y 2 dy
2
du
2y dy
1
4
sin u du
1 1
4
y
1.0
0
dy
x 0
2
cos 1
y dy
0.5
1 du
2
1 cos u
4
0.0
0.0
0.5
1.0
x
2
5. Find the point on the curve r t
a.
20, 15
b.
80, 65
c.
80, 65
d.
175, 25
e.
325, 225
Solution:
d|v |
5
2
dt
3t 2 , 25t
50t
4t 2 at which the speed is a minimum.
Correct Choice
v
50
8t
4t
2
6t, 25
40
40t 125
8t
50
|v |
0
t
5
6t
r5
2
25
250
8t
2
75, 125
5 4t 2
40t
100
125
175, 25
6. The surface of the Death star is a sphere of radius 2 with a hole cut out of one end, which we will take
as centered at the south pole. It may be parametrized by
R ,
2 sin cos , 2 sin sin , 2 cos
2
for 0
. Find the surface area.
3
a. A
12
b. A
6
c. A
3
d. A
2
Correct Choice
e. A
Solution:
We first find the normal and its length:
î
N
e
e
k
e
2 cos cos
2 cos sin
2 sin
e
2 sin sin
2 sin cos
0
î 4 sin 2 cos
4 sin 2 sin
4 sin cos sin 2
k 4 sin cos cos 2
4 sin 2 cos , 4 sin 2 sin , 4 sin cos
2
4 sin 2 cos
N
16 sin 4
4 sin 2 sin
16 sin 2 cos 2
2
4 sin cos
2
4 sin
The surface area is
A
2
2 /3
0
0
1 dS
8
4 sin d d
cos 2
3
cos 0
8
2
cos
1
2
1
2 /3
0
12
3
7. Consider the solid below the cone given in cylindrical coordinates by z
r above the xy-plane.
4
Find the z-component of its centroid.
a.
b. 1
2
c. 1
Correct Choice
d. 32
3
e. 64
3
Solution:
The volume is
4
2
4 r
4
1 r dz d dr
V
0
0
2r 2
2
2
rz
0
0
r3
3
4
2
32
4
rz
2
2
r 2 dr
4r
0
64
3
64
3
0
4
4 r
0 dr
The z-moment is
4
Vz
2
4 r
z r dz d dr
0
0
2
0
4
16r
0
8r 2
r 3 dr
8r 2
0
4 r
2
4
dr
r4
r 2 dr
0
0
3
r4
4
8r
3
4
8 42
0
3
44
4
84
3
64
81 1
3
3
So the z-component of the centroid is
43 2
z
8.
Compute
Vz
V
1
ye x
f ds for f
y
R
3
counterclockwise around
2
the polar curve
1
r
3
cos 6
-4
shown at the right.
-1
2
4
x
-2
Hint: Use a Theorem.
a. 0
-2
-3
Correct Choice
b. 6e 3
c. 3e 6
d. 6e 6
3e 3
e. 3e 3
6e 6
Solution: By the FTCC,
f ds
fB
fA
0 because B
A no matter what point you start at.
R
4
y
9.
Compute
F ds for F
5x 3
2
6y 4
7y, 4x
3
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. 20
b. 3
c.
3
d.
30
e.
60
Correct Choice
Solution:
P 5x 3 7y
By Green’s Theorem,
Q
F ds
R
10. Compute
6y 4
4x
xQ
yP
xQ
dx dy
yP
4
3
3 area
3 dx dy
R
7
3 20
60
R
F dS over the cone z
x2
y 2 for z
4 oriented down and out for
C
F
y z, x z, z .
Hint: Use a Theorem.
a. 4
b. 8
c. 16
d. 32
Correct Choice
e. 64
Solution:
By Stokes’ Theorem,
At the boundary, z
x2
y2
F ds oriented clockwise.
F dS
C
C
4. So the boundary is the circle of radius r
4 at height z
4,
parametrized by r
4 cos , 4 sin , 4 . The vector field is F
8 sin , 8 cos , 2 .
The tangent vector is v
4 sin , 4 cos , 0 . Reversed v
4 sin , 4 cos , 0 .
So F v 16 sin 2
16 cos 2
16. And the integral is
2
F ds
C
2
F v dt
0
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
x 3 z, y 3 z, x 2
for the vector field F
x2
between the paraboloid z
y2
3
and the solid
y 2 and the plane z
4.
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:
4
2
z
F dV
0
V
0
4
3
z 3 dz
2 0
Right Hand Side:
3zr 2 r dr d dz
4
2
0
0
3 z4
2 4
4
3z r
4
z
dz
r 0
4
96
0
The boundary surface consists of a paraboloid P and a disk D with appropriate orientations.
d. Parametrize the disk D:
R r,
r cos , r sin , 4
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
3
on the disk:
4r 3 cos 3 , 4r 3 sin 3 , r 6
R r,
h. Compute the dot product:
F N
r7
i. Compute the flux through D:
2
2
F dS
D
0
0
r 7 dr d
2
r8
8
2
64
0
6
r cos , r sin , r 2
The paraboloid P may be parametrized by R r,
j. Compute the tangent vectors:
er
cos ,
e
sin ,
r sin ,
2r
r cos ,
0
k. Compute the normal vector:
N
î 2r 2 cos
2r 2 sin
k r cos 2
r sin 2
2r 2 cos , 2r 2 sin , r
2r 2 cos , 2r 2 sin , r
This is oriented up. Need down. Reverse: N
x 3 z, y 3 z, x 2
l. Evaluate F
F
y2
3
on the paraboloid:
r 5 cos 3 , r 5 sin 3 , r 6
R r,
m. Compute the dot product:
2r 7 sin 4
2r 7 cos 4
F N
r7
r 7 2 cos 4
2 sin 4
1
n. Compute the flux through P:
Hints: Use these integrals as needed, without proof:
2
2
sin 2 d
0
0
2
2
F dS
0
P
2
cos 2 d
2
sin 4 d
0
r 7 2 cos 4
2 sin 4
r8
8
1 dr d
0
cos 4 d
0
2
2
2
0
0
2
3
4
cos 4 d
sin 2 cos 2 d
0
2
2
sin 4 d
0
1
4
2
1d
0
3
25 3
2
32
2
2
o. Compute the TOTAL right hand side:
F dS
V
F dS
D
F dS
64
32
96
which agrees with (c).
C
7
12.
y
(5 points) Sketch the region of integration
1.0
for the integral
1
1
0
x
0.5
x sin y 2 dy dx.
2
Shade in the region.
0.0
You computed its value in problem (4).
0.0
0.5
1.0
x
13. (20 points) Ham Duet is flying the Millenium Eagle through the galaxy. His current galactic position
is P
1, 4, 4 lightyears. He is passing through a deadly polaron field whose density is
polarons/lightyear 3 .
a. If his current velocity is v
0. 3, 0. 2, 0. 1
x2
yz
lightyears/year, at what rate does he see the polaron
density changing?
Solution:
d
v
dt
2x, z, y
0. 3 2
P
2, 4, 4
P
0. 2 4
0. 1 4
0. 6 polarons/lightyear 3 /year.
b. Ham decides to change his velocity to get out of the polaron field. If the Millenium Eagle’s
maximum speed is 0. 9 lightyears/year, with what velocity should Ham travel to reduce the
polaron’s density as fast as possible?
Solution:
The direction of maximum increase is
The direction of maximum decrease is
The length is
P
4
16
16
P
P
2, 4, 4 .
2, 4, 4 .
6.
So the unit vector of maximum decrease is v
P
1
6
2, 4, 4
1, 2, 2 .
3 3 3
P
Since the maximum speed is |v | 0. 9, Ham’s velocity should be
v |v |v 0. 9 1 , 2 , 2
0. 3, 0. 6, 0. 6 .
3 3 3
8