Name
ID
MATH 221
1-9
/45
Fall 2012
10
/15
P. Yasskin
11
/25
12
/15
Total
/100
Final Exam
Sections 502
Solutions
Multiple Choice: (5 points each. No part credit.)
1. If a
3, 3, 1
and b
then
2, 5, 1
a
3b
a. 26
Correct Choice
b. 13
c. 0
d.
13
e.
26
SOLUTION:
a 3b
3, 3, 1 3 2, 5, 1
a 3b
9 144 16
3
169
6, 3
13
15, 1
x2
2
2. Find the point on the elliptic paraboloid z
a.
3, 12, 4
y2
4
2, 1, 2 .
3 3 3
1 where a unit normal is
2, 4, 7
b.
2, 4, 7
c.
4, 2, 10
d.
1, 1, 7
4
e.
3
Correct Choice
1, 1, 7
4
2
SOLUTION:
f z x
2
y
x k2,
k1,
3
2
3
2
y2
x
z 1
1 1
2
4
2
y2
4
1
1
4
f
k2
3
x,
k
y
,1
2
3
2
k 2, 1, 2
3 3 3
x
k2
1
3
y
k2
3
1
7
4
1
t, t 2 , 2 t 3
3
3. Find the point on the curve r t
1, 1, 1
2 4 12
1, 1, 2
3
2 , 2, 4 2
3
2, 4, 16
3
1, 1, 2
3
a.
b.
c.
d.
e.
SOLUTION:
r
4. Compute
1, 2 2 , 4
5
5
5
.
Correct Choice
1, 2t, 2t 2
vt
1 4t 2
1, 2 2 , 4
5
5
5
|v |
1 , 2t , 2t 2
1 2t 2 1 2t 2 1 2t 2
2
2 , 2, 4 2
3
T
where the unit tangent is T
8
2
0
y 1/3
4t 4
2t 2
1
2t 2
1
5
t
2
cos x 4 dx dy
HINT: Plot the region of integration and reverse the order of integration.
a. 1 sin 64
4
1
4
1
4
1
4
1
4
b.
c.
d.
e.
cos 64
sin 16
cos 16
sin 16
1
4
1
4
Correct Choice
SOLUTION:
0
y
y 1/3
y
8
x
8
6
2
or
4
0
x
2
0
y
x3
2
0
0.0
8
2
2
x3
0
y
0
0
cos x 4 dx dy
1/3
2
0
x 3 cos x 4 dx
cos x 4 dy dx
2
2
0
1.0
1.5
2.0
x
x3
y cos x 4
0
1 sin x 4
4
0.5
dx
0
1 sin 16
4
2
Find the y-component of the centroid (center of mass with density 1)
5.
of the hyperbolic cosine curve y
cosh x for 0
x
2.
t, cosh t .
HINTS: Parametrize the curve as r t
x
x
x
x
e e
e e
cosh x
sinh x
cosh 2 x sinh 2 x
2
2
cosh 0 1
sinh x
cosh x
cosh x
sinh 0 0
2
2
cosh
2x
1
cosh
2x
1
sinh x
cosh x
2
2
4 sinh 4
4 sinh 2
1 cosh 4
2 sinh 2
4 cosh 4
2 sinh 2
cosh 4 1
2 sinh 2
1 sinh 4
2 sinh 2
a. y
b. y
c. y
d. y
e. y
SOLUTION:
ds
v
1, sinh t
|v | dt
cosh t dt
2
y
2.0
sinh x
1.5
1.0
0
1
2
x
2
sinh t
cosh 2 t dt
0
sinh 4 1
4
sinh 2
cosh t
sinh 2
0
y|v | dt
0
Lx
L
2.5
1
2
0
y ds
sinh 2 t
1
|v |
2
0
Lx
3.0
Correct Choice
2
L
y 3.5
2
0
cosh 2t
2
1 dt
1
2
sinh 2t
2
t
2
0
sinh 4
4
1
4 sinh 4
4 sinh 2
6. Find the line perpendicular to the surface y 2 z
2xz
6 at the point
1, 2, 4 .
This line intersects the xy-plane at
a.
17, 30, 0
b.
17, 30, 0
c.
15, 34, 0
d.
15, 34, 0
Correct Choice
e. The line does not intersect the xy-plane.
SOLUTION:
f y 2 z 2xz
f
2z, 2yz, y 2 2x
The direction of the line is the normal to the plane at P
v N
f
8, 16, 2
1, 2, 4 :
1,2,4
Line:
rt
P tv
1, 2, 4 t 8, 16, 2
1 8t, 2 16t, 4 2t
This line intersects the xy-plane when z 0 4 2t. So t
2.
at the point r 2
1 8 2 , 2 16 2 , 4 2 2
17, 30, 0
3
7. Let L
e
lim
x,y
x2 y2
x
0,0
2
1
y
2
a. L does not exist by looking at the paths y
b. L exists and L
x and y
1 by looking at the paths y
x.
mx.
c. L does not exist by looking at polar coordinates.
d. L exists and L
1 by looking at polar coordinates.
e. L exists and L
0 by looking at polar coordinates.
SOLUTION:
Along y
mx, we have L
for all m including 1 and
In polar coordinates, L
8. Compute
lim e
x 0
Correct Choice
1 m2 x2
1 m 2 2x
m 2 2x
1,
1, which proves the limit exists and
1.
1 m2 x2
1
1
l’H
m2 x2
lim
e
1
x 0
1 which proves nothing.
lim e
r2
r 0
1
r2
x 2 y, xy 2
F ds for F
2
r
lim e 2r
r 0
2r
l’H
along the semicircle y
from 4, 0 to
4, 0 followed by the line segment from
HINT: Use Green’s Theorem.
16
x2
back to
4, 0
4, 0 .
a. 0
b. 64
5
c. 64
3
d. 32
e. 64
Correct Choice
SOLUTION:
F ds
4
0
4 x2
4
xQ
F ds for F
yP
x 2 y and Q
Q dy with P
4 x2
4
F ds
9. Compute
P dx
dy dx
4
y2, x2
2xy
y2
4
x 2 dy dx
0
2xy
xy 2 . By Green’s Theorem,
0
r4
4
r 2 r dr d
0
along the parabola y
x 2 from
4
64
0
1, 1
to
2, 4 .
HINT: Find a scalar potential.
a. 11
b. 32
c. 46
Correct Choice
d. 50
e. 92
SOLUTION:
F
2xy y 2 , x 2
By the F.T.C.C.
2xy
f for f
x2y
xy 2 since
xf
2xy
2,4
F ds
f ds
f 2, 4
f 1, 1
2
2
4
y 2 and
2 4
2
x2
yf
1
2
1
2xy
1 1
2
46
1,1
4
Work Out: (Points indicated. Part credit possible. Show all work.)
10.
(15 points) A rectangle sits on the xy-plane with its
2
y2
top 4 vertices on the elliptic paraboloid x
z 6.
4
9
Find the dimensions and volume of the largest such box.
L
2x
W
2y
H
z
Maximize V
LWH
4xyz
METHOD 1: Lagrange Multipliers:
V LWH 4xyz
V
4yz, 4xz, 4xy
2
2
y
y
x
g
z
g
2 x ,2 ,1
4 9
4
9
y
x
V
g
4yz
2
4xz
2
4xy
4
9
2
2
2
y
y
x2
z
4xyz
2x
2
z
4
9
4
9
2
y2
x2
z
z z 2z
Constraint:
6
z
4
9
2
2
y2
x2
z
3
z
3
x
6
4
9
2
2
2
2
Dimensions are
L 2x 2 6
W 2y 3 6
H
Volume:
V
2 63 63
V
2 63 63
3
3 6
2
y
z
3
108
METHOD 2: Eliminate a Variable:
2
2
y2
z 6 x
V LWH 4xyz 4xy 6 x
4
9
4
4
4
2
3
2
2
V x 24y 3x y
y
y 24 3x
y
0
y
9
9
0
x
V y 24x x 3 4 xy 2 x 24 x 2 4 y 2
3
3
3(1)-(2):
48 8x 2 0
x2 6
3(2)-(1):
48 4y 2 4 y 2 0
32y 2 48
9
2
y2
48
z 6 x
6 6
3
4
9
4
32
Dimensions are
L 2x 2 6
W 2y 3 6
Volume:
z
y2
9
24xy
x3y
0
24
3x 2
0
24
x2
x
4 xy 3
9
4 y2 0
9
4 y2 0
3
(1)
(2)
6
9
y
H
z
3 6
2
3
108
5
11.
(25 points) Verify Gauss’ Theorem
F dV
F dS
V
V
2
for the vector field F
cylinder x 2
y2
2
xz , yz , z x
4 for
3
2
z
y
2
and the solid
3.
Be careful with orientations. Use the following steps:
First the Left Hand Side:
a. Compute the divergence:
z2
F
z2
x2
y2
x2
y2
2z 2
b. Express the divergence and the volume element in the appropriate coordinate system:
r2
F
2z 2
dV
r dr d dz
c. Compute the left hand side:
3
2
2
F dV
3
V
2
4z
0
r2
3
2z 2 r dr d dz
2
0
3
r4
4
z2r2
2
3
dz
0
2
4
4z 2 dz
3
3
4z 3
3
4 12
36
192
3
Second the Right Hand Side:
The boundary surface consists of the cylindrical sides C, a disk T at the top and a disk B
at the bottom with appropriate orientations.
d. Parametrize the cylinder C:
R ,z
2 cos , 2 sin , z
e. Compute the tangent vectors:
e
2 sin ,
2 cos ,
0
ez
0,
0,
1
f. Compute the normal vector:
N
î 2 cos
2 sin
k0
2 cos , 2 sin , 0
This is oriented correctly outward.
g. Evaluate F
F
R r,
xz 2 , yz 2 , z x 2
y2
on the cylinder:
2 cos z 2 , 2 sin z 2 , 4z
h. Compute the dot product:
F N
4 cos 2 z 2
4 sin 2 z 2
4z 2
i. Compute the flux through C:
2
3
4z 2 dz d
F dS
C
0
3
2
4z 3
3
3
144
3
6
j. Parametrize the TOP disk T:
R r,
r cos , r sin , 3
k. Compute the tangent vectors:
er
cos ,
sin ,
0
e
r sin ,
r cos ,
0
l. Compute the normal vector:
î0
N
r sin 2
k r cos 2
0
0, 0, r
This is oriented correctly upward.
xz 2 , yz 2 , z x 2
m. Evaluate F
F
y2
on the top disk:
9r cos , 9r sin , 3r 2
R ,
n. Compute the dot product:
F N
0
r4
0
3r 3
o. Compute the flux through T:
2
2
F dS
0
T
3r 3 dr d
3r 4
4
2
0
2
24
0
p. Parametrize the BOTTOM disk B:
R r,
r cos , r sin , 3
q. Compute the tangent vectors:
er
cos ,
sin ,
e
r sin ,
r cos ,
0
0
r. Compute the normal vector:
N
î0
r sin 2
k r cos 2
0
0, 0, r
This is upward. Need down. Reverse: N
s. Evaluate F
F
2
2
xz , yz , z x
2
y
2
0, 0, r
on the bottom disk:
9r cos , 9r sin , 3r 2
R ,
t. Compute the dot product:
F N
3r 3
u. Compute the flux through B:
2
2
0
0
F dS
B
3r 3 dr d
24
v. Compute the right hand side:
F dS
V
F dS
C
F dS
T
F dS
144
24
24
192
which agrees with (i).
B
7
12.
(15 points) Compute
F dS for F
the hyperbolic paraboloid z
x2
for r
y 2 parametrized by
sin 2
r cos , r sin , r 2 cos 2
R r,
over
y, x, z
2 oriented upward.
HINTS: Use Stokes Theorem.
What is the value of r on the boundary?
SOLUTION:
Boundary: r
2 cos , 2 sin , 4 cos 2
r
4 sin 2
4 cos 2
cos sin 3
64 cos 3 sin
2
F dS
F ds
2
F vd
4
0
4
16
cos 4
sin 4
sin 2
v
2 sin , 2 cos , 16 cos sin
sin 2
2 sin , 2 cos , 4 cos 2
Fr
F v
2
4
64 cos 3 sin
64 cos 3 sin
cos sin 3
cos sin 3
d
0
2
8
0
8