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Ma 227
ID#:________________________
Final Exam Solutions
12/17/07
Name:
Lecture Section:
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Directions: Answer all questions. The point value of each problem is indicated. If you need more work
space, continue the problem you are doing on the other side of the page it is on.
There is a table of integrals at the end of the exam.
Score on Problem #1 ________
#2 ________
#3 ________
#4 ________
#5 ________
#6 ________
#7 ________
#8 ________
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1
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Problem 1
a) (13 points)
Find the eigenvalues and eigenvectors of
2 2
A
2 2
Solution:
2−r
2
2
2−r
 2 − r 2 − 4  r 2 − 4r  rr − 4  0
Thus the eigenvalues are r  0, 4. The system A − rIX  0 is
2 − rx 1  2x 2  0
2x 1  2 − rx 2  0
For r  0 we have
x 1  x 2  0 or x 2  −x 1
1
so the eigenvector corresponding to r  0 is
.
−1
For r  4 we have
− 2x 1  2x 2  0 or x 1  x 2
1
Thus the eigenvector for r  4 is
SNB check
2 2
2 2
1
−1
, eigenvectors:
↔ 0,
1
1
1
b) (12 points)
Solve the nonhomogeneous problem
e −t
x ′ t  Axt 
2e −t
where A is the matrix above in 1a).
Solution:
x h  c 1 e 0t
−1
 c 2 e 4t
1
Let
xp 
ae −t
be −t
Then the DE implies
2
1
1
↔4
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−ae −t
−be −t

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2 2
ae −t
2 2
be −t
e −t

2e −t
or
−ae −t
−be −t
e −t  2ae −t  2be −t

2e −t  2ae −t  2be −t
Therefore
− a  2a  2b  1
− b  2a  2b  2
or
3a  2b  −1
2a  3b  −2
, Solution is: a  15 , b  − 45
xp 
1 e −t
5
− 45 e −t
and
xt  x h  x p  c 1
−1
1
 c2
e 4t
1
1

1 e −t
5
− 45 e −t
SNB check
x ′1  2x 1  2x 2  e −t
x ′2  2x 1  2x 2  2e −t
, Exact solution is:
x 1 t  C 3  15 e −t  C 4 e 4t , x 2 t  C 4 e 4t − 45 e −t − C 3
Problem 2
a) (12 points)
Give an integral in cylindrical coordinates for the volume of the solid region bounded by the cylinder
x 2  y 2  4 and the planes z  0 and y  z  3. Sketch the solid region. Do not evaluate this integral.
Solution:
x2  y2  4
3
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Volume  rdzdrd 
V
2 2 3−r sin 
0 0 0
rdzdrd
b) (13 points)
 
ndS 
Evaluate the surface integral  ∇  F
S
 x, y, z 
F

, where
  dS
∇F
S
xy
e cos zi  x 2 zj 
xyk
and S is the hemisphere x  1 − y 2 − z 2 oriented in the direction of the positive x −axis.
Solution: We use Stokes theorem. Thus

 
∇F
ndS 
S
The boundary of S is the circle y 2  z 2
C : x  0,
so
rt 


  dS
∇F
S
 F  dr
C
 1, x  0. Therefore
yt  cos t, zt  sin t 0 ≤ t ≤ 2
0i  cos tj  sin tk  r ′ t  − sin tj  cos tk
 0, cos t, sin t  cossin ti  0j  0k
F
so
2
 F  dr   0
0dt  0
C
Problem 3
a) (12 points)
Evaluate
1 1
 0  x cos
y 2 dydx
Solution: We reverse the order of integration. Now y goes from y  x to y  1 and x goes from 0 to 1.
4
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Therefore, the region of integration is
x
y
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
x
Therefore
1 1
 0  x cos
y 2 dydx 

1 y
 0  0 cos
1
 0 y cos
y 2 dxdy
y2
1
sin y 2
dy 
2
0
 1 sin 1
2
b) (13 points)
Give two different integral expressions for the area of the region bounded by the parabolas x  y 2 and
x  8 − y 2 . Do not evaluate these expressions. Be sure to sketch the area.
Solution:
x  y2
y
5
4
3
2
1
0
2
4
-1
-2
-3
-4
5
6
8
x
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ID#:________________________
The curves intersect when y 2  8 − y 2 that is, when y  2. Thus at the points 4, 2 and 4, −2
Therefore
Area 
 dA
R
2
8−y 2
4
x

 −2  y 2

0 −
x
dxdy
dydx  
8

4 −
8−x
8−x
dydx
Problem 4
a) (10 points)
Evaluate the line integral
C zdx  xdy  ydz
where C is given by x  t 2 , y  t 3 , z  t 2 , 0 ≤ t ≤ 1.
Solutions: Here
rt  t 2i  t 3j  t 2k  r ′ t  2ti  3t 2j  2tk
 x, y, z  zi  xj  yk
F
so
 t  t 2i  t 2j  t 3k
F
1
C zdx  xdy  ydz   0

1
0
t 2 2t  t 2 3t 2  t 3 2t dt
1
4
2t 3  5t 4 dt  t  t 5
 3
2
2
0
b) (15 points)
Find the inverse of the matrix
1 0 1
A
0 2 1
1 1 1
3
and then use it solve the system AX 
−2
.
2
Solution:
1 0 1 1 0 0
0 2 1 0 1 0
1 1 1 0 0 1
1 0 1
→ −R 1 R 3
0 2 1
1
0
0 0
1 0
0 1 0 −1 0 1
6
1 0 1
→ R 2 ↔R 3
1
0 0
0 1 0 −1 0 1
0 2 1
0
1 0
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1 0 1
→ −2R 2 R 3
1
0
0
0 1 0 −1 0
0 0 1
2
ID#:________________________
→ −R 3 R 1
1
1 −2
1 0 0 −1 −1
2
0 1 0 −1
0
1
0 0 1
1
−2
2
Therefore
A −1
1 0 1
SNB check:
0 2 1
, inverse:

−1 −1
2
−1
0
1
2
1
−2
−1 −1
2
−1
0
1
2
1
−2
1 1 1
3
We note that the solution of AX 
−2
3
is X  A −1
2
−2

2
−1 −1
2
3
−1
0
1
−2
2
1
−2
2

3
−1
0
Problem 5
a) (15 points)
Verify the divergence theorem for the vector field F  x i  y j  z k over the sphere
x2  y2  z2  a2.
Solution: We have to show that
 F  dS   F  N ds   div Fdv
S
V
S
We begin with the surface integral and use spherical coordinates to parametrize it. The   a and
r ,   x i  y j  z k  a sin  cos  i  a sin  sin  j  a cos  k
0 ≤  ≤ , 0 ≤  ≤ 2
Therefore
r   a cos  cos  i  a cos  sin  j − a sin  k
and
r   −a sin  sin  i  a sin  cos  j  0 k
7
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i
N  r  r 
ID#:________________________
j
k
a cos  cos  a cos  sin  −a sin 
−a sin  sin  a sin  cos  0
 a 2 sin 2  cos  i  a 2 sin 2  sin  j  a 2 cos  sin  cos 2   a 2 cos  sin  sin 2  k
 a 2 sin 2  cos  i  a 2 sin 2  sin  j  a 2 cos  sin  k
Note that at

2
.
 N  a 2 j which points outward, so we use this N
, 2
 ,   a sin  cos  i  a sin  sin  j  a cos  k
F
2 
 F  dS   F  N ds   0  0
S
a 3 sin 3  cos 2   a 3 sin 3  sin 2   a 3 cos 2  sin  dd
S
2 
2
 a 3   sin 3   cos 2  sin  dd  a 3 
0
0
0
3
1 cos 3 − 3 cos  − cos 
4
12
3
2
2
− 1  3  1 − 1  3  1 d  2a 3  d
 a3 
4
4
12
3
12
3
0
0
3
 4a
Also
 3
div F
so
2  a
 div Fdv   0  0  0
3 2 sin  ddd
V
2 
 3a 
3 0
 4a 3
3
2
 0 sin dd  a 3  0
b) (10 points)
Let
D
−2 0
0
0
4
0
0
0 −1
Find e Dt .
Solution:
e Dt

e −2t
0
0
0
e 4t
0
0
0
e −t
8
− cos  0 d

d
0
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SNB check: e
−2 0
0
0
4
0
0
0 −1
t

ID#:________________________
e −2t
0
0
0
e 4t
0
0
0
e −t
Problem 6
a) (10 points)
Find a function x, y, z such that
 x, y, z  sin yi  x cos yj  z − sin zk
∇  F
Solution:
 x  sin y  y  x cos y  z  z − sin z
 x  sin y
   x sin y  gy, z
Thus
∂g
 x cos y
∂y
 y  x cos y 
∂g
Thus ∂y  0 and g  hz so
  x sin y  hz
Then we have
 z  h ′ z  z − sin z
2
so hz  z2  cos z  K. Hence
2
  x sin y  z  cos z  K
2
6 b) (15 points)
Verify that Green’s Theorem is true for the line integral
C ydx 
x  y 2 dy
where C is the ellipse 4x 2  9y 2  36 with counterclockwise orientation.
Solution:
 Px, ydx  Qx, ydy  P y − Q x dA
R
C
x2
To calculate the line integral around the ellipse 9 
0 ≤ t ≤ 2. Then dx  −3 sin tdt, dy  2 cos tdt.
9
y2
4
 36 we let xt  3 cos t and yt  2 sin t,
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C ydx 
2
x  y 2 dy 
0

0

0
2
2
ID#:________________________
2 sin t−3 sin t  3 cos t  4 sin 2 t 2 cos t dt
−6 sin 2 t  6 cos 2 t  8 sin 2 t cos t dt
6 cos 2 t − sin 2 t  8 sin 2 t cos t dt
2
 3 sin 2t  8 sin 3 t
0
3
0
Also P y  1  Q x so
P y − Q x dA   0dA  0
R
R
Problem 7
a) (13 points)
Consider the two curves x 2  y 2  2y and x 2  y 2  2x. Give an integral or integrals in polar
coordinates for the area between the two curves. Be sure to sketch the area between the two curves. Do
not evaluate the integral or integrals.
Solution:
The two curves are given by
r  2 sin 
and
r  2 cos 
The graphs are given below.
y
2
1
-1
1
2
x
-1
A
 R rdrd
where R is the region common to both circles. The two circles intersect when
2 cos   2 sin 
or when
tan   1
10
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That is, at    . r goes from 0 to the circle r  2 sin , for 0 ≤  ≤  and from 0 to r  2 cos  for
4
4
 ≤  ≤  . Thus we need two integrals to express the area.
2
4

 2 sin 
2 2 cos 
A4 
rdrd    
rdrd
0
0
4 0
b) (12 points)
Evaluate
 zdV
E
where E lies between the spheres  
 1 and x 2  y 2  z 2  4 in the first octant.
Solution: We use spherical coordinates. The equation of the unit sphere is   1 whereas the equation
of the other sphere is   2. Since E is in the first octant, then 0 ≤  ≤ 2 and 0 ≤  ≤ 2 . Thus
x2
y2
z2




2
 zdV   02  02  1  cos  2 sin ddd
E
2
 02  02
4
4
 16 − 1
4

2

cos  sin dd
1

 02
cos  sin dd

2
sin 2 
 15
8
2
 15
16
0
Problem 8
a) (13 points)
It can be shown that the Cauchy-Euler system
tx ′ t  Axt t  0
where A is a constant matrix, has a nontrivial solutions of the form
xt  t r u
if and only if r is an eigenvalue of A and u is a corresponding eigenvector. Use this information to
solve the system
tx ′ t 
Solution: We first find the eigenvectors of
−4 − r
2
2
−1 − r
−4
2
2
−1
−4
2
2
−1
xt t  0
.
 4  r1  r − 4  r 2  5r
so the eigenvalues are r  0, −5. The system A − rIx  0 is
−4 − rx 1  2x 2  0
2x 1  −1 − rx 2  0
11
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ID#:________________________
1
r  0 yields −4x 1  2x 2  0 or 2x 1  x 2 . This gives the eigenvector
−2
equation x 1  −2x 2 and hence the eigenvector
xt  c 1 t 0
2
. r  −5 leads to the
. Thus
1
1
2
 c 2 t −5
−2
1
b) (12 points)
Rewrite the scalar equation
d3y
dy
−
 y  cos t
3
dt
dt
as a first order system in normal form. Express the system in the matrix form x ′  Ax  f.
Solution: Let
′′
x 1 t  yt, x 2 t  y ′ t, x 3 t  y t
Then
x ′1 t  y ′ t  x 2 t
x ′2 t  y ′′ t  x 3 t
x ′3 t  y ′′′ t  y ′ − y  cos t  x 2 − x 1  cos t
Thus
x ′1
x ′2
x ′3

0
1 0
x1
0
0 1
x2
−1 1 0
x3
12
0

0
cos t
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ID#:________________________
Table of Integrals
 sin 2 xdx  −
1
2
cos x sin x  12 x  C
 cos 2 xdx 
1 cos x sin x  1 x  C
2
2
3
1
3
 sin xdx  12 cos 3x − 4 cos x − 23 cos x
1 sin 3x  C :
 cos 3 xdx  34 sin x  12
 cos 2 x − sin 2 x dt  12 sin 2x  C
13
C