Math B6C – Chapter 13 Quiz – SOLUTIONS
* Fall 2001 *
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x
 xy e
1. Let D denote the rectangle 1 < x < 3, 0 < y < 1. Then
2  y2
dxdy  ?
D
 xy e
13
x2  y 2
dxdy   xy e
D
01
1
x2  y 2
13
dxdy   xy e x e y dxdy
2
2
01
3
  y e dy  x e x dx
y2
0
1
u = y2, du = 2y dy, to get
Next, make the substitution (same for both integrals):
1
4
1
0
9
eu du
1
eu du


2
1
 1 eu
4
9
eu
0
1

10
2
9
 1 e1  e0 e9  e1  e  e  e  e
4
4
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2. If R denotes the region described by 0 < x < /2, y < 2x, y > x, then

R sin  x  cos  y  dxdy  ?
2 2x
R sin  x  cos  y  dxdy  0 x sin  x  cos  y  dydx


2x
2
0 sin  x  x cos  y  dydx  0 sin  x  sin  y 




0 sin  x   2sin  x  cos  x   sin( x)  dx

2
0
For the first integral, let u=sin(x), so du=cos(x).
For the second integral, use the trig identity sin 2  x  
0
x
2
 2  sin 2  x  cos  x  dx 

 2 u du 
dx
2

2
2x
0 sin  x  sin  2 x   sin( x)  dx

1
2
2
0
2
2
0 sin  x  dx
1  cos(2 x)
. Then we get
2
3
1  cos  2 x 
dx  2  u
2
3
1
0

1
2
2
0

1
dx 
2
2
0 cos  2x  dx
2  1 sin  2 x 
  
3 4 2 2

2

0
2  8  3
 
3 4
12
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3. Let B be the rectangular solid defined by 0 < x < /2, 0 < y < 1, 0 < z < .

1 2
xyz sin  x

B
2
Then,
 cos  z  dx dy dz     xyz sin  x2  cos  z 2  dx dy dz
2
00 0

  z cos
0

  dz  y dy  x sin  x  dx
1
z2
0
2
2
0
2
For the first and third of these integrals, let u=z , du=2dz, to obtain:
2
2
1
4
1

cos u du y dy sin u du
4
0
1
4
0
2
y2
sin u
2
0
0
2
1
  cos u 
0
4
0


 2 
  cos 
  cos 0 
2 
 4 

 1  sin  2   sin 0  1
4
sin  2  1  cos  



 
 4 


2
8
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4. Let W be the region 0 < x < 1, x < y < 2x, xy < z < 2xy, then
1 2 x 2 xy
 x  y  z  dx dy dz      x  y  z  dz dy dx

W
0 x xy
1 2 x 2 xy
1 2 x 2 xy
0
1
1 2x

x xy x dz dy dx  0 x xy y dz dy dx  0 x xy z dz dy dx
2 x 2 xy
x 
x
0
1
2x

0
0
2x
x2
x
1 2 x 2 xy
1 2x
x
1
2 xy
xy dz dy dx  0 x y xy dz dy dx  0 x xy z dz dy dx
  x   2 xy  xy  dy dx  
0
1 2 x 2 xy
1 2x
y dy dx   x 
0
x
y2
1 2x
x y  y  dy dx  0 x
1 2x 


0 x
dy dx  1 
2
z2
2
2 xy
dy dx
xy

2
2
xy

xy
 dy dx







2
2x
1
2
  x2 y
2
0
1
x
1
y3
dx   x
3
0
2x
x
1
2x
dx  3  x2  y 2 dy dx
2
x
0
1
1
2
3
2
3
1 2
1 
3 2 y3
  x  2 x   x   dx   x  2 x   x   dx   x
2
3 
2
3



 
0
 
0
1
1
1
0
0
0
0
2x
dx
x
3

3
 3  x4dx  7  x4dx  1  x2  2 x   x   dx
2
3
2


 
1
1
 3 7  x5
7 5
  
 x dx   9 14  1  7  1 
2
 2 3 5
 6 5 2 6
0
0

27
20
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5. If S denotes the unit-hemisphere above the xy-plane, then
x

S
2
 y 2  z 2  dxdy dz  ?
Upon changing to spherical coordinates, this integral becomes:


2 2 1
2
1
2
2
2
   sin  d  d d  d sin 
0 0 0
0
0
0
 


2
 2  sin 
0
1
5
5
d  2
0



4
d  d
sin  d  2   cos 

5 0
5
2

2
0
 2
5
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R xy dxdy  ?
6. If R denotes the region y > 2(x–1), y < 2x + 1, y >1–x, y < 2–x, then
( Hint: Change coordinates to u = y – 2x, v = x + y )
Solving for x and y, in terms of u and v, we get: x = (v-u)/3, y = (u+2v)/3. The Jacobian determinant of this
coordinate transformation is then
1 1
| J |  3 3   2  1  1.
9 9 3
1 2
3 3
Thus, dx dy = 1/3 du dv, and the above integral becomes:
2 1
2 1
1 2
v  u  u  2v  1 dudv  1
3
3 3
27
2
1
2

1 2  2v
1
2  u 2  uv  dudv


2
1
 2  v2dv  du  1  dv  u 2du  1  vdv  udu
27 1
27 1 2
27 1
2
2
2
1
2 v3
1 u3

1 (2)  27 3
27 3
2
1

81 
2
1 v
27 2
2
1
2
108 
3
 2 23 13  3  1 13   2   1

81 

u2
2
2
1
2

2
12 12   2  

 14  1  1  53
27
9
12
108
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7. Let R denote the region y > x, y < x + 1, y >1/x, y < 2/x. Then
(Hint: Change coordinates to u = y – x, v = xy )
R
 y  x
2
 4 xy dxdy  ?
First lets look at the graph of region R:
Following the hint, we need to solve for x and y, in terms of u and v, in order to calculate the Jacobian determinant
(which we need to express dx dy in terms of du dv.
v  xy  y 
v
x
Substituting this expression for y into the equation u = y – x, we obtain
v
u  x,
x
and multiplying both sides times x we get
ux  v  x 2
x 2  ux  v  0.
The quadratic formula then allows us to solve for x in terms of u and v:
x
u  u 2  4v
.
2
As can be seen from the graph of region R, x is positive throughout, and so we must have that
x
Then y = v/x, so we get
u  u 2  4v
.
2
y
2v
.
u  u 2  4v
x
u
Now we calculate the Jacobian determinant of this coordinate transformation, | J | 
y
u

1
1 2
 1  u  4v
2
2

1/ 2


| J | 2v   1  1  u 2  4v 
1/ 2
2



2
 u  u  4v

1
4
 2u 


 2u 

1/ 2 


2
 u  u  4v

2
| J | 2v   1  u 



 u  A
uA
2A
A
2
 2v  u  A
A A  u 
2


1/2


4v  2v  u  A 
 A 
A
2
2
 u  A
 Au
2u  2 A 

1/ 2 
A  u 2  4v
uA
2A
1
A
4
1 2
  2  2v   2 u  4v


1/ 2 

2
 u  u  4v



To simplify this, it will help (relieve writer’s cramp!) if we substitute
1
u
 1  
2
A

u 2  4v
x
v .
y
v

1/ 2

 4

2


into the above expression:
1
A
 2uA  2 A2  4v 


A


 Au
2
1
A
u  A 2uA  2 A2  4v 2v  u  A 1




2
2 A
2uA  2 A2  4v
2A
A A  u 
A A  u 
2
A A  u 
 A  u  A
u  A

1


2
2

uA  A  2v  2v   2
uA  A  

2
2
2
A  Au
A  Au
A  Au
 Au  1  1

A A  u  A
u 2  4v

Thus,
dxdy 
dudv ,
u 2  4v
2 1
1 0

and the above integral becomes
2 1
1
u  4v 
dudv  
u 2  4v
1
2
0
dudv  (2 1)(1 0) 
1
.
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1
8. Change from Cartesian to polar coordinates to compute
1 x2
  1x
1
2
1 x
2
2
y
2

2
dy dx .
First visualize the region we’re integrating over:
Since
dxdy  rdrd , the above integral becomes
2
2 1
2
1
1
0 0 1 r 2 2 r dr d   d  1 r 2 2 dr  2  1 r 2 2 dr
0
0

2r



0
2r


Let u = 1 + r2, so du = 2r dr, and the integral becomes:
2
1
2
1 du  2  1 


2
0 1 r 2 2 dr  2 1 u

2r


u

2
1
 2  1  1   
2
1
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9. Evaluate
 xyz dxdy dz
over the solid ellipsoid
x2 y 2 z 2
  1 .
a 2 b2 c 2
( Hint: let x = au, y = bv, z = cw, then integrate over an appropriate region in uvw–space )
The Jacobian determinant of this suggested coordinate transformation is
a 0 0
| J |  0 b 0  abc.
0 0 c
Thus, dx dy dz = abc du dv dw, and the above integral becomes:

unit
au  bv  cw  abc dudvdw  8  abc 
sphere
where S is the part of the unit sphere in Octant I.
2
S uvw dudvdw
If we convert this integral to spherical coordinates, it becomes:

8  abc 
2
2

2
1
2
0 0 0   sin  cos   sin  sin    cos   sin  d  d d


2
1
2
=8  abc   cos sin d  sin
2
3
 cos d   5 d 
0
0
Let u=sint, du = cos t dt, in the first and second integrals, and we get
8  abc 
2
1
1
2
 u du  u du
0
0
61

6
=8  abc 2 16  u2
2
0
1 41
0
u
4
0
0
=8  abc 2 16  u2
1 41
2
0
u
4

 abc
0
2
6
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3
10. Let D be the unit circle in the xy–plane. Then
2
2
 ( x  y ) dxdy  ?
D
Changing to polar coordinates, the integral becomes:
2 1
0 0  
r2
3
r dr d 
2
1
1
0 d 0
r7
8
r
dr  2

8
4
0
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 x  u3  v2

11. Let 
define a coordinate transformation in some region of the plane.
2
4
 y  u  v
Then
3u 2 2v
dxdy 
dudv  3u 2 4v3  2u 2v dudv
3
2u 4v




dxdy  12u 2v3  4uv dudv
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12. Find the volume of the finite region enclosed by the two paraboloids,
f ( x, y)  x2  y 2 1, g ( x, y)  1  x 2  y 2 . 
By symmetry, the volume of this region will be twice the volume of the area bounded by g and the xy-plane, which
is the integral of g over the unit circle. In polar coordinates then, our desired volume is given by:
2 1
2
0
0 
1 r 2
2
1
0
0
 r dr d  2  d  
r  r3
 dr  4
 r2

 2

1
r4 
   
4
0
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 x  euv
13. Let 
u v define a coordinate transformation in some region of the plane.
 y  e
euvv euvu
Then
dxdy  uv u v dudv   euvveu v  eu veuvu  dudv
e
e
dxdy  euvuv  v  u  dudv