Review for MA227 Final
Chapter 12 - Multiple Integration
1.
1 1
0 
Evaluate the iterated integral:
3
x
e y dydx.
Solution:
We evaluate by switching the order of integration. The region in the xy-plane is above the curve
y  x , below the line y  1 and to the right of x  0.
y x
y
1.0
0.8
0.6
0.4
0.2
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
So
1 1
0 
3
x
e y dydx 
1 y2
0 0
3
e y dxdy 
1
1
 0 xe y 3 | y0 dy   0 y 2 e y 3 dy 
2
1 e y 3 | 1  1 e − 1
0
3
3
(using
u-substitution).
2.a)
4
2
 −2  x 2 x 2 ydydx.
Sketch the region of integration:
Solution:
y
4
3
2
1
-2
-1
0
1
2
x
1
b)
Reverse the order of integration.
Solution:
y  x 2  x   y , 0 ≤ y ≤ 4, so
2
4
4
 −2  x 2 x 2 ydydx   0  −
y
y
x 2 ydxdy.
3.
Evaluate   ydA, where D is the region in the first quadrant that lies above the hyperbola xy  1
D
and the line y  x and below the line y  2.
Solution:
y
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
1
x
2.0
x
For this region, it’s easiest to integrate first with respect to x. We have 1y ≤ x ≤ y, and 1 ≤ y ≤ 2, so
3
2 y
2
2
 D ydA   1  1 ydxdy   1 y y − 1y dy   1 y 2 − 1 dy  y3 − y| 21  43 .
y
If we integrate first with respect to y, we have to break it into 2 regions:
1 ≤ x ≤ 1, 1 ≤ y ≤ 2, and 1 ≤ x ≤ 2, x ≤ y ≤ 2
x
2

1
2
 D ydA   1  1
2
4.
x
ydydx  
2 2

1 x
ydydx  4
3
2 2
0 y
Consider   fx, ydxdy.
a)
Sketch the region of integration.
Solution:
0 ≤ y ≤ 2, y ≤ x ≤ 2
2
y
2.0
1.5
1.0
0.5
0.0
0.0
b)
0.5
1.0
1.5
2.0
x
Reverse the order of integration:
Solution:
0 ≤ x ≤ 2, 0 ≤ y ≤ x 
2 2
2 x
 0  y fx, ydxdy   0  0 fx, ydydx
c)
Rewrite the integral in terms of polar coordinates.
Solution:
In polar coordinates, the line y  x is given by    . Noting that the polar equation of
4
the line x  2 is r cos   2 or r  2 sec , we have
/4 2 sec 
0 0
5.
fr cos , r sin rdrd.
Evaluate   x x 2  y 2 dA, where D is the closed disk with radius 1 and center 0, 1.
D
Solution:
Since we’re integrating over a circle it makes sense to use polar coordinates. The disk with radius
1 and center 0, 1 is given in rectangular coordinates by x 2  y − 1 2  1. Convert to polar:
x 2  y − 1 2  1  x 2  y 2 − 2y  1  1  x 2  y 2  2y
 r 2  2r sin   r  2 sin .
The entire circle is traced out as  ranges from 0 to , so we have
 2 sin 
 D x x 2  y 2 dA   0  0
 2 sin  3
r cos drd
r cos  r 2 rdrd   
0 0

  4 sin 4  cos d  4 sin 5 | 0  0
0
5
(using u-substitution)


0
r 4 cos | 2 sin  d
0
4
6.
Evaluate    ydV, where E is the tetrahedron bounded by the planes x  0, y  0, z  0 and
E
2x  y  z  2.
3
Solution:
2 − 2x − y
In the xy-plane we have
y
2.0
1.5
1.0
0.5
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
,
where 0 ≤ y ≤ 2 − 2x, 0 ≤ x ≤ 1
1 2−2x 2−2x−y
1 2−2x
2−2x−y
0
ydzdydx   
yz| 0
dydx
0 0
0 0
1 2−2x
1
y 3 2−2x
0 0
2y − 2yx − y 2 dydx   y 2 − y 2 x −
| 0 dx
0
3
so    ydV   
E


4
− 4x  4x 2 − 43 x 3 dx  4 x − 2x 2  4 x 3 − x | 10
3
3
3
 4 −2 4 − 1  1.
3
3
3
3
1
0
4
3
7.
Give an expression for the volume inside the sphere x 2  y 2  z 2  4 and outside the cylinder
2
2
x  y  1.
Solution:
4
2
1
-2
1
2
-2
z
yx
1
-1 2
-2
We’ll consider half of the figure and then multiply our answer by 2. Take the top half of the
figure. Then we have 0 ≤ z ≤ 4 − x 2 − y 2 . The region over which we integrate in the xy-plane is the
annular region between the circles x 2  y 2  1 and x 2  y 2  4 :
y
2
1
-2
-1
1
2
x
-1
-2
.
We’ll use cylindrical coordinates to find the volume. Recall that in cylindrical coordinates,
x  r cos , y  r sin , z  z, and
dV  rdzdrd. Our z limits become 0 ≤ z ≤

V  2
4 − r 2 , and we have 1 ≤ r ≤ 2, 0 ≤  ≤ 2
2 2
0
1 0
4−r 2
rdzdrd
.
8.
Find the volume of the region of the ball x 2  y 2  z − 1 2  1 cut out by the cone z 2  x 2  y 2
using spherical coordinates.
Solution:
5
1.5
1z
0.5
-2
-1
-2
-1
0
y x
1
1
2
2
Recall that in spherical coordinates, x   sin  cos , y   sin  sin , z   cos ,
dV   2 sin .
 x 2  y 2  z − 1 2  1   sin  cos  2   sin  sin  2   cos  − 1 2  1
  2  2 cos     2 cos .
The equation of the cone becomes  cos  2   sin  cos  2   sin  sin  2
 tan 2   1     .
4
So
V
2

4
0
2 cos 
0  0
 2 sin ddd  
9.
Find the volume of the solid bounded by the plane z  0 and the paraboloid z  1 − x 2 − y 2 .
Sketch the volume.
Solution:
1 − x2 − y2
The paraboloid z  1 − x 2 − y 2 intersects the x, y −plane on the circle x 2  y 2  1. Let D denote
the inside of the circle. Then the volume is
6
V
1−x 2 −y 2
 D  0
dzdA
Using cylindrical coordinates x  r cos , y  r sin , z  z we have,
V
2 1 1−r 2
0 0 0
r dzdrd 
2 1
0 0
1 − r 2 rdrd  
2
7