CalculusVolume3-OP_mktoy8b-104

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5.9
5.10
Figure 5.20 Breaking the region into three subregions makes
it easier to set up the integration.
Here D1 is Type I and D2 and D3 are both of Type II. Hence,
∬
D
(2x + 5y)dA = ∬
D1
(2x + 5y)dA + ∬
D2
(2x + 5y)dA + ∬
D3
(2x + 5y)dA
= ∫
x = −2
x = 0
∫
y = 0
y = (x + 2)2
(2x + 5y)dy dx + ∫
y = 0
y = 4
∫
x = 0
x = y − (1/16)y3
(2 + 5y)dx dy + ∫
y = −4
y = 0
∫
x = −2
x = y − (1/16)y3
(2x + 5y)dx dy
= ∫
x = −2
x = 0
⎡
⎣
1
2(2 + x)2(20 + 24x + 5x2)⎤
⎦ + ∫
y = 0
y = 4
⎡
⎣
1
256y6 − 7
16y4 + 6y2⎤
⎦
+ ∫
y = −4
y = 0
⎡
⎣
1
256y6 − 7
16y4 + 6y2 + 10y − 4⎤
⎦
= 40
3 + 1664
35 − 1696
35 = 1304
105 .
Now we could redo this example using a union of two Type II regions (see the Checkpoint).
Consider the region bounded by the curves y = ln x and y = ex in the interval [1, 2]. Decompose the
region into smaller regions of Type II.
Redo Example 5.14 using a union of two Type II regions.
Changing the Order of Integration
As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation
that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is
important to learn to recognize when a change in order will simplify our work.
Example 5.15
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Changing the Order of Integration
Reverse the order of integration in the iterated integral ∫
x = 0
x = 2
∫
y = 0
y = 2 − x2
xex2
dy dx. Then evaluate the new
iterated integral.
Solution
The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type
II. Refer to Figure 5.21.
Figure 5.21 Converting a region from Type I to Type II.
We can see from the limits of integration that the region is bounded above by y = 2 − x2 and below by y = 0,
where x is in the interval ⎡
⎣0, 2⎤
⎦. By reversing the order, we have the region bounded on the left by x = 0 and
on the right by x = 2 − y where y is in the interval [0, 2]. We solved y = 2 − x2 in terms of x to obtain
x = 2 − y.
Hence
∫
0
2
∫
0
2 − x2
xex2
dy dx = ∫
0
2
∫
0
2 − y
xex2
dx dy
Reverse the order of
integration then use
substitution.
= ∫
0
2 ⎡
⎣
⎢1
2ex2|0
2 − y⎤
⎦
⎥dy = ∫
0
2
1
2
⎛
⎝e
2 − y − 1⎞
⎠dy = −1
2
⎛
⎝e
2 − y + y⎞
⎠|02
= 1
2
⎛
⎝e2 − 3⎞
⎠.
Example 5.16
Evaluating an Iterated Integral by Reversing the Order of Integration
Consider the iterated integral ∬
R
f (x, y)dx dy where z = f (x, y) = x − 2y over a triangular region R that has
sides on x = 0, y = 0, and the line x + y = 1. Sketch the region, and then evaluate the iterated integral by
Chapter 5 | Multiple Integration 509
5.11
a. integrating first with respect to y and then
b. integrating first with respect to x.
Solution
A sketch of the region appears in Figure 5.22.
Figure 5.22 A triangular region R for integrating in two
ways.
We can complete this integration in two different ways.
a. One way to look at it is by first integrating y from y = 0 to y = 1 − x vertically and then integrating x
from x = 0 to x = 1:
∬
R
f (x, y)dx dy = ∫
x = 0
x = 1
∫
y = 0
y = 1 − x
⎛
⎝x − 2y⎞
⎠dy dx = ∫
x = 0
x = 1
⎡
⎣xy − 2y2⎤
⎦y = 0
y = 1 − x
dx
= ∫
x = 0
x = 1
⎡
⎣x(1 − x) − (1 − x)2⎤
⎦dx = ∫
x = 0
x = 1
⎡
⎣−1 + 3x − 2x2⎤
⎦dx = ⎡
⎣−x + 3
2x2 − 2
3x3⎤
⎦x = 0
x = 1
= − 1
6.
b. The other way to do this problem is by first integrating x from x = 0 to x = 1 − y horizontally and then
integrating y from y = 0 to y = 1:
∬
R
f (x, y)dx dy = ∫
y = 0
y = 1
∫
x = 0
x = 1 − y
⎛
⎝x − 2y⎞
⎠dx dy = ∫
y = 0
y = 1
⎡
⎣
1
2x2 − 2xy⎤
⎦x = 0
x = 1 − y
dy
= ∫
y = 0
y = 1
⎡
⎣
1
2
⎛
⎝1 − y⎞
⎠
2 − 2y⎛
⎝1 − y⎞
⎠
⎤
⎦dy = ∫
y = 0
y = 1
⎡
⎣
1
2 − 3y + 5
2y2⎤
⎦dy
= ⎡
⎣
1
2y − 3
2y2 + 5
6y3⎤
⎦y = 0
y = 1
= − 1
6.
Evaluate the iterated integral ∬
D
⎛
⎝x2 + y2⎞
⎠dA over the region D in the first quadrant between the
functions y = 2x and y = x2. Evaluate the iterated integral by integrating first with respect to y and then
integrating first with resect to x.
Calculating Volumes, Areas, and Average Values
We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same
as those in Double Integrals over Rectangular Regions, but without the restriction to a rectangular region, we can
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now solve a wider variety of problems.
Example 5.17
Finding the Volume of a Tetrahedron
Find the volume of the solid bounded by the planes x = 0, y = 0, z = 0, and 2x + 3y + z = 6.
Solution
The solid is a tetrahedron with the base on the xy -plane and a height z = 6 − 2x − 3y. The base is the region D
bounded by the lines, x = 0, y = 0 and 2x + 3y = 6 where z = 0 (Figure 5.23). Note that we can consider
the region D as Type I or as Type II, and we can integrate in both ways.
Figure 5.23 A tetrahedron consisting of the three coordinate planes and the plane z = 6 − 2x − 3y, with
the base bound by x = 0, y = 0, and 2x + 3y = 6.
First, consider D as a Type I region, and hence D =
⎧
⎩
⎨(x, y)|0 ≤ x ≤ 3, 0 ≤ y ≤ 2 − 2
3x
⎫
⎭
⎬.
Therefore, the volume is
V = ∫
x = 0
x = 3
∫
y = 0
y = 2 − (2x/3)
(6 − 2x − 3y)dy dx = ∫
x = 0
x = 3⎡
⎣
⎢⎛
⎝6y − 2xy − 3
2y2⎞
⎠|y = 0
y = 2 − (2x/3)⎤
⎦
⎥dx
= ∫
x = 0
x = 3
⎡
⎣
2
3(x − 3)2⎤
⎦dx = 6.
Now consider D as a Type II region, so D =
⎧
⎩
⎨(x, y)|0 ≤ y ≤ 2, 0 ≤ x ≤ 3 − 3
2y
⎫
⎭
⎬. In this calculation, the
volume is
Chapter 5 | Multiple Integration 511
5.12
V = ∫
y = 0
y = 2
∫
x = 0
x = 3 − (3y/2)
(6 − 2x − 3y)dx dy = ∫
y = 0
y = 2⎡
⎣
⎢⎛
⎝6x − x2 − 3xy⎞
⎠|x = 0
x = 3 − (3y/2)⎤
⎦
⎥dy
= ∫
y = 0
y = 2
⎡
⎣
9
4(y − 2)2⎤
⎦dy = 6.
Therefore, the volume is 6 cubic units.
Find the volume of the solid bounded above by f (x, y) = 10 − 2x + y over the region enclosed by the
curves y = 0 and y = ex, where x is in the interval [0, 1].
Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have
seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the
area of a general nonrectangular region, as stated in the next definition.
Definition
The area of a plane-bounded region D is defined as the double integral ∬
D
1dA.
We have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by
using double integrals, which can be very useful, as we will see in the later sections of this chapter.
Example 5.18
Finding the Area of a Region
Find the area of the region bounded below by the curve y = x2 and above by the line y = 2x in the first quadrant
(Figure 5.24).
Figure 5.24 The region bounded by y = x2 and y = 2x.
Solution
512 Chapter 5 | Multiple Integration
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