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(x + 2)(x ’ 5)
dx
(b)
(x + 1)(x 2 + 1)
dx
(c)
x 3 ’ 2x 2 ’ 5x + 6
x dx
(d)
x4 ’ 1
dx
(e)
x 3 ’ x 2 ’ 8x + 12
x+1
(f)
x3 ’ x2 + x ’ 1
3. Use the method of u-substitution to evaluate each of the following inde¬nite
integrals.

(1 + sin2 x)2 2 sin x cos x dx
(a)

sin x

(b) dx
x
cos(ln x) sin(ln x)
(c) dx
x

etan x sec2 x dx
(d)

sin x
(e) dx
1 + cos2 x
sec2 x
(f) dx
1 ’ tan2 x
4. Evaluate each of the following inde¬nite trigonometric integrals.

sin x cos2 x dx
(a)

sin3 x cos2 x dx
(b)

tan3 x sec2 x dx
(c)
216 CHAPTER 7 Methods of Integration

tan x sec3 x dx
(d)

sin2 x cos2 x dx
(e)

sin x cos4 x dx
(f)

5. Calculate each of the following de¬nite integrals.
1
ex sin x dx
(a)
0
e
x 2 ln x dx
(b)
1
(2x + 1) dx
4
(c)
x3 + x2
2
π
sin2 x cos2 x dx
(d)
0
π/3
(e) tan x sec x dx
π/4
π/4 tan x
(f) dx
cos x
0
CHAPTER 8



Applications of
the Integral
8.1 Volumes by Slicing
8.1.0 INTRODUCTION
When we learned the theory of the integral, we found that the basic idea was that
one can calculate the area of an irregularly shaped region by subdividing the region
into “rectangles.” We put the word “rectangle” here in quotation marks because the
region is not literally broken up into rectangles; the union of the rectangles differs
from the actual region under consideration by some small errors (see Fig. 8.1). But
the contribution made by these errors vanishes as the mesh of the rectangles become
¬ner and ¬ner.
We will now implement this same philosophy to calculate certain volumes. Some
of these will be volumes that you have heard about (e.g., the sphere or cone), but
have never known why the volume had the value that it had. Others will be entirely
new (e.g., the paraboloid of revolution). We will again use the method of slicing.



8.1.1 THE BASIC STRATEGY
Imagine a solid object situated as in Fig. 8.2. Observe the axes in the diagram, and
imagine that we slice the ¬gure with slices that are vertical (i.e., that rise out of
the x-y plane) and that are perpendicular to the x-axis (and parallel to the y-axis).
Look at Fig. 8.3. Notice, in the ¬gure, that the ¬gure extends from x = a to x = b.

217
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
218 CHAPTER 8 Applications of the Integral


y = f (x)




a b




Fig. 8.1




Y
FL
AM

Fig. 8.2
TE




Fig. 8.3


If we can express the area of the slice at position x as a function A(x) of x,
then (see Fig. 8.4) the volume of a slice of thickness x at position x will be about
A(x) · x. If P = {x0 , x1 , . . . , xk } is a partition of the interval [a, b] then the
volume of the original solid object will be about

V= A(xj ) · x.
j
219
CHAPTER 8 Applications of the Integral




Fig. 8.4

As the mesh of the partition becomes ¬ner and ¬ner, this (Riemann) sum will tend
to the integral
b
A(x) dx.
a
We declare the value of this integral to be the volume V of the solid object.

8.1.2 EXAMPLES
EXAMPLE 8.1
Calculate the volume of the right circular cone with base a disc of radius 3
and height 6.

SOLUTION
Examine Fig. 8.5. We have laid the cone on its side, so that it extends from x =
0 to x = 6. The upper edge of the ¬gure is the line y = 3 ’ x/2. At position x,

y



3




x
6




Fig. 8.5
220 CHAPTER 8 Applications of the Integral

the height of the upper edge is 3 ’ x/2, and that number is also the radius of
the circular slice at position x (Fig. 8.6). Thus the area of that slice is
2
x
A(x) = π 3 ’ .
2




thickness ,x
_
x/2
3




Fig. 8.6

We ¬nd then that the volume we seek is
6
2(3 ’ x/2)3
6 6 2
x
V= A(x) dx = π 3’ dx = ’π = 18π.
2 3
0 0 0

You Try It: Any book of tables (see [CRC]) will tell you that the volume of a right
circular cone of base radius r and height h is 1 π r 2 h. This formula is consistent
3
with the result that we obtained in the last example for r = 3 and h = 6. Use the
technique of Example 8.1 to verify this more general formula.
EXAMPLE 8.2
A solid has base the unit disk in the x-y plane. The vertical cross section at
position x is an equilateral triangle. Calculate the volume.

SOLUTION
Examine Fig. 8.7. The unit circle has equation x 2 +y 2 = 1. For our purposes,
this is more conveniently written as

y = ± 1 ’ x 2. ()
Thus the endpoints of the base of the equilateral triangle at position x are the

points (x, ± 1 ’ x 2 ). In other words, the base of this triangle is

b = 2 1 ’ x 2.
Examine Fig. 8.8. We see that an equilateral triangle of side b has height
√ √2
3b/2. Thus the area of the triangle is 3b /4. In our case then, the equilateral
221
CHAPTER 8 Applications of the Integral




Fig. 8.7




√3
b b b
2




b

Fig. 8.8

triangular slice at position x has area


3 2
A(x) = · 2 1 ’ x 2 = 3(1 ’ x 2 ).
4
Finally, we may conclude that the volume we seek is
1
V= A(x) dx
’1
1√
= 3(1 ’ x 2 ) dx
’1
√ 1
x3
= 3 x’
3 ’1
√ ’1
1
= 3 1’ ’ (’1) ’
3 3

43
= .
3
222 CHAPTER 8 Applications of the Integral

EXAMPLE 8.3
A solid has base in the x-y plane consisting of a unit square with center at
the origin and vertices on the axes. The vertical cross-section at position x
is itself a square. Refer to Fig. 8.9. What is the volume of this solid?




Fig. 8.9

SOLUTION
It is suf¬cient to calculate the volume of the right half of this solid, and to

double the answer. Of course the extent of x is then 0 ¤√ ¤ 1/ 2. At position
x
x, the height of the upper edge√ the square base is 1/ 2 ’ x. So the base of
of
the vertical square slice is 2(1/ 2 ’ x) (Fig. 8.10). The area of the slice is then
√ √
2 2
A(x) = 2 1/ 2 ’ x = 2 ’ 2x .




2 (1/√2 _ x)




2(1/√2 _ x)

Fig. 8.10

It follows that

1/ 2
V =2· A(x) dx
0


1/ 2
2
=2 2 ’ 2x dx
0
223
CHAPTER 8 Applications of the Integral

√ 1/ 2
3
2 ’ 2x
=2 ’
6
0

03 22
=2 ’ ’ ’
6 6

22
= .
3

You Try It: Calculate the volume of the solid with base in the plane an equilateral
triangle of side 1, with base on the x-axis, and with vertical cross-section parallel
to the y-axis consisting of an equilateral triangle.

EXAMPLE 8.4
Calculate the volume inside a sphere of radius 1.

SOLUTION
It is convenient for us to think of the sphere as centered at the origin in the
x-y plane. Thus (Fig. 8.11) the slice at position x, ’1 ¤ x ¤ 1, is a disk. Since
we are working with base the unit circle, we may calculate ( just as in Example
√ √
8.2) that the diameter of this disk is 2 1 ’ x 2 . Thus the radius is 1 ’ x 2 and
the area is

2
A(x) = π · 1 ’ x2 = π · (1 ’ x 2 ).




Fig. 8.11

In conclusion, the volume we seek is

1
V= π(1 ’ x 2 ) dx.
’1
224 CHAPTER 8 Applications of the Integral

We easily evaluate this integral as follows:
1
x3
V =π· x’
3 ’1
’1
1
=π · 1’ ’ ’1 ’
3 3
4
= π.
3
You Try It: Any book of tables (see [CRC]) will tell you that the volume inside a
sphere of radius r is 4π r 3 /3. This formula is consistent with the answer we obtained
in the last example for r = 1. Use the method of this section to derive this more
general formula for arbitrary r.



8.2 Volumes of Solids of Revolution
8.2.0 INTRODUCTION
A useful way”and one that we encounter frequently in everyday life”for generat-

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