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ing solids is by revolving a planar region about an axis. For example, we can think
of a ball (the interior of a sphere) as the solid obtained by rotating a disk about
an axis (Fig. 8.12). We can think of a cylinder as the solid obtained by rotating
a rectangle about an adjacent axis (Fig. 8.13). We can think of a tubular solid as
obtained by rotating a rectangle around a non-adjacent axis (Fig. 8.14).




Fig. 8.12




Fig. 8.13
225
CHAPTER 8 Applications of the Integral




Fig. 8.14

There are two main methods for calculating volumes of solids of revolution:
the method of washers and the method of cylinders. The ¬rst of these is really an
instance of volume by slicing, just as we saw in the last section. The second uses a
different geometry; instead of slices one uses cylindrical shells. We shall develop
both technologies by way of some examples.

8.2.1 THE METHOD OF WASHERS
EXAMPLE 8.5
A solid is formed by rotating the triangle with vertices (0, 0), (2, 0), and (1, 1)
about the x-axis. See Fig. 8.15. What is the resulting volume?




(1,1)




(0,0) (2,0)


Fig. 8.15


SOLUTION
For 0 ¤ x ¤ 1, the upper edge of the triangle has equation y = x. Thus
the segment being rotated extends from (x, 0) to (x, x). Under rotation, it will
generate a disk of radius x, and hence area A(x) = π x 2 . Thus the volume
generated over the segment 0 ¤ x ¤ 1 is
1
V1 = π x 2 dx.
0
Similarly, for 1 ¤ x ¤ 2, the upper edge of the triangle has equation y =
2 ’ x. Thus the segment being rotated extends from (x, 0) to (x, 2 ’ x). Under
226 CHAPTER 8 Applications of the Integral

rotation, it will generate a disk of radius 2’x, and hence area A(x) = π(2’x)2 .
Thus the volume generated over the segment 1 ¤ x ¤ 2 is
2
V2 = π(2 ’ x)2 dx.
1

In summary, the total volume of our solid of revolution is

V = V1 + V2
1 2
’(2 ’ x)3
x3
=π +
3 3
0 1
1 1
=π ’ 0 + ’0 ’ ’
3 3

= .
3
EXAMPLE 8.6
The portion of the curve y = x 2 between x = 1 and x = 4 is rotated about
the x-axis (Fig. 8.16). What volume does the resulting surface enclose?




Fig. 8.16


SOLUTION
At position x, the curve is x 2 units above the x-axis. The point (x, x 2 ), under
rotation, therefore generates a circle of radius x 2 . The disk that the circle bounds
227
CHAPTER 8 Applications of the Integral

has area A(x) = π · (x 2 )2 . Thus the described volume is
4
4 x5 1023π
V= π · x dx = π · =
4
.
5 5
1 1

Math Note: The reasoning we have used in the last two examples shows this:
If the curve y = f (x), a ¤ x ¤ b, is rotated about the x-axis then the volume
enclosed by the resulting surface is
b
V= π · [f (x)]2 dx.
a

You Try It: Calculate the volume enclosed by the surface obtained by rotating

the curve y = x + 1, 4 ¤ x ¤ 9, about the x-axis.
EXAMPLE 8.7
The curve y = x 3 , 0 ¤ x ¤ 3, is rotated about the y-axis. What volume
does the resulting surface enclose?

SOLUTION
It is convenient in this problem to treat y as the independent variable and x as
the dependent variable. So we write the curve as x = y 1/3 . Then, at position y,
the curve is distance y 1/3 from the axis so the disk generated under rotation will
have radius y 1/3 (Fig. 8.17). Thus the disk will have area A(y) = π · [y 1/3 ]2 .
Also, since x ranges from 0 to 3 we see that y ranges from 0 to 27. As a result,




Fig. 8.17
228 CHAPTER 8 Applications of the Integral

the volume enclosed is
27
27 y 5/3 729π
V= π ·y dy = π · =
2/3
.
5/3 5
0 0

Math Note: The reasoning we have used in the last example shows this: If the
curve x = g(y), c ¤ y ¤ d, is rotated about the y-axis then the volume enclosed
by the resulting surface is
d
V= π · [g(y)]2 dy.
c

You Try It: Calculate the volume enclosed when the curve y = x 1/3 , 32 ¤ x ¤
243, is rotated about the y-axis.
EXAMPLE 8.8
Set up, but do not evaluate, the integral that represents the volume gener-
ated when the planar region between y = x 2 + 1 and y = 2x + 4 is rotated


Y
about the x-axis.
FL
SOLUTION
When the planar region is rotated about the x-axis, it will generate a donut-
AM

shaped solid. Notice that the curves intersect at x = ’1 and x = 3; hence
the intersection lies over the interval [’1, 3]. For each x in that interval, the
segment connecting (x, x 2 + 1) to (x, 2x + 4) will be rotated about the x-axis.
It will generate a washer. See Fig. 8.18. The area of that washer is
TE



A(x) = π · [2x + 4]2 ’ π · [x 2 + 1].
[Notice that we calculate the area of a washer by subtracting the areas of two
circles”not by subtracting the radii and then squaring.]
It follows that the volume of the solid generated is
3
V= π · [2x + 4]2 ’ π · [x 2 + 1] dx.
’1


8.2.2 THE METHOD OF CYLINDRICAL SHELLS
Our philosophy will now change. When we divide our region up into vertical strips,
we will now rotate each strip about the y-axis instead of the x-axis. Thus, instead
of generating a disk with each strip, we will now generate a cylinder.
Look at Fig. 8.19. When a strip of height h and thickness x, with distance r
from the y-axis, is rotated about the y-axis, the resulting cylinder has surface area
2πr · h and volume about 2π r · h · x. This is the expression that we will treat
in order to sum up the volumes of the cylinders.
229
CHAPTER 8 Applications of the Integral




Fig. 8.18

y




h


r x

Fig. 8.19

EXAMPLE 8.9
Use the method of cylindrical shells to calculate the volume of the solid
enclosed when the curve y = x 2 , 1 ¤ x ¤ 3, is rotated about the y-axis.
SOLUTION
As usual, we think of the region under y = x 2 and above the x-axis as
composed of vertical segments or strips. The segment at position x has height
x 2 . Thus, in this instance, h = x 2 , r = x, and the volume of the cylinder is
2π x · x 2 · x. As a result, the requested volume is
3
V= 2π x · x 2 dx.
1
We easily calculate this to equal
3
3 x4 34 14
V = 2π · x dx = 2π = 2π ’ = 40π.
3
4 4 4
1 1
230 CHAPTER 8 Applications of the Integral

EXAMPLE 8.10
Use the method of cylindrical shells to calculate the volume enclosed when
the curve y = x 2 , 0 ¤ x ¤ 3, is rotated about the x-axis (Fig. 8.20).


y




x




Fig. 8.20


SOLUTION
We reverse, in our analysis, the roles of the x- and y-axes. Of course y
ranges from 0 to 9. For each position y in that range, there is a segment stretch-
√ √
ing from x = y to x = 3. Thus it has length 3 ’ y. Then the cylinder
generated when this segment (thickened to a strip of width y) is rotated
about the x-axis has volume

V (y) = 2πy · 3 ’ y y.

The aggregate volume is then


9
V= 2πy · 3 ’ y dy
0
9
= 2π · 3y ’ y 3/2 dy
0
9
3y 2 y 5/2
= 2π · ’ dy
2 5/2 0
231
CHAPTER 8 Applications of the Integral

243 2 · 243 00
= 2π · ’ ’ ’
2 5 25
243
= 2π ·
10
243π
= .
5

You Try It: Use the method of cylindrical shells to calculate the volume enclosed
when the region 0 ¤ y ¤ sin x, 0 ¤ x ¤ π/2, is rotated about the y-axis.


8.2.3 DIFFERENT AXES
Sometimes it is convenient to rotate a curve about some line other than the
coordinate axes. We now provide a couple of examples of that type of problem.

EXAMPLE 8.11
Use the method of washers to calculate the volume of the solid enclosed

when the curve y = x, 1 ¤ x ¤ 4, is rotated about the line y = ’1. See
Fig. 8.21.


y




y = √x


x




Fig. 8.21


SOLUTION
√ The key is to notice that, at position x, the segment to be rotated has height

x ’ (’1)”the distance from the point (x, x) on the curve to the line

y = ’1. Thus the disk generated has area A(x) = π · ( x + 1)2 . The resulting
232 CHAPTER 8 Applications of the Integral

aggregate volume is

4
2
V= π· x + 1 dx
1

4
=π x + 2 x + 1 dx
1
4
x 2 2x 3/2
=π + +x
2 3/2 1
2·8 12 2 · 1
42
=π· + +4 ’π · + +1
2 3/2 2 3/2
119
= π.
6

You Try It: Calculate the volume inside the surface generated when y = x +x
3

is rotated about the line y = ’3, 1 ¤ x ¤ 4.
EXAMPLE 8.12
Calculate the volume of the solid enclosed when the area between the
curves x = (y ’ 2)2 + 1 and x = ’(y ’ 2)2 + 9 is rotated about the line
y = ’2.

y




x = (y _ 2)2 + 1 x = _(y _ 2)2 + 9




x


Fig. 8.22


SOLUTION
Solving the equations simultaneously, we ¬nd that the points of intersec-
tion are (5, 0) and (5, 4). The region between the two curves is illustrated in
Fig. 8.22.
At height y, the horizontal segment that is to be rotated stretches from
((y ’ 2)2 + 1, y) to (’(y ’ 2)2 + 9, y). Thus the cylindrical shell that is
233
CHAPTER 8 Applications of the Integral

generated has radius y ’ 2, height 8 ’ 2(y ’ 2)2 , and thickness y. It therefore
generates the element of volume given by
2π · (y ’ 2) · [8 ’ 2(y ’ 2)2 ] · y.
The aggregate volume that we seek is therefore
4
V= 2π · (y ’ 2) · [8 ’ 2(y ’ 2)2 ] dy
0
4
= 16π(y ’ 2) ’ 4π(y ’ 2)3 dy
0
4
= 8π(y ’ 2)2 ’ π(y ’ 4)4 0
= 256π.
You Try It: Calculate the volume enclosed when the curve y = cos x is rotated
about the line y = 4, π ¤ x ¤ 3π .



8.3 Work
One of the basic principles of physics is that work performed is force times distance:
If you apply force F pounds in moving an object d feet, then the work is
W =F ·d foot-pounds.
The problem becomes more interesting (than simple arithmetic) if the force is
varying from point to point. We now consider some problems of that type.
EXAMPLE 8.13
A weight is pushed in the plane from x = 0 to x = 10. Because of a
prevailing wind, the force that must be applied at point x is F (x) = 3x 2 ’
x + 10 foot-pounds. What is the total work performed?




0 10

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