Is whose bases are bounded by thex2 - 1, with the indicated cross

:o the x-axis.(b) Rectangles of height 1

1 2"~\x

:Is whose bases are bounded by the~e indicated cross sections taken

(b) Equilateral triangles

(d) Isosceles right triangles

of intersection (the solid common:cular cylinders of radius r whosee figure).

Solid of intersection

sections (taKen perpenmcmar to me y-axis): (a) squares,(b) semicircles, (c) equilateral triangles, and (d) semiellipseswhose heights are twice the lengths of their bases.

75. A manufacturer drills a hole through the center of a metalsphere of radius R. The hole has a radius r. Find the volume ofthe resulting ring.

76. For the metal sphere in Exercise 75, let R = 6. What value of rwill produce a ring whose volume is exactly half the volume ofthe sphere?

The region bounded by the graphs of y = 8x/(9 + x2), y = 0,x = 0, and x = 5 is revolved about the x-axis. Use a graphingutility and Simpson’s Rule (with n = 10) to approximate thevolume of the solid.

78. The solid shown in the figure has cross sections bounded by thegraph of ]xl~’ + ]yl~’ = 1, where 1 _< a -< 2.

(a) Describe the cross section when a = 1 and a = 2.

(b) Describe a procedure for approximating the volume of thesolid.

]xf + 13,l~ = 1 Ix[2 + 13,12= 1

79. Two planes cut a right circular cylinder to form a wedge. Oneplane is perpendicular to the axis of the cylinder and the secondmakes an angle of 0 degrees with the first (see figure).

(a) Find the volume of the wedge if 0 = 45°.

(b) Find the volume of the wedge for an arbitrary angle 0.Assuming that the cylinder has sufficient length, how doesthe volume of the wedge change as 0 increases from 0°

to 90°?Y

Figure for 79 Figure for 80

80. (a) Show that the volume of the torus shown in the figure isgiven by the integral 8~rRf~v/~_y2dy, whereR>r>O.

(b) Find the volume of the torus.

h

P +w2

Axis of revolution

Figure 7.27

In this section, you will study an alternative method for finding the volume of a solidof revolution. This method is called the shell method because it uses cylindricalshells. A comparison of the advantages of the disk and shell methods is given later inthis section.

To begin, consider a representative rectangle as shown in Figure 7.27, where w isthe width of the rectangle, h is the height of the rectangle, and p is the distancebetween the axis of revolution and the center of the rectangle. When this rectangle isrevolved about its axis of revolution, it forms a cylindrical shell (or tube) of thicknessw. To find the volume of this shell, consider two cylinders. The radius of the largercylinder corresponds to the outer radius of the shell, and the radius of the smallercylinder corresponds to the inner radius of the shell. Because p is the average radiusof the shell, you know the outer radius is p + (w/2) and the inner radius is p - (w/2).

Outer

So, the volume of the shell is

Volume of shell = (volume of cylinder) - (volume of hole)

= 2rrphw= 2 rr (average radius) (height) (thickness).

/~ (y)

Plane region

Figure 7.28

Axis ofrevolution

Solid of revolution

You can use this formula to find the volume of a solid of revolution. Assume thatthe plane region in Figure 7.28 is revolved about a line to form the indicated solid. Ifyou consider a horizontal rectangle of width Ay, then, as the plane region is revolvedabout a line parallel to the x-axis, the rectangle generates a representative shell whosevolume is

AV = 2rr[p(y)h(y)] Ay.

You can approximate the volume of the solid by n such shells of thickness Ay, heighth(yi), and average radius p(yi).

Volume of solid ~ ~ 2yr[p(yi)h(Yi)]Ay = 2rr~ [p(yi)h(yi)] Ay

i=1 i=1

This approximation appears to become better and better as IIAII -+ 0oo)o So,the volume of the solid is

Volume of solid =ItAIl-+olim 2rr~.~l[p(y,)h(y,)]Ay.:

= 2rr £a[p(y)h(y)] dy.

(~, o)

Volume = V = 2"n Jc p(y)h(y) dy Volume = V = 2"rr Ja p(x)h(x) dx

h(y)

Horizontal axis of revolutionFigure 7.29

p(y)

~Ax

p(x)

Vertical axis of revolution

h(x)

EXAMPLE ~’~ Using the Shell Method to Find Volume

Find the volume of the solid of revolution formed by revolving the region bounded by

y=x--x3

and the x-axis (0 _< x _< 1) about the y-axis.

Solution Because the axis of revolution is vertical, use a vertical representativerectangle, as shown in Figure 7.30. The width Ax indicates that x is the variable ofintegration. The distance from the center of the rectangle to the axis of revolution isp(x) = x, and the height of the rectangle is

h(x) = x - x3.

Because x ranges from 0 to 1, the volume of the solid is

x(x - x3) dxV = 2rr p(x)h(x) dx = 2rr

= 2’rr £1

Apply ,daelI method.

(--X4 -{- X2) dx Simpli(.v.

= 20r -~-+ 3Jo

1+

4vr15

1 i e-y2

Ay

p(y) = 3’~ h(Y) = e-Y-____

rAe;ioSlu°ifio n

Figure 7.31

Y

c

Vertical axisof revolution

Disk method: Representative rectangle isperpendicular to the axis of revolution.Figure 7.32

S01uti0n Because the axis of revolution is horizontal, use a horizontal representativerectangle, as shown in Figure 7.31. The width Ay indicates that y is the variable ofintegration. The distance from the center of the rectangle to the axis of revolution isp(y) = y, and the height of the rectangle is h(y) = e-y2. Because y ranges from 0 to1, the volume of the solid is

V = 2~ p(y)h(y) dy = 2vr ye-Y2dy .\ppb’ ,,hell melhod.

=--rr[e-y2]

1.986.

Integrate.

To see the advantage of using the shell method in Example 2, solve the equationfor 3’.

=L~’l~-ln 0 _< x _< 1/eYx, 1/e<x<_ 1

Then use this equation to find the volume using the disk method.

Comparison of Disk and Shell MethodsThe disk and shell methods can be distinguished as follows. For the disk method, therepresentative rectangle is always perpendicular to the axis of revolution, whereasfor the shell method, the representative rectangle is always parallel to the axis ofrevolution, as shown in Figure 7.32.

V = 2rC j ~c Ph (c~ "~

Horizontal axisof revolution

Vertical axisof revolution

Shell method: Representative rectangle isparallel to the axis of revolution.

Horizontal axisof revolution

y=x2 + 1, y= O, x= O, and x= 1

about the y-axis.

Solution In Example 4 in the preceding section, you saw that the washer methodrequires two integrals to determine the volume of this solid. See Figure 7.33(a).

V = ~r (12 - 02) dy ÷ rr [12 __ (~)2] dy i\pp!.x \\asher method.

£1 £2= ~ 1 dy + ~ (2 - y)dy Nimplit5.

= ~ y + ~ 2y- Integrate,0 2Jl ~

3~2

In Figure 7.33(b), you can see that the shell method requires only one integral to findthe volume.

b

V = 2rr p(x)h(x) dx

£1= 2rr Y(X2 + 1)&

2/0

3-n-2

Suppose the region in Example 3 were revolved about the vertical line x -- 1.Would the resulting solid of revolution have a greater volume or a smaller volume thanthe solid in Example 3? Without integrating, you should be able to reason that theresulting solid would have a smaller volume because "more" of the revolved regionwould be closer to the axis of revolution. To confirm this, try solving the followingintegral, which gives the volume of the solid.

£1V = 2rr (1 - x)(x2 + 1) dx /,,.,-, -- I

FOR FURTHER INFORMATION To learn more about the disk and shell methods, see thearticle "The Disk and Shell Method" by Charles A. Cable in The American MathematicalMonthly. To view this article, go to the website wwmmatharticles.com. ~!

Y

t ,’(x) = 0x2

3 ~- R(x) = 1

-4-3 -2 -1 I 1 2 3 4

(a) Disk method

3’

~ p(y) = y

-4 -3 -2 -1 I 1 2 3 4

(b) Shell methodFigure 7.35

Axis ofrevolution

p(x) = 2 - x ,It(x) = x3 + x + 1 - 1 ’

’, I1 2

Figure 7.36

about the x-axis, where x and 3’ are measured in feet. Find the volume of the pontoon.

Solution Refer to Figure 7.35(a) and use the disk method as follows.

V= 7r 1 - 16,/ dx

= "r; 1---~-k~ dx4x-- ~ + 1280j_4

64~: ~ ~ 13.4 cubic feet15

Apply dixk incthod.

Simi,lif\.

Try using Figure 7.35(b) to set up the integral for the volume using the shell method.Does the integral seem more complicated? ¯

To use the shell method in Example 4, you would have to solve for x in terms ofy in the equation

y-- 1,- (X2/16).

Sometimes, solving for x is very difficult (or even impossible). In such cases youmust use a vertical rectangle (of width Ax), thus making x the variable of integration.The position (horizontal or vertical) of the axis of revolution then determines themethod to be used. This is shown in Example 5.

EXAMPLE ~ Shell Method Necessary

Find the volume of the solid formed by revolving the region bounded by the graphs ofy = x~ + x + 1, y = 1, andx = 1 about the linex = 2, as shown in Figure 7.36.

Solution In the equation 3’ = x~ + x + 1, you cannot easily solve for x in terms ofy. (See Section 3.8 on Newton’s Method.) Therefore, the variable of integration mustbe x, and you should choose a vertical representative rectangle. Because the rectangleis parallel to the axis of revolution, use the shell method and obtain

V= 2rr p(x)h(x)dx= 2rr (2-x)(x3 ÷x÷ 1 - 1) dx .~pp!>,l~cllmetl>d.

f01= 27r (--X4 ÷ 2X3 -- X2 ÷ 2x) dx Simplil5.

2 7"/" [X5 X4 X3

k

27r(_ 1 1 1~÷~--~÷ 1)

29’n-15