Volume__The_Shell_Method - Is whose bases are bounded by the x2 1 with the indicated cross:o the x-axis(b Rectangles of height 1 1 2\"~ x:Is whose bases

# Volume__The_Shell_Method - Is whose bases are bounded by...

• Notes
• 3

This preview shows page 1 out of 3 pages. Unformatted text preview: Is whose bases are bounded by the x2 - 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 sections (taKen perpenmcmar to me y-axis): (a) squares, (b) semicircles, (c) equilateral triangles, and (d) semiellipses whose heights are twice the lengths of their bases. 75. A manufacturer drills a hole through the center of a metal sphere of radius R. The hole has a radius r. Find the volume of the resulting ring. 76. For the metal sphere in Exercise 75, let R = 6. What value of r will produce a ring whose volume is exactly half the volume of the 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 graphing utility and Simpson’s Rule (with n = 10) to approximate the volume of the solid. 78. The solid shown in the figure has cross sections bounded by the graph 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 the solid. h P +w2 Axis of revolution Figure 7.27 Outer So, the volume of the shell is (b) Equilateral triangles Volume of shell = (volume of cylinder) - (volume of hole) = 2rrphw = 2 rr (average radius) (height) (thickness). Ix[2 + 13,12= 1 xf + 13,l~ = 1 (d) Isosceles right triangles In this section, you will study an alternative method for finding the volume of a solid of revolution. This method is called the shell method because it uses cylindrical shells. A comparison of the advantages of the disk and shell methods is given later in this section. To begin, consider a representative rectangle as shown in Figure 7.27, where w is the width of the rectangle, h is the height of the rectangle, and p is the distance between the axis of revolution and the center of the rectangle. When this rectangle is revolved about its axis of revolution, it forms a cylindrical shell (or tube) of thickness w. To find the volume of this shell, consider two cylinders. The radius of the larger cylinder corresponds to the outer radius of the shell, and the radius of the smaller cylinder corresponds to the inner radius of the shell. Because p is the average radius of the shell, you know the outer radius is p + (w/2) and the inner radius is p - (w/2). 79. Two planes cut a right circular cylinder to form a wedge. One plane is perpendicular to the axis of the cylinder and the second makes 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 does the volume of the wedge change as 0 increases from 0° to 90°? You can use this formula to find the volume of a solid of revolution. Assume that the plane region in Figure 7.28 is revolved about a line to form the indicated solid. If you consider a horizontal rectangle of width Ay, then, as the plane region is revolved about a line parallel to the x-axis, the rectangle generates a representative shell whose volume is /~ (y) AV = 2rr[p(y)h(y)] Ay. Y Plane region of intersection (the solid common :cular cylinders of radius r whose e figure). You can approximate the volume of the solid by n such shells of thickness Ay, height h(yi), and average radius p(yi). Axis of revolution 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 -+ 0 the volume of the solid is Figure for 79 Solid of intersection Figure for 80 80. (a) Show that the volume of the torus shown in the figure is given by the integral 8~rRf~v/~_y2dy, where R>r>O. (b) Find the volume of the torus. Volume of solid =ItAIl-+olim 2rr~.~l[p(y,)h(y,)]Ay.: Solid of revolution Figure 7.28 = 2rr £a[p(y)h(y)] dy. oo)o So, Volume = V = 2"n Jc p(y)h(y) dy Volume = V = 2"rr Ja p(x)h(x) dx S01uti0n Because the axis of revolution is horizontal, use a horizontal representative rectangle, as shown in Figure 7.31. The width Ay indicates that y is the variable of integration. The distance from the center of the rectangle to the axis of revolution is p(y) = y, and the height of the rectangle is h(y) = e-y2. Because y ranges from 0 to 1, the volume of the solid is h(y) ~Ax 1 i e-y2 Ay h(x) V = 2~ p(y)h(y) dy = 2vr ye-Y2dy p(y) = 3’~ h(Y) = e-Y-____ =--rr[e-y2] p(y) .\ppb’ ,,hell melhod. Integrate. rAe;ioSlu°ifio n p(x) Horizontal axis of revolution Figure 7.29 1.986. Figure 7.31 Vertical axis of revolution To see the advantage of using the shell method in Example 2, solve the equation for 3’. EXAMPLE ~’~ Using the Shell Method to Find Volume Y =L~’l~-ln 0 _< x _< 1/e x, 1/e<x<_ 1 Find the volume of the solid of revolution formed by revolving the region bounded by Then use this equation to find the volume using the disk method. y=x--x3 and the x-axis (0 _< x _< 1) about the y-axis. Comparison of Disk and Shell Methods Solution Because the axis of revolution is vertical, use a vertical representative rectangle, as shown in Figure 7.30. The width Ax indicates that x is the variable of integration. The distance from the center of the rectangle to the axis of revolution is p(x) = x, and the height of the rectangle is The disk and shell methods can be distinguished as follows. For the disk method, the representative rectangle is always perpendicular to the axis of revolution, whereas for the shell method, the representative rectangle is always parallel to the axis of revolution, as shown in Figure 7.32. h(x) = x - x3. Y Because x ranges from 0 to 1, the volume of the solid is V = 2rr p(x)h(x) dx = 2rr x(x - x3) dx = 2’rr £1 (--X4 -{- X2) dx V = 2rC j ~c Ph (c~ "~ Apply ,daelI method. Simpli(.v. c (~, o) = 20r -~-+ 3Jo 1+ 4vr 15 Vertical axis of revolution Disk method: Representative rectangle is perpendicular to the axis of revolution. Figure 7.32 Horizontal axis of revolution Vertical axis of revolution Shell method: Representative rectangle is parallel to the axis of revolution. Horizontal axis of revolution y=x2 + 1, y= O, x= O, and x= 1 about the x-axis, where x and 3’ are measured in feet. Find the volume of the pontoon. about the y-axis. Y V = ~r (12 - 02) dy ÷ rr [12 __ (~)2] dy 3 ~- = ~ 1 dy + ~ (2 - y)dy Nimplit5. =~y Integrate, ~ 0 + ~ 2y- 2Jl R(x) = 1 1 2 3 3~ 2 3’ I 1 p(y) = y 2 3 1---~-k~ dx Simi,lif\. 4 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 of y in the equation (b) Shell method Figure 7.35 y-- 1,- (X2/16). Sometimes, solving for x is very difficult (or even impossible). In such cases you must 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 the method to be used. This is shown in Example 5. 1)& 2/0 EXAMPLE ~ Shell Method Necessary 3-n2 Axis of revolution 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 than the solid in Example 3? Without integrating, you should be able to reason that the resulting solid would have a smaller volume because "more" of the revolved region would be closer to the axis of revolution. To confirm this, try solving the following integral, which gives the volume of the solid. V = 2rr (1 - x)(x2 + 1) dx Apply dixk incthod. 4 V = 2rr p(x)h(x) dx Y(X2 + 1 - 16,/ dx 64~ : ~ ~ 13.4 cubic feet 15 b £1 V= 7r x-- ~ + 1280j_4 -4 -3 -2 -1 In Figure 7.33(b), you can see that the shell method requires only one integral to find the volume. Solution Refer to Figure 7.35(a) and use the disk method as follows. = "r; 4 (a) Disk method ~ = 2rr x2 i\pp!.x \\asher method. -4-3 -2 -1 I £1 £2 ,’(x) = 0 t Solution In Example 4 in the preceding section, you saw that the washer method requires two integrals to determine the volume of this solid. See Figure 7.33(a). V= 2rr p(x)h(x)dx= 2rr (2-x)(x3 ÷x÷ 1 - 1) dx .~pp!>,l~cllmetl>d. = /,,.,-, -- I £1 FOR FURTHER INFORMATION To learn more about the disk and shell methods, see the article "The Disk and Shell Method" by Charles A. Cable in The American Mathematical Monthly. To view this article, go to the website wwmmatharticles.com. ~! Find the volume of the solid formed by revolving the region bounded by the graphs of y = 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 of y. (See Section 3.8 on Newton’s Method.) Therefore, the variable of integration must be x, and you should choose a vertical representative rectangle. Because the rectangle is parallel to the axis of revolution, use the shell method and obtain It(x) = x3 + x + 1 - 1 p(x) = 2 - x ’, 1 Figure 7.36 , ’ I 2 f01 27r (--X4 ÷ 2X3 -- X2 ÷ 2x) dx 2 7"/" [ X5 X4 X3 k 27r(_ 1 1 1 ~÷~--~÷ 1) 29’n15 Simplil5. ...
View Full Document

• • • 