450❙❙❙❙CHAPTER 6APPLICATIONS OF INTEGRATIONSOLUTIONFigure 11 shows a horizontal cross-section. It is a washer with inner radiusand outer radius , so the cross-sectional area isThe volume isWe now ﬁnd the volumes of three solids that are notsolids of revolution.EXAMPLE 7Figure 12 shows a solid with a circular base of radius 1. Parallel cross-sectionsperpendicular to the base are equilateral triangles. Find the volume of the solid.Let’s take the circle to be . The solid, its base, and a typical cross-section at a distance from the origin are shown in Figure 13.FIGURE 13yy60°60°BACœ„3y(c) A cross-sectionxy0yxAB(x, y)y=œ„„„„„„1-≈(b) Its base(a) The solid0xyABC1_1xx2±y2±1FIGURE 11x0x=œ„yx=yx=_1yy11+y1+œ„y±²±4y3²23³y22³y33³01±2±y10(2sy³y³y2)dy±y10[(1±sy)2³´1±yµ2]dyV±y10A´yµdy±(1±sy)2³´1±yµ2A´yµ±´outer radiusµ2³´inner radiusµ21±sy1±yFIGURE 12Computer-generated pictureof the solid in Example 7yx
Since lies on the circle, we have and so the base of the triangleis . Since the triangle is equilateral, we see from Figure 13(c) that itsheight is . The cross-sectional area is thereforeand the volume of the solid isEXAMPLE 8Find the volume of a pyramid whose base is a square with side and whoseheight is .SOLUTIONWe place the origin at the vertex of the pyramid and the -axis along its cen-tral axis as in Figure 14. Any plane that passes through and is perpendicular to the -axis intersects the pyramid in a square with side of length , say. We can express interms of by observing from the similar triangles in Figure 15 thatand so . [Another method is to observe that the line has slope andso its equation is .] Thus, the cross-sectional area isThe pyramid lies between and , so its volume isNOTE■We didn’t need to place the vertex of the pyramid at the origin in Example 8. Wedid so merely to make the equations simple. If, instead, we had placed the center of thebase at the origin and the vertex on the positive -axis, as in Figure 16, you can verify thaty±L2h2x33±0h±L2h3V±yh0A²x³dx±yh0L2h2x2x±hx±0xyOxhFIGURE 14sxhLxyOPFIGURE 15A²x³±s2±L2h2x2y±Lx´²2h³L´²2h³OPs±´hxh±s´2L´2±sLxssxxPxxOhL±2 y10s3²1±x2³±2s3µx±x33±01±4s33V±y1±1A²x³±y1±1s3²1±x2³A²x³±12±2s1±x2±s3s1±x2±s3²1±x2³s3y±s3s1±x2¶AB¶±2s1±x2ABCy±s1±x2BSECTION 6.2VOLUMES❙❙❙❙451xyh0yFIGURE 16Resources / Module 7/ Volumes / Start of Mystery of theTopless Pyramid
452❙❙❙❙CHAPTER 6APPLICATIONS OF INTEGRATIONwe would have obtained the integralEXAMPLE 9A wedge is cut out of a circular cylinder of radius 4 by two planes. One planeis perpendicular to the axis of the cylinder. The other intersects the ﬁrst at an angle of
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