Calculus_205th_20Edition_20-_20James_20Stewart_Part19

# Calculus_205th_20Edition_20-_20James_20Stewart_Part19 - 450...

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450 ❙❙❙❙ CHAPTER 6 APPLICATIONS OF INTEGRATION SOLUTION Figure 11 shows a horizontal cross-section. It is a washer with inner radius and outer radius , so the cross-sectional area is The volume is We now ﬁnd the volumes of three solids that are not solids of revolution. EXAMPLE 7 Figure 12 shows a solid with a circular base of radius 1. Parallel cross-sections perpendicular 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 13 yy 60° 60° B A C œ„3y (c) A cross-section x y 0 y x A B(x, y) y=œ„„„„„„ 1-≈ (b) Its base (a) The solid 0 x y A B C 1 _1 x x 2 ± y 2 ± 1 FIGURE 11 x 0 x=œ„ y x=y x=_1 y y 1 1+y 1+œ„ y ± ² ± 4 y 3 ² 2 3 ³ y 2 2 ³ y 3 3 ³ 0 1 ± 2 ± y 1 0 (2 s y ³ y ³ y 2 ) dy ± y 1 0 [(1 ± s y ) 2 ³ ´ 1 ± y µ 2 ] dy V ± y 1 0 A ´ y µ dy ± (1 ± s y ) 2 ³ ´ 1 ± y µ 2 A ´ y µ ± ´ outer radius µ 2 ³ ´ inner radius µ 2 1 ± s y 1 ± y FIGURE 12 Computer-generated picture of the solid in Example 7 y x

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Since lies on the circle, we have and so the base of the triangle is . Since the triangle is equilateral, we see from Figure 13(c) that its height is . The cross-sectional area is therefore and the volume of the solid is EXAMPLE 8 Find the volume of a pyramid whose base is a square with side and whose height is . SOLUTION We 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 in terms of by observing from the similar triangles in Figure 15 that and so . [Another method is to observe that the line has slope and so its equation is .] Thus, the cross-sectional area is The pyramid lies between and , so its volume is NOTE We didn’t need to place the vertex of the pyramid at the origin in Example 8. We did so merely to make the equations simple. If, instead, we had placed the center of the base at the origin and the vertex on the positive -axis, as in Figure 16, you can verify that y ± L 2 h 2 x 3 3 ± 0 h ± L 2 h 3 V ± y h 0 A ² x ³ dx ± y h 0 L 2 h 2 x 2 x ± h x ± 0 x y O xh FIGURE 14 s x h L x y O P FIGURE 15 A ² x ³ ± s 2 ± L 2 h 2 x 2 y ± Lx ´² 2 h ³ L ´² 2 h ³ OP s ± ´ h x h ± s ´ 2 L ´ 2 ± s L x s s x x P x x O h L ± 2 y 1 0 s 3 ² 1 ± x 2 ³ ± 2 s 3 µ x ± x 3 3 ± 0 1 ± 4 s 3 3 V ± y 1 ± 1 A ² x ³ ± y 1 ± 1 s 3 ² 1 ± x 2 ³ A ² x ³ ± 1 2 ± 2 s 1 ± x 2 ± s 3 s 1 ± x 2 ± s 3 ² 1 ± x 2 ³ s 3 y ± s 3 s 1 ± x 2 AB ± 2 s 1 ± x 2 ABC y ± s 1 ± x 2 B SECTION 6.2 VOLUMES ❙❙❙❙ 451 x y h 0 y FIGURE 16 Resources / Module 7 / Volumes / Start of Mystery of the Topless Pyramid
452 ❙❙❙❙ CHAPTER 6 APPLICATIONS OF INTEGRATION we would have obtained the integral EXAMPLE 9 A wedge is cut out of a circular cylinder of radius 4 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the ﬁrst at an angle of

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## This note was uploaded on 09/14/2010 for the course MATH 114 taught by Professor Temkin during the Fall '07 term at UPenn.

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Calculus_205th_20Edition_20-_20James_20Stewart_Part19 - 450...

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