Calculus_205th_20Edition_20-_20James_20Stewart_Part19
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Calculus_205th_20Edition_20-_20James_20Stewart_Part19

Course Number: MATH 114, Fall 2010

College/University: UPenn

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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 1 y and outer radius 1 Ay sy, so the cross-sectional area is outer radius 2 inner radius 1 y 2 2 (1 The volume is V sy )2 y 1 0 A y dy y [(1 0 1 sy )2 1 y 2 ] dy y (2 sy 0 1 y y2 2 y 2 ) dy y3 3 1 4y 3 2 3 0 2 y 1+ y 1+y 1 x= y y x=y 0 x FIGURE 11 x=_1 We...

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6 450 CHAPTER APPLICATIONS OF INTEGRATION SOLUTION Figure 11 shows a horizontal cross-section. It is a washer with inner radius 1 y and outer radius 1 Ay sy, so the cross-sectional area is outer radius 2 inner radius 1 y 2 2 (1 The volume is V sy )2 y 1 0 A y dy y [(1 0 1 sy )2 1 y 2 ] dy y (2 sy 0 1 y y2 2 y 2 ) dy y3 3 1 4y 3 2 3 0 2 y 1+ y 1+y 1 x= y y x=y 0 x FIGURE 11 x=_1 We now nd 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. SOLUTION Lets take the circle to be x 2 y y 2 1. The solid, its base, and a typical crosssection at a distance x from the origin are shown in Figure 13. y y C x 1- y= B(x, y) y C FIGURE 12 B _1 0 1 x Computer-generated picture of the solid in Example 7 0 x x y 3 A 60 y 60 y B A A FIGURE 13 (a) The solid (b) Its base (c) A cross-section SECTION 6.2 VOLUMES 451 Since B lies on the circle, we have y s1 x 2 and so the base of the triangle ABC is AB 2 s1 x 2. Since the triangle is equilateral, we see from Figure 13(c) that its height is s3 y s3 s1 x 2. The cross-sectional area is therefore Ax 1 2 2 s1 x 2 s3 s1 x2 s3 1 x2 and the volume of the solid is V y 1 1 A x dx 1 y 1 1 s3 1 x 2 dx x3 3 1 2 y s3 1 0 x 2 dx 2 s3 x 0 4 s3 3 Resources / Module 7 / Volumes / Start of Mystery of the Topless Pyramid EXAMPLE 8 Find the volume of a pyramid whose base is a square with side L and whose height is h. SOLUTION We place the origin O at the vertex of the pyramid and the x-axis along its central axis as in Figure 14. Any plane Px that passes through x and is perpendicular to the x-axis intersects the pyramid in a square with side of length s, say. We can express s in terms of x by observing from the similar triangles in Figure 15 that x h s2 L2 s L and so s L x h. [Another method is to observe that the line OP has slope L 2h and so its equation is y L x 2h .] Thus, the cross-sectional area is Ax y s2 y L2 2 x h2 P x O h x O s x h L x y h FIGURE 14 FIGURE 15 The pyramid lies between x V 0 and x h, so its volume is y h y 0 A x dx h y h 0 L2 2 x dx h2 L2 x 3 h2 3 0 x s 0 L2h 3 FIGURE 16 NOTE We didnt 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 the on positive y-axis, as in Figure 16, you can verify that 452 CHAPTER 6 APPLICATIONS OF INTEGRATION we would have obtained the integral V y h 0 L2 h h2 y 2 dy L2h 3 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 rst at an angle of 30 along a diameter of the cylinder. Find the volume of the wedge. SOLUTION If we place the x-axis along the diameter where the planes meet, then the base of the solid is a semicircle with equation y s16 x 2, 4 x 4. A crosssection perpendicular to the x-axis at a distance x from the origin is a triangle ABC, as shown in Figure 17, whose base is y s16 x 2 and whose height is BC y tan 30 s16 x 2 s3. Thus, the cross-sectional area is C 0 y Ax B y= 16- 1 2 s16 x2 1 s16 s3 x2 A 4 x 16 x 2 2 s3 and the volume is V C y 4 4 A x dx y 4 4 16 x 2 dx 2 s3 1 16 x s3 x3 3 4 30 A FIGURE 17 y B 1 s3 y 4 0 16 x 2 dx 0 128 3 s3 For another method see Exercise 62. |||| 6.2 118 |||| Exercises 10. y 11. y 12. y 13. y 14. y 15. x 16. y 17. y 18. y s s Find the volume of the solid obtained by rotating the region bounded by the given curves about the specied line. Sketch the region, the solid, and a typical disk or washer. 1. y 2. y 3. y 4. y 5. y 6. x 7. y 8. y 9. y 2 x 2 3, x x, y x 2, y x 4, y 1 x, y y 2, x x, y x,x x, y s 1, y sx; 4; 1; 0; about the y-axis 1 4 2 3; about y 1 2 1 4; s about y about y about y 1, x x 2, x e, y 1 x, x sx x 2, 0 y x 1, y 0, x 1, x 1, x x 0; about the x-axis 0, x 2, y 2, x 2, y 1; about the x-axis 0; about the x-axis 5, y 4, x 0; about the x-axis 0; about the y-axis 0, x 1; sx; y; 0, x s 1 about x about x about x 2, x s y 2, x 0; about the y-axis x 2, y 2 sec x, y x, x x; about the x-axis 1, x 1, x 1; about the x-axis 2 2 about x s s 1 s s s s 2y; about the y-axis S ECTION 6.2 VOLUMES 453 1930 |||| Refer to the gure and nd the volume generated by rotating the given region about the specied line. y 43. s y s 1 0 y4 s y 8 dy s s s 44. s y 2 0 s 1 s cos x s 2 12 dx s s C(0,1) y= x T T T y= O B(1,1) 45. A CAT scan produces equally spaced cross-sectional views of a A(1,0) x human organ that provide information about the organ otherwise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas, in square centimeters, are 0, 18, 58, 79, 94, 106, 117, 128, 63, 39, and 0.

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