HW15solutions

HW15solutions - HW 15 Solutions Math 1A with Prof. Stankova...

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HW 15 Solutions Math 1A with Prof. Stankova UC-Berkeley, Spring 2009 6.2: Volumes For these solutions, I will graph some of the regions, but not all of them. Drawing regions and the corresponding volumes is very important, and should not be skipped; drawing such things on the computer, however, is very slow. The illustrators for Stewart’s Calculus are rather incredible. 2. We graph the curves: - 2 - 1 1 2 - 1 1 2 0 From the picture we immediately see that the curve y = 1 - x 2 is on top. The two curves intersect at ( - 1 , 0) and at (1 , 0). For rotations around the x -axis, we work with respect to x , whence the volume is: Volume = π Z 1 - 1 ( 1 - x 2 ) 2 dx = π Z 1 - 1 ( 1 - 2 x 2 + x 4 ) dx = π ± x - 2 3 x 3 + 1 5 x 5 ²³ ³ ³ ³ 1 - 1 = π ±± 1 - 2 3 + 1 5 ² - ± ( - 1) - - 2 3 + - 1 5 ²² = 16 π 15 4. For rotations around the x -axis, we work with respect to x . The volume is: Volume = π Z 4 2 ´ p 25 - x 2 µ 2 dx = π Z 4 2 ( 25 - x 2 ) dx = π ± 25 x - x 3 3 ²³ ³ ³ ³ 4 2 = π ±± 25 · 4 - 4 3 3 ² - ± 25 · 2 - 2 3 3 ²² = π ± 50 + 56 3 ² = π · 68 1 3 writing the answer as a mixed number since it is so large. 10. Here is the graph: - 2 - 1 1 2 - 1 1 2 0 1
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For rotating around the y -axis, we work with respect to y , whence the interesting curves are x = 2 for the outer radius and x = 2 y for the inner radius. The domain of y is [0 , 1]. Then the volume is Volume = π Z 1 0 ( 2 2 - (2 y ) 2 ) dy = π Z 1 0 (4 - 4 y ) dy = π ( 4 y - 2 y 2 ± 1 0 = 2 π 12. We rotate around y = 2, which means that we should work with respect to dx — we always work with the coordinate parallel to the line of rotation, so that the slices are perpendicular to the rotational axis. The curves y = e - x and y = 1 intersect that x = 0, and the other boundary point value is x = 2. The inner radius is r = | 1 - 2 | = 1, and the outer radius is R = | e - x - 2 | = 2 - e - x . Then the volume is: Volume = π Z 2 0 ² ( 2 - e - x ) 2 - 1 2 ³ dx = π Z 2 0 ( 3 - 4 e - x + e - 2 x ) dx = π ´ 3 x + 4 e - x - 1 2 e - 2 x µ± ± ± ± 2 0 = π ´ 3(2 - 0) + 4( e - 2 - 1) - 1 2 ( e - 4 - 1) µ = π ´ 3 2 + 4 e - 2 - 1 2 e - 4 µ 14. The axis y = - 1 is horizontal. We compute: Volume = π Z 3 1 ´ 1 x + 1 µ 2 - 1 2 ! dx = π Z 3 1 ( x - 2 + 2 x - 1 ) dx = π ( - x - 1 + ln x ± 3 1 = π ´ 2 3 + ln 3 µ 16. The axis x = 2 is vertical, so we work with respect to y . The curves are x = y and x = y 2 , which intersect at y = 0 and y = 1. The inner radius is 2 - y and the outer radius is 2 - y 2 , since y 2 y for y [0 , 1]. Thus, the volume is: Volume = π Z 1 0 ² ( 2 - y 2 ) 2 - (2 - y ) 2 ³ dy = π Z 1 0 ( y 4 - 3 y 2 + 2 y ) dy = π ´ 1 5 - 3 3 + 2 2 µ = π 5 18. We graph the figure. The thick dotted line is the axis of rotation. The thin dashed line is where we divide the region into two easier regions: - 1 1 2 3 4 5 - 1 1 2 3 4 5 0 2
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Reading from the figure it is clear that the inner axis is only piecewise-defined, which is why we have divided the region in two. Then considering the volumes swept out be each region separately, we see that: Volume = π Z 2 0 ( 3 2 - 1 2
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HW15solutions - HW 15 Solutions Math 1A with Prof. Stankova...

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