271_HW6.4 - MAT 271 Summer 2011 Calculus II Scott Zinzer...

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MAT 271, Summer 2011 Calculus II, Scott Zinzer 6.4 HW 6.4. Volume by Shells. 12. Let R be the region bounded by the curves y = 8, y = 2 x + 2, x = 0 and x = 2. Use the shell method to find the volume of the solid generated when R is revolved about the x -axis. Solution: We use the shell method and thus slice parallel to the x -axis. We will be integrating with respect to y so we need to solve for x in terms of y in each of the above equations. y = 2 x + 2 becomes x = y/ 2 - 1. We will need two separate integrals. For 0 y 6, the region in question is the region bounded by y = 6, x = y/ 2 - 1, and x = 0. A typical shell has width Δ y , radius y , and height y/ 2 - 1. The volume of such a shell is 2 π ( y )( y/ 2 - 1)Δ y . For 6 y 8, the region in question is the region bounded by y = 6, y = 8, and x = 2. A typical shell has width Δ y , radius y and height 2. The volume of such a shell is 2 π ( y )(2)Δ y . Putting this all together, the volume of the solid is Z 6 2 2 π ( y )( y/ 2 - 1) dy + Z 8 6 2 π ( y )(2) dy = 2 π Z 6 2 y 2 / 2 - y dy + 4 π Z 8 6 y dy = 2 π 1 6 y 3 - 1 2 y 2 6 2 + 4 π 1 2 y 2 8 6 = 2 π 1 6 (6) 3 - 1 2 (6) 2 - 1 6 (2) 3 - 1 2 (2) 2 + 4 π 1
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