Shell Method Help - 24pi(-2/5y^2+7/5y^3-y^4)dy

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SUMMARY

The discussion focuses on the application of the shell method to calculate the volume of a solid of revolution defined by the function X=12(y^2-y^3) when rotated around the line y=-2/5. The volume formula used is V=2π∫(shell radius)(shell height) dy, leading to the integral 24π(-2/5+y)(y^2-y^3)dy. The user attempts to integrate from 0 to 1, arriving at a volume of 2/5π, while the textbook solution states the volume is 2π, indicating a potential misunderstanding of the problem setup or boundaries.

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Homework Statement


X=12(y^2-y^3) Rotation is around y=-2/5


Homework Equations


V=2pi from the integral of a to b (shellradius)(shell height) dx


The Attempt at a Solution


2pi (-2/5+y)(12)(y^2-Y^3)dy
24pi (-2/5+y)(y^2-y^3)dy
24pi (-2/5y^2+7/5y^3-y^4)dy
24pi (-8/60y^2+21/60y^4-12/60y^5)
answer i get is 2/5pi when i integrat from 0 to 1 the book answer is 2pi
 
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Please post the entire problem statement. Is the region being rotated bounded by some line, such as the y-axis?
 

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