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"Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis."

I have two equations : x= 2y^(2)-y^(3) and x= 0. I am supposed to rotate about x axis. I'm stuck on what region I am looking at for rotation. I got x=0 being the y axis, and the other equation to look something like a sin(x) curve. The area rotate is not very intuitive for me.

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Maybe with cylindrical shells you are accustomed to $y=f(x)$ rotate about $y$ axis. If it will make you feel more comfortable, rotate region bounded by $y=2x^2-x^3$ and the $y$-axis, about the $y$-axis. –  André Nicolas Oct 11 '11 at 22:52

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Your cylinders lie along the $x$ axis. At a given $y$, the minimum $x$ is $0$ and the maximum is $2y^2-y^3$. The upper plot in this Alpha page shows it well (but note the axes are switched as André Nicolas suggests). You are rotating the region above the horizontal axis around the vertical axis. This would give $\int_0^2 (2y^2-y^3) 2\pi y\;dy$ where the height of the shell is $x=2y^2-y^3$, the radius is $y$, and the thickness of the shell is $dy$.

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