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I'm hoping to get some help with finding the volume using cylindrical shells of the problem below:

Use the method of shells to find the volume generated when the area bounded by
$x=\frac{1}4y^4-\frac{1}{2}y^2$ and $x=\frac{1}{2}y^2$ is revolved about the line $y=-\frac{5}{8}$

This is what I have set up but I feel like something isn't right. Could someone tell me if I'm on the right track or point me in the right direction? I feel pretty confident about the radius but the height of the cylinder is where I'm getting confused. Not sure if it only depends on one of the curves or both, and in which order one should be subtracted from the other. Any help would be greatly appreciated!

$2\pi\int_0^2 [((\frac{1}{2}y^2) - (\frac{1}{4}y^4-\frac{1}{2}y^2))(y+\frac{5}{8})]dx $

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    The region intersects the axis of revolution. Is that correct? Or do you only want the portion of the region where $y\ge0$? – 2017-02-07
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    Your integral is set up correctly to find the volume of the solid of revolution if you are using only the "top half" of the region, that is, that portion where $y\ge0$. You just need to simplify the integrand. It will be a fifth degree polynomial in $y$ so it should be straightforward. – 2017-02-07
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    Yes, it's only for the portion where $y\ge0$. Thank you so much for your reply! – 2017-02-07

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