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How do I find the volume of this region using the method of cylindrical shells?

The region is enclosed by the given curves, $y=x^2$ and $y=2-x^2$

The volume of the region is found by rotating the region enclosed by those curves, about the axis x=1

Specifically, what I am stuck on with this problem is finding the radius of the cylindrical shell that would be formed by rotating this region around x=1. I understand that typically, one would just make the radius 1-x, but since this region crosses over to the negative side of the x values, I'm not sure what to do.

So far, I have pictured the x-y plane graph of the bounded region:

enter image description here

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1 Answer 1

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You found the intersections of the two curves at $x=-1$ and $x=1$. Since $x\le1$, you have the radius $r=1-x$ varying between $+2$, at $x=-1$ and $0$ at $x=1$. The height of the cylindrical shell is $h=2-x^2-x^2=2(1-x^2)$. In the interval from $-1$ to $1$ the height is always non-negative. It is $0$ only at the $x=\pm1$

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