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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. $y = x^2, y = 1$; about $y = 2$

I know that the function will be rotated around the $x$-axis, but I am having problems with understanding getting the radius since the function rotates around $y = 2$. Any help would be appreciated. Thanks!

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    $\begingroup$ Have you visualized the shape? It has a cylindrical hole in it of radius $1$. Before we drill out the hole, the radius "at" $x$ is $2-y$, that is, $2-x^2$. $\endgroup$ – André Nicolas Jan 7 '15 at 4:31
  • $\begingroup$ I guess the problem is that I'm not drawing it correctly. Thank you for the help. I was thinking I was using the disk method, not the washer method. $\endgroup$ – Sherbs Jan 7 '15 at 4:33
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    $\begingroup$ Usually in these volume problems once you have a good picture and visualization, the volume writes itself. For success, the first thing one needs to do is to have a clear idea of the shape. $\endgroup$ – André Nicolas Jan 7 '15 at 4:47
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$x^2=1 \to x =\pm 1 \to V = \displaystyle \int_{-1}^1 \pi \left((2-x^2)^2 - (2-1)^2\right)dx$

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