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Compute the volume of the solid of revolution that results from revolving $f(x)=\cos(x)$ between $x=-\pi/2$ and $x=\pi/2$ around $y=-1$. I know how to do so around the $x$-axis or the $y$-axis, however, I am not sure what to do in this case.

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    $\begingroup$ If you shift the coordinate axis down, you can think of the rotation surounding the line y=0 and f(x)=cos(x)+1. Can you take it from there? $\endgroup$ – Fede Poncio Dec 18 '18 at 18:44
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    $\begingroup$ Yes, thank you. $\endgroup$ – M.Papapetros Dec 18 '18 at 18:49
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If you draw the graph, you can see that it is the same volume as revolving $ f(x) = cos(x) + 1$ around the x-axis. You just translate the whole thing up with 1.

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