use cylindrical shells to find the volume $V$ of the heavy torus (the donut-shaped solid shown in the figure) with radii $r$ and also $R$.

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This is as much I have actually come. How can I resolve further?

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To solve $int_-r^rsqrtr^2-t^2dt$, you can use the substitution $t=rsin heta$. Then if $t=r$, $ heta=pi/2$; $t=-r$, $ heta=-pi/2$. Also, $dt=rcos heta d heta$ and also $sqrtr^2-t^2=rcos heta$. Hence, us have$$int_-r^rsqrtr^2-t^2dt=int_-pi/2^pi/2r^2cos^2 heta d heta=r^2int_-pi/2^pi/2cos^2 heta d heta.$$I will leave the remaining component to you.


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