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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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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