Use cylindrical shells to find the volume $V$ of the solid torus (the donut-shaped solid shown in the figure) with radii $r$ and $R$.
This is as far I have come. How can I solve further?
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That last integral smells like trig substitution. Try something like $t = r\cos\theta$. – Neal Jan 25 '13 at 02:49
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This is a lot easier with Pappus Throrem, volume=area*length of path of centroid. – Maesumi Jan 25 '13 at 03:26
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To solve $\int_{-r}^r\sqrt{r^2-t^2}dt$, you can use the substitution $t=r\sin\theta$. Then if $t=r$, $\theta=\pi/2$; $t=-r$, $\theta=-\pi/2$. Also, $dt=r\cos\theta d\theta$ and $\sqrt{r^2-t^2}=r\cos\theta$. Hence, we have $$\int_{-r}^r\sqrt{r^2-t^2}dt=\int_{-\pi/2}^{\pi/2}r^2\cos^2\theta d\theta =r^2\int_{-\pi/2}^{\pi/2}\cos^2\theta d\theta.$$ I will leave the remaining part to you.
Paul
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