Volume of the torus with an integral

I am stuck on the following question:

Find the volume of the torus obtained by rotating the circle $(x-a)^2+y^2=b^2$, where $a>b$, around the y-axis.

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– Jonas Meyer Feb 2 '12 at 7:21
VolumeOfTorus – pedja Feb 2 '12 at 7:24

One way is to slice it circularly like an onion. This way, the cross-section corresponding to the distance of $x$ from the origin on the $x$-axis (from $a-b$ to $a+b$) has area given by the circum-ference of the circle made ($C=2\pi x$) times the height of the torus at that point, which is given by the formula $h=2y=2\sqrt{b-(x-a)^2}$. Thus we have
$$V=\int_{a-b}^{a+b}2 \pi \cdot 2x\sqrt{b^2-(x-a)^2}dx.$$
To evaluate this, consider a change of variables $x=bv+a$ first. We get
$$=4\pi b^2\int_{-1}^1 (bv+a)\sqrt{1-v^2}dv=4\pi b^2\left(b\cdot0+a\frac{\pi}{2}\right)=2\pi^2ab^2.$$
Here we use the fact that $v\sqrt{1-v^2}$ is an odd function (so that integrating it over a symmetric integral evaluates to zero) and $\displaystyle\int_{-1}^1\sqrt{1-v^2}dv$ describes half the area of the unit circle.