# How do you use an integral to find the volume of a solid torus?

Aug 30, 2014

If the radius of its circular cross section is $r$, and the radius of the circle traced by the center of the cross sections is $R$, then the volume of the torus is $V = 2 {\pi}^{2} {r}^{2} R$.

Let's say the torus is obtained by rotating the circular region ${x}^{2} + {\left(y - R\right)}^{2} = {r}^{2}$ about the $x$-axis. Notice that this circular region is the region between the curves: $y = \sqrt{{r}^{2} - {x}^{2}} + R$ and $y = - \sqrt{{r}^{2} - {x}^{2}} + R$.

By Washer Method, the volume of the solid of revolution can be expressed as:
$V = \pi {\int}_{- r}^{r} \left[{\left(\sqrt{{r}^{2} - {x}^{2}} + R\right)}^{2} - {\left(- \sqrt{{r}^{2} - {x}^{2}} + R\right)}^{2}\right] \mathrm{dx}$,
which simplifies to:
$V = 4 \pi R \setminus {\int}_{- r}^{r} \sqrt{{r}^{2} - {x}^{2}} \mathrm{dx}$
Since the integral above is equivalent to the area of a semicircle with radius r, we have
$V = 4 \pi R \cdot \frac{1}{2} \pi {r}^{2} = 2 {\pi}^{2} {r}^{2} R$