Question

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

Answer 1

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=2pi^2r^2R`.

Let's say the torus is obtained by rotating the circular region `x^2+(y-R)^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[(sqrt{r^2-x^2}+R)^2-(-sqrt{r^2-x^2}+R)^2]dx`,
which simplifies to:
`V=4piR\int_{-r}^r sqrt{r^2-x^2}dx`
Since the integral above is equivalent to the area of a semicircle with radius r, we have
`V=4piRcdot1/2pi r^2=2pi^2r^2R`