Question

A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (> r) from the center of the circle. Find the volume of the torus.

Answer 1

Consider a circle with radius
`r` centered at some point
`(R+r,0)` on the
`x`-axis. This circle has equation

`(x-(R+r))^2+y^2=r^2`

Revolve the region bounded by this circle across the
`y`-axis to get a torus. Using the shell method, the volume of the resulting torus is

`\displaystyle2\pi\int_R^(R+2r)2xy\,\mathrm dx`

where
`2y=√(r^2-(x-(R+r))^2)-(-√(r^2-(x-(R+r))^2))=2√(r^2-(x-(R+r))^2)`.

So the volume is

`\displaystyle4\pi\int_R^(R+2r)x√(r^2-(x-(R+r))^2)\,\mathrm dx`

Substitute

`x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt`

and the integral becomes

`\displaystyle4\pi r^2\int_(-\pi/2)^(\pi/2)(R+r+r\sin t)\cos^2t\,\mathrm dt`

Notice that
`\sin t\cos^2t` is an odd function, so the integral over
`\left[-\frac\pi2,\frac\pi2\right]` is 0. This leaves us with

`\displaystyle4\pi r^2(R+r)\int_(-\pi/2)^(\pi/2)\cos^2t\,\mathrm dt`

Write

`\cos^2t=\frac{1+\cos(2t)}2`

so the volume is

`\displaystyle2\pi r^2(R+r)\int_(-\pi/2)^(\pi/2)(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}`