I want to find the volume of a torus with a given thickness and a given radius.
Let r be the radius of a circle with its midpoint at $M(0|b)$ ($b \geq r$). Now I want to rotate this circle about the x-axis, that is to say about a circular path which has the length $2 \pi \cdot |b|$. So I thought I'd simply integrate:
$V = \int\limits_0^{2 \pi \cdot |b|} \pi r^2 dz = 2 \pi \cdot r^2 \cdot |b|$, which turns out to be the correct result.
However, I don't find it trivial that the volume of this torus is the same as the volume of a cylinder with the corresponding height. I read the article on Wikipedia about the torus and it said that this was due to Cavalieri's theorem, which to my mind doesn't really have a lot to do with the torus vs. the cylinder…
Is there some easy way to prove that a torus has the same volume as a cylinder with the height equal to the torus' perimeter?
Best Answer
(i) Slice the torus into a million near-disks.
(ii) Rotate every second disk through $180^\circ$.
(iii) Stick them all together again.
You get a near-cylinder whose height is nearly $2\pi b$. Now let a million tend to infinity.