# How to prove that a torus has the same volume as a cylinder (with the height equal to the torus' perimeter)

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?

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Pappus's centroid theorem. It's hard to find a proof of this theorem online, though – kahen Nov 24 '10 at 19:14
I'll search for one. At least, the theorem is available on Wikipedia so I know I'm on the right path... – Huy Nov 24 '10 at 19:56
It's basically a oneliner, but surprisingly hard to find on the web! It's written out (sketchily) on German Wikipedia: de.wikipedia.org/wiki/Guldinsche_Regeln#Zweite_Regel. – Hans Lundmark Nov 24 '10 at 19:58

(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.

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Ah, a three-dimensional version of the rearrangement argument. Nice. – Rahul Nov 24 '10 at 20:33
This is a great example of where formal mathematical language would create a headache, whereas communicating the key ideas gets the message through straightaway. – P-i- Sep 2 '14 at 11:54

Perhaps I don't understand the question, but if you have a cylinder, you can imagine bending it so that the top and bottom are connected. Then you have a torus.

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It's not obvious that this preserves the volume! The inner parts are compressed and the outer parts extended, and it's a bit surprising that these effects precisely cancel. – Hans Lundmark Nov 24 '10 at 19:38
My thoughts exactly! – Huy Nov 24 '10 at 19:55
@Hans, @Huy: TonyK’s answer can be seen exactly as an argument for why they do. – Peter LeFanu Lumsdaine Nov 24 '10 at 20:29

To use Cavalieri's theorem, lay the torus and cylinder on a table and slice them with planes parallel to the table. Then it suffices to show that the torus slices (annuli) have the same area as the cylinder slices (rectangles).

At height $h$ (measured from the centre of the torus or cylinder), the annulus has inner and outer radii $$R_\pm = |b| \pm \sqrt{r^2-h^2},$$ so its area is $$A_1 = \pi R_+^2 - \pi R_-^2 = \pi(R_+ + R_-)(R_+ - R_-) = \pi \cdot 2|b| \cdot 2\sqrt{r^2-h^2}.$$ The rectangle has width $2\sqrt{r^2-h^2}$ and length $2\pi |b|$, so its area is $$A_2 = 2\sqrt{r^2-h^2} \cdot 2\pi|b|.$$ The areas are equal, so we're done!

I prefer Pappus's centroid theorem though...

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