Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

3 Answers 3

up vote 15 down vote accepted

(i) Slice the torus into a million near-disks.
(ii) Rotate every second disk through 180 degrees.
(iii) Stick them all together again.
You get a near-cylinder whose height is nearly 2πb. Now let a million tend to infinity.

share|improve this answer
3  
Ah, a three-dimensional version of the rearrangement argument. Nice. –  Rahul Nov 24 '10 at 20:33

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...

share|improve this answer

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.

share|improve this answer
10  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.