# Calculate Volume of Torus Given Circumferences

What, if any, is the formula to calculate the volume of a torus given the circumference of the tube and the outer circumference of the ring?

• en.wikipedia.org/wiki/Torus – Will Jagy Aug 21 '16 at 21:11
• The formula you're looking for is $2\pi^2 r^2 R$ where $r$ is the radius of of the tube, and R is the distance from the origin to the center of the tube. The parameterization is given in the wikipedia article, and the integral is fairly easy to compute from that information. – JAustin Aug 21 '16 at 21:13
• mathworld.wolfram.com/PappussCentroidTheorem.html – Jack D'Aurizio Aug 21 '16 at 21:21

By Pappus' centroid theorem, the volume of a torus generated by the rotation of a circle with radius $r$ with its centre on a circle with radius $R$ is just given by $2\pi^2 r^2 R$. In our case we have $\color{purple}{l}=2\pi r$ and $\color{green}{L}=2\pi(R+r)$, hence:

$$\boxed{ \color{red}{V} = \frac{1}{4\pi}\color{purple}{l}^2(\color{green}{L}-\color{purple}{l})}$$

• Jack's answer is extremely clear and concise. I upvoted it. One thing perplexing is that the first formula for area not volume in the mathworld. So if one scroll down, one will find the formula for volume. – Zack Ni Aug 21 '16 at 23:49

Since one can describe the volume of a torus using the radii of the two circles, the answer is yes. Suppose the circumference of the the larger ring is given by $C$ and the circumference of the smaller ring is given by $c$.

Similarily, let $R$ represent the center radius of the larger ring and $r$ represent the radius of the smaller ring. This is given by the picture

Thus the circumference is related to the radius by

$$C=2\pi (R+r)$$ $$c=2\pi r$$

Solving these for $R$ and $r$ yields

$$R=\frac{C-c}{2\pi}$$ $$r=\frac{c}{2\pi}$$

Using the formula for the volume of a torus $V=\pi r^2 \cdot 2 \pi R$ we obtain

$$V=\pi \left(\frac{c}{2\pi}\right)^2\cdot 2\pi \frac{C-c}{2\pi}=\frac{c^2(C-c)}{4\pi}.$$

This expresses the volume of a torus using the circumference.

• I think maybe you want $V=\pi r^2\cdot 2\pi(R-r)$; see Jack D'Aurizio's answer. – user84413 Aug 21 '16 at 21:42
• Yup, my bad. I've edited the post accordingly. – William T. Aug 21 '16 at 21:55

There are already two answers here, but I went ahead and computed the volume using the parametric equation of the torus, given by

\begin{align}&x=(a+r\cos v)\cos u\\&y=(a+r\cos v)\sin u\\&z= r\sin v\end{align}

where $a$ is the distance from the origin to the center of the 'tube', $r$ is the radius of the 'tube' itself, and $u$ and $v$ are two parameters corresponding to the central angle and the circular angle inside the 'tube'.

The laboriously calculated Jacobian (giving the magnitude of the volume element at every point inside the torus) is simply $ar + r^2\cos v\,dr\,dv\,du$, so the volume of the torus is given by the integral

$$\int_0^{2\pi}\int_0^{2\pi}\int_0^r ar + r^2\cos v\,dr\,du\,dv=2\pi^2r^2R$$

as desired.

Edit: Since I'm doubly bored, I went ahead and found the volume using a solid of revolution method. The torus can be considered the solid constructed by rotating a circle around the $z$ axis. The volume is given by the double integral

$$2\pi\int_{-r}^r\int_{-\sqrt{r^2-x^2}+R}^{\sqrt{r^2-x^2}+R}y\ dy\,dx=2\pi^2 r^2 R$$

Technically, the variables should be reversed, but the answer is the same.

• This answer would probably get upvoted if you (a) explained what the circmferences are in the problem, and (b) how your technique works. The parameterization also falls from the sky, so OP probably does not see how you actually solved the problem, just the answer. – Alfred Yerger Aug 21 '16 at 21:47
• Thanks for your comment. Updated with at least some exposition. – JAustin Aug 21 '16 at 22:00