Deriving the volume of an elliptic torus

Question

Following this question, I was looking forward to derive the volume of the elliptic torus using: $$V = \int_\Omega rdrd\theta dz$$ in the cylindrical coordinates. where the origin of the system will lie at the tangent to the inner circles as shown in the figure, then, and if rotational symmetry, the volume is: $$ V = \int\limits_{0}^{2\pi}d\theta \int\limits_{R_0-b}^{R_0+b}\int\limits_{0}^{2a} rdrdz $$. I need a help in choosing the right boundaries and if there is need to use the implicit equation to pass through the derivation. This looks weird for me, and I guess the answer should look like: $$V=2\pi R_0 \times \pi ab$$

Answer 1

Why not plainly use the Pappus Guldin theorem ?

Here, the sectional area (area of an ellipse with semiaxes $a$ and $b$) is $\pi a b$ to be multiplied by the trajectory of the center of gravity : $2 \pi R_0$ giving (indeed !) the volume :

$$V=2 \pi^2 a b R_0$$

Answer 2

If you want to treat the torus like a cylinder, you could consider the following: A torus is a cylinder, where the two ends meet. For the volume, this periodicity can basically be ignored. So we integrate over the disk (one slice of the torus is a disk):

$$ \int^{2\pi}_0\!\mathrm{d}\theta\int^{r}_0\!r\prime\,\mathrm{d}r\prime $$

Then we have to stack the disks on top of each other. The total 'height' of the torus is equal to the circumference, i.e. $2\pi(R+r)=2\pi R_0$.

This gives us the volume of the torus:

$$ \int^{R_0}_0\!\mathrm{d}z\int^{2\pi}_0\!\mathrm{d}\theta\int^{r}_0\!r\prime\,\mathrm{d}r\prime=2\pi^2r^2R_0. $$