Question

A ring torus has an ellipse of semi axes a and b as cross section. The radius to the axis of the torus from its center is c. Without integration for solid of revolution, how do you prove that the volume is `2pi^2abc`?

Answer 1

See proof below

Explanation:

When a ring torus of radius `c` & cross section as an ellipse of semi-axes `a` & `b`, is cut off from a cross-section & unfolded, torus becomes like a straight cylinder of length `2\pi c` & cross section as an ellipse with semi-axes `a` & `b` if the volume if kept constant.

Hence, the volume of given ring torus

`=\text{volume of cylinder of length }2\pi c \ \ & \ \ \text{cross-section area} \ \ \pi ab`

`=2\pi c(\pi ab)`

`=2\pi^2abc`