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#?

1 Answer

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#