# How to compute the max/min surface area of a donut-shape solid generated by a revolved 2D circle, as the volume of the solid doesn't change?

A donut shape solid is generated by revolving a circle $$(x-a)^2+y^2=b^2$$ around the y-axis. $$a$$ is the distance from the center of the hole of the donut to the center of the circle revolved, and $$b$$ is the radius of the circle revolved. I'm trying to compute the minimized surface area and the maximized surface area (if the solid has) with the value of $$a$$ and $$b$$, while the volume of the solid doesn't change (which is $$90\pi^2$$). Thanks a lot if someone can help me :)

Let's call, more conventional, the small and big radius r and R, where R>=r. Then the volume and area are:

V== 2 Pi^2 r^2 R
A== 2 Pi^2 r R


You specify V= 90 Pi^2. Then A can be written as a function of only one variable, e.g. r:

A = =V/r
R == V/(2 Pi^2 r^2)


Therefore, the largest value of A of infinity is reached for r->0.

The smallest value of A is reached if r == R:

rmin = V^(1/3)/(2^(1/3) \[Pi]^(2/3))


In your case for V= 90 Pi^2