Question

Let `b > a > 0` be constants. Find the area of the surface generated by revolving the circle `(x − b)^2 + y^2 = a^2` about the y-axis?

Let `b > a > 0` be constants. Find the area of the surface generated by revolving the circle `(x − b)^2 + y^2 = a^2` about the y-axis?

Answer 1

`4pi^2ab`

Explanation:

Being `ds = a d theta` the length element in the circle with radius `a`, having the vertical axis as rotation center and the circle origin at a distance `b` from de axis of rotation, we have

`S=int_{0}^{2pi} 2 pi (b + a cos theta)a d theta = 4pi^2ab`