I have a problem where i dont even know where to begin. Usually, when i post problems here i have something to show but this time i dont. Im sorry for that.

here is the problem:

FInd the surface area of the torus (or donut) obtained by rotating the circle (x - b)2 + y2 = a2; (a < b) about the

y-axis.

would i do something like the Area of a surface of revolution theorem?:

$\displaystyle S = 2 \pi \int_a^b f(t) \sqrt {(dx/dt)^2 + (dy/dt)^2} dt$

if so i am not even sure what to do to get it into the form to put it into that formula?

Any help is appreciated since i am lost!!

Thank you