Let $A$ and $B$ be points on a circle centered at $O$ with radius $R,$ and let $\angle AOB = 2 \alpha \le \pi.$ Minor arc $AB$ is rotated about chord $\overline{AB}.$ Find the surface area of the resulting solid in terms of $R$ and $\alpha.$
I first let $\overline{AB}$ be the $x$-axis. I know that I need to apply the formula for the surface area of a three-dimensional solid, which is $$2\pi \int_a^b f(x)\sqrt{1+(f'(x))^2}\,dx,$$ where $f(x)$ here is the equation of minor arc $AB$ and $a$ and $b$ are the $x$-coordinates of $A$ and $B,$ respectively. However, I am unsure how to find $f(x)$ or how to proceed with this problem.
Hint: Let $f(x) = x^2+y^2$. Then, we take the derivative of this, which gets us $$f'(x) = -\frac{x}{y}.$$ Now plug this into the integral. Finally, try and represent x and y in terms of R and $\alpha$, then plug THAT into the equation. To get rid of A and B, use the R and $\alpha$ versions of it as well.