Finding the surface area of a minor arc rotated over a chord

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 $\alpha.$

This has been asked here. However, it has not received an answer yet. I found the distance between the arc and the chord to be $\sqrt{R^2 - x^2} - R \cos \alpha.$ From this, using the formula for surface area in Calculus, I simplified the answer to, $$2\pi\int_{-R\sin\alpha}^{R\sin\alpha}\left(\sqrt{R^2 - x^2} - R\cos\alpha\right)\cdot R\sqrt{\frac{1}{R^2 - x^2}}\, dx.$$

I do not know how to simplify this further, or whether it is even correct so far. Can someone please help me out with the answer? Thanks!

Solution 1:

Let's compute the formula you gave.

Noting that $$\int_{-R\sin(\alpha)}^{R\sin(\alpha)}\sqrt{R^2-x^2}\,R\,\sqrt{\frac{1}{R^2-x^2}}\,dx=2R^2\sin(\alpha),$$ and $$\int_{-R\sin(\alpha)}^{R\sin(\alpha)}R\cos(\alpha)\,R\,\sqrt{\frac{1}{R^2-x^2}}\,dx=R^2\cos(\alpha)\int_{-R\sin(\alpha)}^{R\sin(\alpha)}\frac{1}{\sqrt{R^2-x^2}}\,dx.$$ Using change of variable, we assume $x=R\sin(t)$ where $t\in[-\alpha,\alpha]$, then \begin{align*} \int_{-R\sin(\alpha)}^{R\sin(\alpha)}\frac{1}{\sqrt{R^2-x^2}}\,dx&=\int_{-\alpha}^\alpha\frac{1}{\sqrt{R^2-R^2\sin^2(t)}}\,R\cos(t)\,dt=2\alpha. \end{align*} Combine these result and plug into the formula you gave, we yield, $$S=2\pi\,\left(2R^2\sin(\alpha)-2R^2\alpha\cos(\alpha)\right)=4\pi R^2\big(\sin(\alpha)-\alpha\cos(\alpha)\big).$$