I need to calculate the volume of a circular hole in a cylinder and I've come across a problem. The problem is finding the "cap-volume", which is needed to complete the volume of the hole. I created a quick example of the problem.
Cylinder which will be drilled
Hole. The cap is shown with the magenta coloured curves
Let $R$ and $r$ be the radius of the cylinder and the hole. We will assume $k = \frac{r}{R} < 1$.
We will further assume the hole is drilled toward the center and perpendicular to the axis of the cylinder. Under these assumptions, the volume of the cap is given by
$$\begin{align} \verb/Vol/_{cap} &= 4 \int_0^r \int_0^{\sqrt{r^2-x^2}} \left( \sqrt{R^2-x^2} - \sqrt{R^2 - r^2} \right) dy dx\\ &= 4 \int_0^r \sqrt{r^2 - x^2}\left(\sqrt{R^2 - x^2} - \sqrt{R^2-r^2}\right) dx\\ &= 4Rr^2 \left[ \int_0^1 \sqrt{(1-x^2)(1-k^2x^2)} dx - \frac{\pi}{4}\sqrt{1-k^2}\right]\\ &= 4Rr^2 \left[\frac{(1+k^2)E(k) -(1-k^2)K(k)}{3k^2} - \frac{\pi}{4} \sqrt{1-k^2}\right] \end{align} $$ where
$$\begin{align} K(k) &= \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-k^2\sin^2\theta}} d\theta = \int_0^1 \frac{1}{\sqrt{(1-k^2x^2)(1-x^2)}} dx \\ E(k) &= \int_0^{\frac{\pi}{2}} \sqrt{1-k^2\sin^2\theta} d\theta = \int_0^1 \sqrt{\frac{1-k^2x^2}{1-x^2}} dx \end{align}$$
are the complete elliptic integrals of the first and second kind respectively.
Similarly, the volume of the hole is given by
$$\verb/Vol/_{hole} = 8Rr^2 \left[\frac{(1+k^2)E(k) -(1-k^2)K(k)}{3k^2} \right]$$