The Overlapping Circles Problem

The overlapping section common to two circles is known as a lens . In this case, the area of the lens formed by $C_0$ and $C_1$ (when $x\ge0$ ) is

$f(x)=\pi - 2\tan^{-1} \frac{x}{\sqrt{4-x^2}} - \frac{x \sqrt{4-x^2}}{2}$

or, slightly more conveniently,

$f(x)=\pi - 2\sin^{-1} \frac{x}{2} - \frac{x \sqrt{4-x^2}}{2}$

Since $\frac{\pi}{2}-sin^{-1} \frac{x}{2} = cos^{-1} \frac{x}{2}$ , this is just

$f(x)=2\cos^{-1} \frac{x}{2} - \frac{x \sqrt{4-x^2}}{2}$

Note that this is only valid for $x\ge0$ (this is due to a signed area hidden in the last term). However, we clearly have $f(x)=f(-x)$ , so we need to find

$I=\int_{-2}^2 f(x) dx = 2\int_0^2 f(x) dx = \int_0^2 \left[ 4\cos^{-1} \frac{x}{2} - x \sqrt{4-x^2} \right] dx$

The integration tricks we need are integration of an inverse function (for the first term) and substitution (eg $u=4-x^2$ ) for the second term:

$I=8-\frac{8}{3} = \frac{16}{3}$

leading to the answer $16^3=\boxed{4096}$ .

Consider the diagram above, where $O$ denotes the centre of $C_{0}$ and $O’$ denotes the centre of $C_{1}$ . The area of the sector is equal to $($ sector $OAB) -$ $($ triangle $OAB) +$ $($ sector $O’AB) -$ $($ triangle $O’AB)$ .

Length $OP$ is equal to $\frac{x}{2}$ and since $OA=1$ , $AP=\sqrt{1-\frac{x^{2}}{4}}$ . Angle $AOP=arccos(\frac{x}{2})$ and therefore angle $AOB=2arccos(\frac{x}{2})$ .

So sector $OAB=arccos(\frac{x}{2})$ and triangle $OAB=\frac{x}{2}\sqrt{1-\frac{x^{2}}{4}}$ . The area of the overlapping section, $f(x)$ , is twice the difference between the sector and the triangle and is therefore $2arccos(\frac{x}{2})-x\sqrt{1-\frac{x^{2}}{4}}$ .

Integrate this expression from $0$ to $2$ to get $\frac{8}{3}$ . So the integral from $-2$ to $2$ would be $\frac{16}{3}$ as $f(x)$ is symmetric across the y-axis. This yields $m=16$ and $n=3$ . $m^{n}=16^{3}=\boxed{4096}$ , the desired result.