Expected area of the intersection of two circles If we pick randomly two points inside a circle centred at $O$ with radius $R$, and draw two circles centred at the two points with radius equal to the distance between them, what is the expected area of the intersection of the two cirlces that contain the origin $O$.
 A: Let $\vec{x}_1$ and $\vec{x}_2$ be the two points. Let $r = |\vec{x}_1 - \vec{x}_2|$ be the distance between them. By elementary geometry, if you draw two circle of radius $r$ using these two points as center, the area of their intersection is given by  $(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})r^2$. Notice the picking of two points are independent, we have:
$$E\left[ \vec{x}_1 \cdot \vec{x}_2 \right] = E\left[\vec{x}_1\right] \cdot E\left[\vec{x}_2\right] = \vec{0} \cdot \vec{0} = 0$$
This implies
$$E\left[|\vec{x}_1 - \vec{x}_2|^2\right] = E\left[|\vec{x}_1|^2 + |\vec{x}_2|^2\right] = 2\frac{\int_0^R r^3 dr}{\int_0^R rdr} = R^2$$
As a result, the expected area of the intersection is $(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})R^2$.
Update for those who are curious
Let $\mathscr{C}$ be the set of events such that the intersection contains the origin, then:
$$\begin{align}
\operatorname{Prob}\left[\,\mathscr{C} \right] &= \frac{2\pi + 3\sqrt{3}}{6\pi}\\
E\left[\,|\vec{x}_1 - \vec{x}_2|^2 : \mathscr{C}\right] &= \frac{20\pi + 21\sqrt{3}}{6(2\pi + 3\sqrt{3})}
\end{align}$$
and the expected area of intersection conditional to containing the center is given by:
$$\frac{(4\pi - 3\sqrt{3})(20\pi + 21\sqrt{3})}{36(2\pi + 3\sqrt{3})}$$
To evaluate $E\left[ \varphi(\vec{x}_1,\vec{x}_2) ) : \mathscr{C} \right]$ for any function $\varphi( \vec{x}_1, \vec{x}_2 )$ which is symmetric and rotational invariant w.r.t its argument, you need to compute an integral of the from:
$$\int_{\frac{\pi}{3}}^{\pi} \frac{d\theta}{\pi} \left[2\int_{0}^{R} \frac{2udu}{R^2} \left( \int_{\alpha(\theta)u}^{u}  \frac{2vdv}{R^2} \phi( \vec{x}_1, \vec{x}_2 )  \right) \right] $$
where $u \ge v$ are $|\vec{x}_1|$ and $|\vec{x}_2|$ sorted in descending order.
$\theta$ is the angle between $\vec{x}_1$ and $\vec{x}_2$. The mysterious $\alpha(\theta)$ is $\max(2\cos(\theta),0)$ for $\theta \in [\frac{\pi}{3},\pi]$.
The integral is a big mess and I need a computer algebra system to crank that out.
I won't provide more details on this part not relevant to the main answer.
A: May be you can try to integrate the Circular Segment the half one and then multiply by 2 (since the R of each circles are the same).
Since minimum angle when  the intersection happen exactly at each center point, so the minimum angle is 120 degree, and the maximum angle is exactly 180 degree.
Based on that, we  just put that into integration :
$$E[A] =2\int_{2\pi/3}^\pi \frac{R^2}2(\theta-sin\theta)d\theta$$
Where E[A] is expected area, then just integrate that formula like basic integral.
thanks. hope that can help, if not someone can fix that.
thanks
