Convexity Coefficient

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Convexity Coefficient
$\chi(D)$是连接区域 $D$ 中两个随机点的线段完全包含在 $D$ 内的概率。对于凸区域，$\chi(D)=1$.
如何证明 $\chi={2a^2\cos^{-1}(b/a)-2b\sqrt{a^2-b^2}\over\pi(a^2-b^2)} $ 对于内径为$b$, 外径为$a$的圆环〇

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\begin{align*}
\chi&=\mathbb E\left[\frac{阴影面积(y,\theta)}{圆环面积}\right]\\
&=\frac1{圆环面积}{\iint \frac{阴影面积(y,\theta)}{圆环面积}dS}\\
&=\frac1{圆环面积^2}{\iint 阴影面积(y,\theta)\,dS}\\
&=\frac1{\pi^2(R^2-r^2)^2}\int^{2\pi}_0d\theta\int_r^R\left[\left(R^{2}-r^{2}\right)\arccos \left(\frac{r}{y}\right)+R^{2} \arccos \left(\frac{r}{R}\right)-r \sqrt{R^{2}-r^{2}}\right]ydy\\
&=\frac1{\pi^2(R^2-r^2)^2}2\pi\int_r^R\left[\left(R^{2}-r^{2}\right)\arccos \left(\frac{r}{y}\right)+R^{2} \arccos \left(\frac{r}{R}\right)-r \sqrt{R^{2}-r^{2}}\right]ydy\\
&=\frac2{\pi(R^2-r^2)^2}\left\{\left(R^{2}-r^{2}\right)\int_r^R\arccos \left(\frac{r}{y}\right)ydy+\frac{R^2-r^2}2\left[R^{2} \arccos \left(\frac{r}{R}\right)-r \sqrt{R^{2}-r^{2}}\right]\right\}\\
&=\frac2{\pi(R^2-r^2)^2}\left\{\left(R^{2}-r^{2}\right)\frac{1}{2} \left[R^2 \arccos\left(\frac{r}{R}\right)-r \sqrt{R^2-r^2}\right]+\frac{R^2-r^2}2\left[R^{2} \arccos \left(\frac{r}{R}\right)-r \sqrt{R^{2}-r^{2}}\right]\right\}\\
&=\frac2{\pi(R^2-r^2)}\left[R^2 \arccos\left(\frac{r}{R}\right)-r \sqrt{R^2-r^2}\right]
\end{align*}推广到3D呢