|
|
\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 |
|
