We want to find the area of a domain with piecewisely smooth boundary by using the coordinates $(p,\theta)$ of the random line:
- It has been known that every straight line $\ell$ on $R^2$ can be uniquely represented by the coordinates $(p,\theta)$, where $p$ is the distance from the line $\ell$ to the origin, $\theta$ is the angle from $x$-axis to the line.
- The area of the domain $D$ bounded by the piecewisely smooth curve can be calculated by the integral $$2\pi\cdot Area(D)= \int\sigma dp\wedge d\theta,$$ where $\sigma=$ the length of the chord segment intersected by the line $\ell$ and $D$; the integral go over all lines such that $\ell\cap D\neq \emptyset$.
Does anyone know how to precisely prove this ?
Roughly speaking, for the integral we fix the angle $\theta$ first, then $\sigma \Delta p$ is the small area of the rectangle varying near the chord segment. When integrate w.r.t. $p\in R$, we have the $Area(D)$, then integrate w.r.t. the angle $\theta\in [0,2\pi]$, we got $2\pi\cdot Area(D)$. But how to write the proof more precisely?
Thanks for your help.