# Is the area the integral of the perimeter?

Let $A$ be a compact subset of the plane with piecewise smooth boundary. Define $A_r$ to be the set $\{x\in \mathbb{R}^2: dist(x,A)\leq r\}$, i.e. the points with distance at most $r$ from $A$. Motivate by the case of a circle, I was wondering if the following formula holds

$$|A_r|=|A|+ \int_{0}^r L(\partial A_s) ds,$$ where $| |$ denotes the area and $L(\partial A_s)$ the perimeter of $A_s$. Does this formula hold in general? What if $A$ is a convex domain or perhaps a finite union of disks?

Let $A$ be compact convex in $\mathbb{R}^2$, and let other notations be as in the question. Then Jacob Steiner (1840) proved:
\begin{align} \left\lvert A_r\right\rvert&=\left\lvert A\right\rvert+L(\partial A)r+\mathrm{\pi}r^2\\ L(\partial A_r)&=L(\partial A) +2\mathrm{\pi}r\text{.} \end{align} For compact convex domains, your equality is an immediate consequence. Nowadays, this Steiner formula is considered to be an early result in integral geometry.
In case $A$ is convex, a formula for $|A_r|$ is $$|A_r| = |A| + r L(\partial A) + \pi r^2$$ and a formula for $L(\partial A_s)$ is $$L(\partial A_s) = L(\partial A) + 2\pi s.$$
If $A$ is a finite union of disks things go bad once the disks comprising $A_r$ start to overlap. In the case of general $A$ I don't think such a formula is valid.