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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?

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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.

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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 $$L(\partial A_s) = L(\partial A) + 2\pi s.$$

Your formula arises quickly from here. These are classical theorems of convex geometry. You should be able to find a citation.

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.

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  • $\begingroup$ Since the disks might overlap the quantity on the right should be an upper bound right? $\endgroup$ Feb 26, 2018 at 1:58
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    $\begingroup$ @ChrisApostol : I vaguely recall that the theory is still tractable for finite unions of closed convex domains. For further reading, see Klain and Rota's Introduction to Geometric Probability, $\endgroup$
    – K B Dave
    Feb 26, 2018 at 22:47

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