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If $n$ circles of radius $r$ are drawn at random in the plane such that their centers lie in some region $S$ (we could take $S$ to be a square), what is the expected area $E$ of their union?

Edit: In light of Noam's comment, it seems like the answer is

$\displaystyle \int \int 1- (1-f(x,y))^n dx dy$

where $f(x,y) = a(B_r(x,y) \cap S)/a(s)$; the probability a random circle contains $(x,y)$.

If the region $S$ is large compared to $r$ and it has a reasonably "out of the way" boundary, then $f(x,y)$ is approximately constant over all of $S$, and is about $\pi r^2/ a(S)$, and so the expected value $E$ for a fixed $n$ is $n\pi r^2 + O(r^4)$.

However, the more interesting cases occur when $r$ is not small or when $\partial S$ gets "in the way" significantly.

Suppose $S$ is a square of side length $s$ and the circles have radius $r=s/2$. How does $E$ grow with $n$ asymptotically? What if $S$ is a circle of radius $s$? A regular $k$-gon inscribed in a circle of radius $s$?

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Compute the probability $p$ that a given point of $S$ is not hit by any of the circles (which is constant away from the edge), and multiply $1-p$ by the area of $S$. – Noam D. Elkies Sep 3 '12 at 15:58
@Noam's comment is obviously the answer, with the one caveat that a non-asymptotic result will be a little tricky. – Igor Rivin Sep 3 '12 at 16:13
You will probably get a more "natural" answer if you choose a "torus", i.e., identify opposite edges of a square, to eliminate edge effects. – Dick Palais Sep 3 '12 at 23:24

The question is already interesting for general $r$ when $S=[0,1]$. Then the area of $E$ is $\pi r^2 + 2 r$. Because the area defect to $E$ of two circles being $d$ apart is $\frac{d^3}{12~ r}$, and $d\approx n^{-1}$, for large $n$ only the distance between minimum $a$ and maximum $b$ of the circle centers in $S$ determines the asymptotics. The independent expectation values for $a$ and $b$ are $1/(n+1)$ and $1-1/(n+1)$ giving an asymptotic area defect to $E$ of $4r/(n+1)$.

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