Expected area covered by circles inside a square $n$ circles of radius $r$ are to be placed randomly inside a square of side length $l$. What is the expected area the circles will enclose?
Edit: As @Mirko suggested, we define random placement of the circles by picking points from a square of side length $l-2r$ concentric to the initial one from a uniform distribution and using each one as a center for a specific circle.
The problem I'm facing is that I do not know how to statistically account for overlaps with the circles. I have tried to divide the square into little square "cells" of side $2r$ such that each circle can exist in one of the cells and obtained an expression for the probability of $k$ circles to exist with no overlap. However, this approach drastically understates the actual probability as it doesn't account for circles existing in between cells.
Any alternative approach would be greatly appreciated.
 A: So we're assuming $l\ge2r$. Let's take the square in Cartesian coordinates to be $S:=[-l/2,\,l/2]^2$. The point $(x,\,y)$ therein falls within a circle centred at $(u,\,v)\in S^\prime:=[-(l/2-r),\,l/2-r]$ iff$$(x-u)^2+(y-v)^2\le r^2.$$Let $f(x,\,y)$ denote the probability of this when we take $U,\,V$ to be Uniform IIDs on their support; the same treatment of $X,\,Y$ makes the probability a random point lies in at least one of the circles$$1-\int_S(1-f)^ndxdy,$$so multiplying this by $\pi l^2$ gives the circles' union's mean area. Note the $n\to\infty$ limit is less than $\pi l^2$ because some points have $f=0\implies\lim_{n\to\infty}(1-f)^n=1$. In particular, a point in $S$ may be as far as $r\sqrt{2}$ from the nearest point in $S^\prime$.
So now we have two tasks: compute $f$, then compute $\int_S(1-f)^ndxdy$. Note $f$ is the proportion of the area of $S^\prime$ within a distance $r$ of $(x,\,y)$. While obtaining $f$ as a piecewise function is feasible, I suspect the integrals are numerical for large $n$.
