Geometric Solution to Calculus Problem Given a rectangle $A$ of sides $L$ and $W$, such that $L$ is less than $W$, what is the average area of all circles entirely contained in $A$?  To be more explicit, what is the average area of all circles of all radii with centers located inside of $A$ such that each circle intersects any given side of A at most once? 

I used calculus to find the solution to this problem and got that this average area is equal to 

$$(4W-3L)(\pi L^2)(72W)^{-1} .$$

However, I was wondering if there exists a simpler, geometric interpretation to this problem.

EDIT:  Thanks to joriki, the 36 was corrected to be 72.
I have not thought about an analytic solution but a simulation approach seems straightforward. 

First, sample a point by drawing a uniform random variable $c_x \sim U(0.L)$ and $c_y \sim U(0,W)$ where $(c_x, c_y)$ represents the center of the circle.

Second, given the center, sample a radius using a uniform distribution as: $r \sim U(0,min(c_x,L-c_x,c_y,W-c_y))$.

Third, compute the area of the circle and store it.

Fourth, repeat steps 1 and 2 lots, lots of times and compute the empirical average.

The empirical average should coincide with your calculus based solution provided there are no mistakes in your math and the simulation algorithm/code.
I have not thought about an analytic solution but a simulation approach seems straightforward. 

First, sample a point by drawing a uniform random variable $c_x \sim U(0.L)$ and $c_y \sim U(0,W)$ where $(c_x, c_y)$ represents the center of the circle.

Second, given the center, sample a radius using a uniform distribution as: $r \sim U(0,min(c_x,L-c_x,c_y,W-c_y))$.

Third, compute the area of the circle and store it.

Fourth, repeat steps 1 and 2 lots, lots of times and compute the empirical average.

The empirical average should coincide with your calculus based solution provided there are no mistakes in your math and the simulation algorithm/code.