[GeeTwo]

Originally Posted by Ether

 mode*     0.48 
*based on 100 bins, max slope of smoothed percentile vs length

I finished the polar integral up to r=1, found the mode to be (4+√(16-3π))/3 = .47859...

So thanks for the sanity check!

Other stats so far:

• The first part of the histogram curve (0..1) is 2πr - 8r² + 2r³.
• I have covered π/4 of the square (78.54%) by area
• I have integrated over π - 13/6 of the segments (97.49%).
• The numerator of the integral is up to 2π/3 - 8/5 = .494395..
• There were only three terms in the integral over Θ, the most complicated one being cos2Θ.
• I have only have four terms to integrate over r=[1..√2]. Two are powers of r, one looks familiar from the rectangular integration, and the last one looks strange, but probably no worse than anything in the rectangular problem.

Definitely an easier way to go unless that last one proves nastier than it looks.

Also, doing the integral this way explains the sudden change in behavior of the histogram at length=1.