Brooks Moses

Hmm. After poking at that in Mathematica a bit to evaluate the integrals, I can confirm that, yes, that does indeed seem to work. (Well, ok, it doesn't work as written because the limit should be (x²+y²)=1 rather than (x²+y²)=0, but close enough.)

Writing the code to do all the special cases for the (x²+y²)=1 arc intersecting the rectangle isn't going to be easy, I suppose, but then it wasn't ever going to be -- and it is at least quite possible.

Thanks muchly, masked commenter, whomever you are!