Maybe I was too fast saying "easiest", arc-length works as well. My approach generalizes to arbitrary dimension though. As for why it's uniform, recall that wh

Maybe I was too fast saying "easiest", arc-length works as well. My approach generalizes to arbitrary dimension though. As for why it's uniform, recall that when X and Y are independent, then P(X and Y) = P(x)P(Y) ~ e^{-x^2-y^2} = e^{-r^2}, which is independent of the angle.