Points in a square puzzle

I'm going to take a random variable + calculus approach here. Let us define $|x|, |y|$ as the distances between the x & y-coordinates of our two randomly chosen points in $(0,1)$ . These random variables then follow a Triangular Distribution (instead of a Uniform Distribution):

$|x|, |y| \Rightarrow f_{|x|} = 2(1-|x|), f_{|y|} = 2(1-|y|)$ for $|x|, |y| \in (0,1)$

These translate into the following double integral:

$A = 4\displaystyle{\int_{0}^{1} \int_{0}^{1} \sqrt{|x|^2 + |y|^2}(1-|x|)(1-|y|) \mathrm{d}|y| \mathrm{d}|x|}$

which can be converted into cylindrical coordinates: $|x| = r cos\theta, |y| = r sin\theta, d|x|d|y| = rdrd\theta, r \in [0, sec\theta], \theta \in[0,\frac{\pi}{4}]$ , or simply:

$A = 8\displaystyle{\int_{0}^{\frac{\pi}{4}} \int_{0}^{sec\theta} r \cdot (1-r cos\theta)(1-r sin\theta) r \mathrm{d}r \mathrm{d}\theta}$ ; (NOTE: we need to double our integral to accommodate both halves of our unit square)

or $8\displaystyle{\int_{0}^{\frac{\pi}{4}} \int_{0}^{sec\theta} r^2 - (cos\theta + sin\theta)r^3 +(cos\theta sin\theta)r^4 \mathrm{d}r \mathrm{d}\theta}$ ;

or $8\displaystyle{\int_{0}^{\frac{\pi}{4}} \frac{sec^{3}\theta}{3} - (cos\theta + sin\theta)\frac{sec^{4}\theta}{4} +(cos\theta sin\theta)\frac{sec^{5}\theta}{5} \mathrm{d}r}$

or $8\displaystyle{\int_{0}^{\frac{\pi}{4}} \frac{sec^{3}\theta}{3} - \frac{sec^{3}\theta}{4} - \frac{tan\theta \sec^{3}\theta}{4} + \frac{tan\theta sec^{3}\theta}{5} \mathrm{d}r}$

or $8\displaystyle{\int_{0}^{\frac{\pi}{4}} \frac{sec^{3}\theta}{12} - \frac{tan\theta sec^{3}\theta}{20} \mathrm{d}r}$

or $8 \cdot [\frac{1}{24} \cdot [tan\theta sec\theta - ln[cos(\frac{\theta}{2}) - sin(\frac{\theta}{2})] + ln[cos(\frac{\theta}{2}) + sin(\frac{\theta}{2})]] - \frac{sec^{3}\theta}{60}]$ for $\theta \in [0, \frac{\pi}{4}];$

or $A = \frac{(2 + \sqrt{2}) + 5ln(1 + \sqrt{2})}{15} \approx \boxed{0.521}.$