Finding solution to a double integral Well, here it is
$$\int\int_{[0,1]^{2}}\sqrt{1+x^2+y^2}\mathrm dx\mathrm dy\tag*{}$$
Maybe there's a really nice way of doing it, hopefully someone knows. Good luck!

We can do this integral in polar coordinates by recognizing a symmetry - divide the square in half by the line $y=x$. The integral on the top half will equal the integral on the bottom half, so we will do one of the integrals and multiply its value by 2.
$$\iint_{[0,1]^2}\sqrt{1+x^2+y^2}dA = \int_0^{\pi/4} \int_0^{\sec\theta}2r\sqrt{1+r^2}drd\theta = \frac{2}{3}\int_0^{\pi/4} (1+\sec^2\theta)^{3/2} - 1 d\theta$$
$$= \frac{2}{3}\int_0^{\pi/4} (1+\sec^2\theta)^{3/2} d\theta - \frac{\pi}{6}$$
Now focusing on the integral left over, do the following substitution $$1+\sec^2\theta = 2\cosh^2 t$$ $$\sec^2\theta \tan\theta d\theta = 2\cosh t \sinh t dt \implies d\theta = \frac{\sqrt{2}\cosh t}{\cosh 2t}dt$$ The integral becomes:
$$\frac{8}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} \frac{\cosh^4 t}{\cosh 2t}dt = \frac{8}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} \frac{\cosh^2 t + \cosh^2 t \sinh^2 t}{\cosh 2t}dt$$ $$ = \frac{2}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} \frac{2+ 2\cosh 2t + \sinh^2 2t}{\cosh 2t}dt 
$$$$= \frac{2}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} 2\text{ sech } 2t + 2 + \tanh 2t \sinh 2tdt$$
Integrating tanh sinh by parts, we get
$$\frac{2}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} (\text{sech } 2t + 2)dt + \frac{1}{3}\sinh 2t = \frac{1}{3}\left[\tan^{-1}(\sinh 2t) + 4t + \sinh 2t \right]_{0}^{\cosh^{-1}(\sqrt{3/2})}$$ $$= \frac{1}{3}\left[\tan^{-1}(2u\sqrt{u^2-1}) + 4\cosh^{-1}(u) + 2u\sqrt{u^2-1}\right]_{1}^{\sqrt{3/2}} = \frac{1}{3}\left[\tan^{-1}(\sqrt{3}) + 4\cosh^{-1}(\sqrt{3/2}) + \sqrt{3}\right]$$
Simplifying and subtracting off the term from earlier, we get 
$$\frac{4}{3}\cosh^{-1}\left(\sqrt{\frac{3}{2}}\right) + \frac{1}{\sqrt{3}} - \frac{\pi}{18}$$
