<s>[INST] Integrate $\sqrt{1+x^2+y^2}$ Calculate $$I=\int_{-1}^1\int_{-1}^1\sqrt{1+x^2+y^2}\,\mathrm{d}y\,\mathrm{d}x.$$
It's a problem from a book about calculus. My attempt:
$$\begin{align}
I &= \int_{-1}^1\int_{-1}^1\sqrt{1+x^2+y^2}\,\mathrm{d}y\,\mathrm{d}x \\
  &= \int_{-1}^1\left.\frac{x}{2}\sqrt{1+x^2+y^2}+\frac{1+y^2}{2}\log\left(x+\sqrt{1+x^2+y^2}\right)\right|_{-1}^{1}\,\mathrm{d}y \\
  &= \int_{-1}^1\sqrt{2+y^2}+\frac{1+y^2}{2}\left(\log \left(\sqrt{2+y^2}+1\right)-\log\left(\sqrt{2+y^2}-1\right)\right)\,\mathrm{d}y \\
  &= \sqrt{3}+2\operatorname{arsinh}\frac{1}{\sqrt{2}}+\int_{-1}^1(y^2+1) \operatorname{arsinh} \frac{1}{\sqrt{y^2+1}}\,\mathrm{d}y \\
  &= \color{red}\ldots \\
  &= -\frac{2}{9} (\pi + 12 \log 2 - 6 \sqrt{3} - 24 \log (1+ \sqrt{3}))
\end{align}$$
(answer taken from the solutions, no idea how to reach it).
[edit] Here is an attempt with polar coordinates. Due to symmetry, it's enough to integrate over $0 \le x \le 1$ and $0 \le y \le x$, 1/8th of the initial square.
$$\begin{align}
I &= 8\int_0^{\pi/4} \int_0^{1/\cos \theta}r \sqrt{1+r^2}\,\mathrm{d}\theta\\
&= 8\int_0^{1/\cos \theta} \frac{(1+1/\cos^2\theta)^{3/2}-1}{3}\,\mathrm{d}\theta\\
&= {?}
\end{align}$$
 [/INST] Not an answer, but a start. Might not work.
In polar coordinates, it is $$4\int_{0}^{\sqrt2} f(r)\sqrt{1+r^2}\,dr$$ where $f(r)=\pi r/2$ for $0\leq r\leq 1.$
When $1<r\leq \sqrt2,$ we need $f(r)$ to be the length of the arc of radius $r$ around $(0,0)$ contained in the square $[0,1]^2.$
This is $$f(r)=r\left(\frac\pi2-2\arctan\left(\sqrt{r^2-1}\right)\right).$$
So the integral is:
$$4\frac{\pi}2\int_0^1 r\sqrt{1+r^2}\,dr+4\int_1^{\sqrt2}r\left(\frac\pi2-2\arctan\left(\sqrt{r^2-1}\right)\right)\sqrt{1+r^2}\,dr$$
The first integral is easy. Don't know what to do about the second, however.
When $x\leq 1,$ $2\arctan x =\arctan\frac{2x}{1-x^2}$ and $\frac{\pi}2-\arctan(y)=\arctan(1/y).$
So $$\frac{\pi}2-2\arctan\left(\sqrt{r^2-1}\right)=\arctan\left(\frac{2-r^2}{2\sqrt{ r^2-1}}\right)$$
If we set $r=\sec\theta,$ then $$\arctan\left(\frac{2-r^2}{2\sqrt{ r^2-1}}\right)=\frac\pi2-2\theta.$$
So the integral becomes:
$$\int_{0}^{\pi/4}\left(\frac\pi2-2\theta\right)\sec^2\theta\tan(\theta)\sqrt{1+\sec^2\theta}\,d\theta$$
</s>