<s>[INST] If $S$ is the unit square with opposite corners $(0, 0)$ and $(1, 1)$, show $\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}=\log(1+\sqrt{2})$ Let $S$ be the unit square with opposite corners $(0, 0)$ and $(1, 1)$. Show $$\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}=\log(1+\sqrt{2}).$$ 
What I tried:
Since $\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}$ converges in $S$. So the integral is convergent independent of choice of ${D_n}$, the let $D_n$ be square with opposite corners $(1/n, 1/n)$ and $(1, 1)$.
Then
\begin{align}
\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}&=\int^1_{1/n}\int^1_{1/n}\frac{y^2}{\sqrt{(x^2+y^2)^3}}
\\&=\int^1_{1/n}y^2dy\int^1_{1/n}\frac{1}{(x^2+y^2)^{2/3}}dx
\\&=\int^1_{1/n}y^2\frac{x}{y^2\sqrt{y^2+x^2}}|^1_{1/n}dy
\\&=\log(y+\sqrt{1+y^2}|^1_{1/n}-\frac{1}{n}\log(y+\sqrt{y^2+(1/n)^2})|^1_{1/n}
\\&=\log(1+\sqrt{2})-0-0+\frac{1}{n}\log\left(\frac{1}{n}+\sqrt{\frac{1}{n}^2+\frac{1}{n}^2}\right)
\end{align}
But $\frac{1}{n}\log(\frac{1}{n}+\sqrt{\frac{1}{n}^2+\frac{1}{n}^2})$ seem to go to infinity.
 [/INST] Under the variable interchange $x\leftrightarrow y$ we have that
$$I = \iint\limits_{[0,1]^2} \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}}dA = \iint\limits_{[0,1]^2} \frac{x^2}{(x^2+y^2)^{\frac{3}{2}}}dA$$
and adding the two integrals gets us
$$2I = \iint\limits_{[0,1]^2} \frac{x^2+y^2}{(x^2+y^2)^{\frac{3}{2}}}dA \implies I = \frac{1}{2} \iint\limits_{[0,1]^2} \frac{1}{\sqrt{x^2+y^2}}dA$$
By symmetry we can do the integral in polar coordinates
$$I = 2\cdot\frac{1}{2}\int_0^1 \int_0^x \frac{1}{\sqrt{x^2+y^2}}dA = \int_0^{\frac{\pi}{4}}\int_0^{\sec\theta}\:dr\:d\theta$$
$$ = \log|\sec\theta+\tan\theta|\Biggr|_0^{\frac{\pi}{4}} = \log(1+\sqrt{2})$$
</s>