<s>[INST] Need help to evaluate $\int\limits_{\frac{1}{\sqrt{2}}}^{1} \int\limits_{\sqrt{1-x^{2}}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$ I have this double integral $\int\limits_{\frac{1}{\sqrt{2}}}^{1} \int\limits_{\sqrt{1-x^{2}}}^{x} \frac{1}{\sqrt{x^2+y^2}}dydx$
I tried to transform into polar coordinates using $x = r\cos \theta$ , $y=r\sin\theta$ with $\left | J \right |= r$. Getting something like
$\int \int \frac{1}{r}rdrd\theta$ , but unable to define the upper and lower bounds of the integral.
Any help with that?
 [/INST] Bounds come from inequalities
$$\left\{\begin{array}{l}
\frac{1}{\sqrt{2}} \leqslant x  \leqslant 1 \\
\sqrt{1-x^{2}} \leqslant y \leqslant x
\end{array}\right\}$$
putting polar coordinates we obtain
$$\left\{\begin{array}{l}
\frac{1}{\sqrt{2}} \leqslant r \cos \theta  \leqslant 1 \\
\sqrt{1-r^{2}\cos^2 \theta} \leqslant r \sin  \theta \leqslant r \cos \theta
\end{array}\right\}$$
from first line we have two inequalities $\frac{1}{\sqrt{2} \cos \theta} \leqslant r $ and $ r \leqslant \frac{1}{\cos \theta}$.  Second line gives $1 \leqslant r$ and $\sin  \theta \leqslant \cos \theta$. Putting together we have
$$\int\limits_{0}^{\frac{\pi}{4}}\int\limits_{1}^{\frac{1}{\cos \theta}}$$
And, of course, it can be solved without polar coordinates using
$$\int \frac{1}{\sqrt{x^2+y^2}}dy=\ln\left(\frac{|y+\sqrt{x^2+y^2}|}{|x|} \right)+C$$
but, I prefer polar.
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