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?

Note that\begin{align}\sqrt{1-x^2}\leqslant y\leqslant x&\iff1-x^2\leqslant y^2\leqslant x^2\\&\iff1\leqslant x^2+y^2\leqslant2x^2.\end{align}In polar coordinates, the final pair of inequalities becomes$$1\leqslant r^2\leqslant2r^2\cos^2\theta.$$So, take $\theta\in\left[0,\frac\pi4\right]$, so that $x,y\geqslant 0$ and that $\cos^2\theta\geqslant\frac12$. You also know that $r\geqslant1$. But you also know that $x\leqslant1$; in other words, $r\leqslant\frac1{\cos\theta}$. So, compute$$\int_0^{\pi/4}\int_1^{1/\cos\theta}1\,\mathrm dr\,\mathrm d\theta.$$
