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?
 A: 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$$\int_0^{\pi/4}\int_1^{1/\cos\theta}1\,\mathrm dr\,\mathrm d\theta.$$
A: 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.
A: you have:
$$D_1=\left\{x,y\in\mathbb R\,|\,1/\sqrt{2}\le x\le 1,\wedge\sqrt{1-x^2}\le y\le x\,\right\}$$
drawing this out notice that which is effectively the same as:
$$D_2=\left\{0\le x\le 1\,\wedge\,0\le y\le x\right\}/\left\{1/\sqrt{2}\le x\le 1\,\wedge\, 0\le y\le \sqrt{1-x^2}\right\}$$
both parts of which are easy  to define in polar coordinates.

The second part will be:
$$0\le \theta\le\pi/4\,\wedge\,0\le r\le 1$$
whilst the first part can be worked out just using some trigonometry, I will leave that to you :)
