# 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?

• I edited bounds for integral in question - look, please, is it what you are asked about? and why $\left | J \right |= 1$? is it polar Jacobian? Aug 14, 2021 at 17:31
• Yes, this is what I asked about. Sorry for typing $\left | J \right |$ value wrong. I found $\left | J \right | = r$ using polar coordinates. Can it be solved without using polar coordinates ? Aug 14, 2021 at 17:39
• Sketch the domain for integral. Bounds are y=x and x=1 lines with circular arc. Aug 14, 2021 at 17:39

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$$\int_0^{\pi/4}\int_1^{1/\cos\theta}1\,\mathrm dr\,\mathrm d\theta.$$

• Thanks for the explanation. Aug 14, 2021 at 18:29

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.

• Thanks for the explanation . Actually , polar made it easier. Aug 14, 2021 at 18:29
• Glad to be useful. Aug 14, 2021 at 18:53

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 :)