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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?

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  • $\begingroup$ 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? $\endgroup$
    – zkutch
    Aug 14, 2021 at 17:31
  • $\begingroup$ 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 ? $\endgroup$
    – Hirak_93
    Aug 14, 2021 at 17:39
  • $\begingroup$ Sketch the domain for integral. Bounds are y=x and x=1 lines with circular arc. $\endgroup$ Aug 14, 2021 at 17:39

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

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  • $\begingroup$ Thanks for the explanation. $\endgroup$
    – Hirak_93
    Aug 14, 2021 at 18:29
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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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  • $\begingroup$ Thanks for the explanation . Actually , polar made it easier. $\endgroup$
    – Hirak_93
    Aug 14, 2021 at 18:29
  • $\begingroup$ Glad to be useful. $\endgroup$
    – zkutch
    Aug 14, 2021 at 18:53
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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 :)

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