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I have to solve the integral of the function:
$$f(x,y)=\frac{1}{x^2+y^2}$$ on this set: $T=\{(x,y): y^2\lt x\lt y\}$
I think to substitute with polar coordinates, but i can't find the integration limits.

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    For a point $(x,y) \in T$, we can arbitrary close to the origin and integrand tends to infinity as we do that. So it would be very plausible that your integral diverges, are you sure everything is written correctly? – 2017-01-15
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    Yes is all right in the question, i know only the result of this, not the method. Anyway the result is $\sqrt{2}-1$ – 2017-01-15
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    @Mazzorca Well, so far three members have told you the integral diverges. Of course, we could be mistaken...but it looks like it is *you* the one who's mistaken, and it really doesn't help you say "the result is $\;\sqrt2-1$" , since if the integral is really divergent then that is **not** the result... – 2017-01-15

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For all $(x,y) \in T$ you have $y^2 \le y$, so necessarily $y \in [0,1]$. As a consequence $$T= \{ (x,y) : y \in [0,1] , x \in [y^2, y] \}$$

Now, your integral is $$\int_0^1 \int_{y^2}^y \frac{1}{x^2+y^2} \mathrm{d}x \ \mathrm{d} y = \int_0^1 \left[ \frac{1}{y} \arctan \left( \frac{x}{y} \right)\right]_{x=y^2}^{x=y} \mathrm{d} y = \int_0^1 \frac{1}{y} \left( \arctan 1 - \arctan y\right) \mathrm{d}y$$

which is a divergent integral.

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    I know only the result, that are $\sqrt{2}-1$ – 2017-01-15
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Make a drawing, and get convinced that either

$$\color{red}{0\le x\le 1\;,\;\;x\le y\le \sqrt x}\;,\;\;\text{or}\;\;\color{green}{0\le y\le1\;,\;\;y^2\le x\le y}$$

Using now polar coordinates you want

$$\frac\pi4\le\theta\le\frac\pi2\;,\;\;\text{since}\;\;r\cos\theta\le r\sin\theta\le\sqrt{r\cos\theta}$$

and the first inequality gives

$$\tan\theta\ge1\implies \theta\ge\frac\pi 4\;,\;\;\text{whereas the second one gives}\;\;$$

$$r^2\cos^2\theta\le r\cos\theta\implies r\le\frac1{\cos\theta}$$

We get a divergent integral, and this can also be clearly seen (imo) in your original one...Are you sure of the integration domain?

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    Yes, only the equal T$=${$(x,y): y^2\lt x \lt y$} – 2017-01-15