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I need to calculate the value of the integral: $$\int_T\frac 1 {\sqrt{x^2+y^2}} \, dx \, dy$$ where $T=\{(x,y) : x\in[-2,2], x^2

Specifically, I need to know how to set integration extremes.

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    $\int_{-2}^2 dx \int_{x^2}^4 dy \frac 1 {\sqrt{x^2+y^2}} $ – 2017-01-09

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The integration extremes are very easy to find:

$$T = \{(x, y): x \in [-2, 2], x^2 < y < 4\}$$

clearly tells you that $-2 \leq x \leq 2$. Therefore $x$ will range from $-2$ to $2$. Now imagine that $x$ is fixed at $x_0$. At that point, $x_0^2 \leq y \leq 4$. This means the integral becomes

$$\int_T\frac 1 {\sqrt{x^2+y^2}} \, dx \, dy = \int_{-2}^{2}\int_{x^2}^{4} \frac 1 {\sqrt{x^2+y^2}} \, dy \, dx$$

i.e. $y$'s extremes must come after $x$'s because the values for $y$ depend on the value of $x$.

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    Your last statement is very confusing: (all that follows i.e.) – 2017-01-09
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    @amWhy would you say it is better now? – 2017-01-09
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The integration limits are $$\int_{-2}^2\int_{x^2}^4\cdots dy\,dx.$$ But maybe changing to polar coordinates is convenient.

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    I try to change to polar coordinates, how change the integral in this case, i have $\int_T \frac{1}{\rho}\rho d\rho d\theta$ but $-2\le\rho cos\theta\ge 2$ and $\rho^2 cos^2\theta\lt\rho sin\theta\gt 4$ – 2017-01-10
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    Sorry, i wrong the verso $-2\le\rho cos\theta\le 2$ and $\rho^2 cos^2\theta\lt\rho sin\theta\lt 4$ – 2017-01-10
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You have to integrate over $y$ first with the bounds between $x^2$ and $4$. After that integrate over $x$ with the bounds between $-2$ and $2$.

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    I believe that is wrong. The order of the extremes cannot be that. – 2017-01-09
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    @RSerrao Even in your set-up, the *evaluation* of the integral is performed from "inside to outside," so that we integrate over y first, and then, we integrate over x (the outer.) There was no reason to downvote this answer. – 2017-01-09