# Calculate double integral

I need some help with this integral:

$$\iint_D \frac{x^2}{x^2 + y^2} \, dx \, dy$$ where $D= \{ (x,y) \in \Bbb R : 0 \leq x\leq 1, x^2\leq y\leq 2-x^2\}$

So first tried doing a variable change to polar coordinates ($x=rcos\theta$ and $y=rsen\theta$) but I got stuck on finding the limits of integration for $\theta$. I also tried ths variable change $x=\sqrt{v-u}$ and $y=\sqrt{v+u}$ but the Jacobian made the new function complicated.

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Welcome to Math.SE! Please consider updating your question with some information about what you have tried or where you are getting stuck. You will find people are much more willing to help if you do! – Nick Peterson Oct 20 '13 at 20:14

We make the transformation $u=x^{2},v=y$. Then the Jacobian is given by $(\frac{1}{2\sqrt{u}},0), (0,1)$ and the determinant is $\frac{1}{2\sqrt{u}}$. So we have

$$\frac{1}{2}\int^{1}_{0}\sqrt{u}\int_{u}^{2-u}\frac{1}{u+v^{2}}dudv$$

The inner integral $\int_{u}^{2-u}\frac{1}{u+v^{2}}dv$ can be computed by $\arctan[v/\sqrt{u}]*\frac{1}{\sqrt{u}}$. So we need to compute

$$\frac{1}{2}\int^{1}_{0}\arctan[(2-u)/\sqrt{u}]-\arctan[\sqrt{u}]$$

Both the integrals has a messy closed form. And an explicit computation can be found at here. I am not sure this is the best way to do this problem, though.

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That should be $dv\,du$, right? – apnorton Nov 8 '13 at 20:06
I guess if you insist one has to write it in a way to specify the order of integration, then yes. – Bombyx mori Nov 18 '13 at 0:03