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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.

Thanks in advance.

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1 Answer 1

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? –  anorton 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
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