How to proceed with this integral? 
Let $ G:= \left\{ (x,y) \in \mathbb{R}^2 : 0 < y,\: x^2 + \frac{y^2}{9} <1\: ,\: x^2+y^2 > 1 \right\} $.
I want to calculate this integral:
$ \displaystyle\int_G x^2\,dxdy $.

I want to try with polar coordinates:
so I set $ (x,y) = (r\cos\phi,r\sin\phi)$, but
I am not sure how to get the right boundaries for $\phi $. Isn't it $ x^2 +y^2 = r^2 $ ?
Any help is very appreciated !
 A: Even though I believe that the other answers are the best way to go, I think that you can use polar coordinates if you wish.
Since $y>0$, we are integrating over the first two quadrants, so that $0 \leq \phi \leq \pi$.
Since $x^2+y^2> 1$, we have $r > 1$, and since $x^2+y^2/9 < 1$, we see that
$$ r^2\cos^2\theta + \frac{r^2\sin^2\theta}{9} < 1 \quad \Rightarrow \quad   r < \frac{3}{\sqrt{8\cos^2\theta+1}}.  $$
Thus
\begin{align}
\int_0^\pi \int_{1}^{\frac{3}{\sqrt{8\cos^2\theta+1}}} r^2\cos^2\theta \cdot r \,\mathrm dr \, \mathrm d\theta &= \frac{81}{4} \int_0^\pi \frac{\cos^2\theta}{(8\cos^2\theta+1)^2} \, \mathrm d\theta - \frac{1}{4} \int_0^\pi \cos^2\theta \, \mathrm d\theta = \frac{\pi}{4}.
\end{align}
A: It is not a good idea to use polar coordinates. The integral can be written as $\int_{-1}^{1}\int_{\sqrt{1-x^{2}}} ^{3\sqrt{1-x^{2}}}x^{2}\, dy\, dx=\int_{-1}^{1}2\sqrt {1-x^{2}}x^{2}\,  dx$. To evaluate this put $x=\sin\, \theta$ and use the formulas $2\sin\, \theta \cos\, \theta =\sin\, 2\theta$, $2\sin^{2}\, 2\theta =1-\cos (4\theta)$. 
A: Why in polar coordinates? Maybe it is easier by cartesian coordinates. Note that in $G$, $-1\leq x\leq 1$ and 
$$\sqrt{1-x^2}<y<3\sqrt{1-x^2}.$$
Therefore
$$\int_G x^2\,dxdy=\int_{x=-1}^1x^2\left(3\sqrt{1-x^2}-\sqrt{1-x^2}\right)dx$$
Can you take it from here?
