Vrftsjtryk

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:
$ displaystyleint_G x^2,dxdy $.

I want to try with polar coordinates:
so I set $ (x,y) = (rcosphi,rsinphi)$, 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 !

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
$$ r^2cos^2theta + frac{r^2sin^2theta}{9} < 1 quad Rightarrow quad r < frac{3}{sqrt{8cos^2theta+1}}. $$

Thus
begin{align}
int_0^pi int_{1}^{frac{3}{sqrt{8cos^2theta+1}}} r^2cos^2theta cdot r ,mathrm dr , mathrm dtheta &= frac{81}{4} int_0^pi frac{cos^2theta}{(8cos^2theta+1)^2} , mathrm dtheta - frac{1}{4} int_0^pi cos^2theta , mathrm dtheta = frac{pi}{4}.
end{align}

It is not a good idea to use polar coordinates. The integral can be written as $int_{-1}^{1}int_{sqrt{1-x^{2}}} ^{3sqrt{1-x^{2}}}x^{2}, dy, dx=int_{-1}^{1}2sqrt {1-x^{2}}x^{2}, dx$. To evaluate this put $x=sin, theta$ and use the formulas $2sin, theta cos, theta =sin, 2theta$, $2sin^{2}, 2theta =1-cos (4theta)$.

Why in polar coordinates? Maybe it is easier by cartesian coordinates. Note that in $G$, $-1leq xleq 1$ and
$$sqrt{1-x^2}<y<3sqrt{1-x^2}.$$
Therefore
$$int_G x^2,dxdy=int_{x=-1}^1x^2left(3sqrt{1-x^2}-sqrt{1-x^2}right)dx$$
Can you take it from here?