mjhjmtu

Let $ G:= left (x,y) in mathbbR^2 : 0 < y,: x^2 + fracy^29 <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
$$ r^2cos^2theta + fracr^2sin^2theta9 < 1 quad Rightarrow quad r < frac3sqrt8cos^2theta+1. $$

Thus
beginalign
int_0^pi int_1^frac3sqrt8cos^2theta+1 r^2cos^2theta cdot r ,mathrm dr , mathrm dtheta &= frac814 int_0^pi fraccos^2theta(8cos^2theta+1)^2 , mathrm dtheta - frac14 int_0^pi cos^2theta , mathrm dtheta = fracpi4.
endalign

It is not a good idea to use polar coordinates. The integral can be written as $int_-1^1int_sqrt1-x^2 ^3sqrt1-x^2x^2, dy, dx=int_-1^12sqrt 1-x^2x^2, dx$. 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
$$sqrt1-x^2<y<3sqrt1-x^2.$$
Therefore
$$int_G x^2,dxdy=int_x=-1^1x^2left(3sqrt1-x^2-sqrt1-x^2right)dx$$
Can you take it from here?