Compute$$\iint_D e^{-(2x^2-2xy+5y^2)}\arctan\left(\frac{x+y}{x-2y}\right)dA$$
over the region $D$, defined by $1\leq 2x^2 -2xy +5y^2\leq 9, (1-\sqrt3)x+(1+2\sqrt3)y\geq 0$ and $\sqrt3(x+y) \geq x-2y$.
My try:
I noted that $2x^2-2xy+5y^2 = (x-2y)^2 + (x+y)^2$. So I let $u=x+y, v=x-2y$, then, I can rewrite the domain $D$: $1 \leq u^2 + v^2 \leq 9, \frac{u}{v} \geq \sqrt{3}, \frac{u}{v} \geq \frac{1}{\sqrt{3}}$ (if $x-2y > 0$). After that, I also rewrite the integrand as follows (omitting the computation of the Jacobian Determinant):
$$\frac{1}{3}\iint_D e^{-(u^2+v^2)}\arctan\left(\frac{u}{v}\right)dA$$
Now, it is clear that I need to use polar coordinates: $u=r\sin(\theta), v=r\cos(\theta)$. Now, I can rewrite the integrand again:
$$\frac{1}{3} \iint e^{-r^2}r\theta \, dr \,d\theta$$
But in the end, I have problems computing the domain (the limits of integration). I know that $1 \leq r \leq 3$, but for the angle I have doubts, I don't know how to prooced. Any hints or corrections in my procedure are appreciated.