I have this domain $A=\{ (x,y) \in R^2 : x^2+y^2 \ge4, x^2+y^2-2x-2y\le0 \}$
It's right the change in polar coordinates : $$\{ (r,\theta): \theta \in [0,\frac{\pi}{2}], r \in [2,2(\sin\theta+\cos\theta)] \}$$
I'm using this for $\int_A {y\over(x^2+y^2)} dx dy$.
Best Answer
See if this helps.
If $x = r\cos \theta$ and $y = r\sin \theta$ then $4 \leq x^2 + y^2 = r^2 $. Moreover
$$r^2 -2r(\cos \theta + \sin \theta) \leq 0 \Rightarrow \Big(r-(\cos\theta + \sin\theta)\Big)^2 \leq (\cos\theta + \sin\theta)^2$$