Domain in polar coordinates

Question

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 $\int_A {y\over(x^2+y^2)} dx dy$.

Answer 1

See if this helps.

If $x = r\cos \theta$ and $y = r\sin \theta$ then $4 \leq x^2 + y^2 = r^2 $. Moreover

$$r^2 -2r(\cos \theta + \sin \theta) \leq 0 \Rightarrow \Big(r-(\cos\theta + \sin\theta)\Big)^2 \leq (\cos\theta + \sin\theta)^2$$

Answer 2

Draw a picture. Your conditions specify points outside a circle $x^2+y^2\geq 2^2$ and inside another circle $(x-1)^2+(y-1)^2\leq (\sqrt{2})^2$. You can determine the intersection of the two circles and find angles from there, or otherwise note that the intersection points are $(2,0)$ and $(0,2)$.

For $\theta\in[0,\pi/2]$ you need to specify bounds on $r$ (between the two circles), the easier one being the lower bound of $r=2$ for the circle centered at the origin. For the circle centered at $(1,1)$, you can look at $$ x^2+y^2-2(x+y)=0 $$ in polar coordinates $$ r^2-2r(\cos\theta+\sin\theta)=0 $$ and solve for $r$ to get $$ r=2(\sin\theta+\cos\theta). $$ So the region is $$ \{(r,\theta) \ : \ \theta\in[0,\pi/2], \ 2\leq r\leq 2(\sin\theta+\cos\theta)\} $$ as you've written.

For the integral, you get $$ \int_A\frac{y}{x^2+y^2}dxdy=\int_{\theta=0}^{\theta=\pi/2}\int_{r=2}^{r=2(\sin\theta+\cos\theta)}\frac{r\sin\theta}{r^2}rdrd\theta=2\int_{0}^{\pi/2}(\sin^2\theta+\sin\theta\cos\theta) d\theta $$ which you can integrate however you want.