Task: find a double integral $$\iint_D (x+y)dxdy,$$ where D is bound by $x^2 + y^2 = x + y$.
What I have done so far: turns out it's a circle $$(x-1)^2 + (y-1)^2 = 2$$
Calculating it as a common double integral is hard because I get something like this: $$\int_{1-\sqrt{2}}^{1+\sqrt{2}} dx \int_{1 – \sqrt{2 – (x-1)^2}}^{1 + \sqrt{2 – (x-1)^2}} (x + y) dy.$$
So, I decided to give up on this. My next idea is to transform it into Polar coordinates. And that's where I got stuck. $$dxdy = rdrd\theta \\ x = r \cos{\theta} \\ y = r \sin{\theta}.$$
What to do next? For me, it looks like $$0 \leq\theta \leq 2\pi \\ 0 \leq r \leq 2\sqrt{2},$$
but this seems like a case when the origin of a circle is $(0, 0)$. I have my circle shifted and there should be some tricks.
Any help would be appreciated.
Best Answer
$\iint_D (x+y)dxdy=\iint_{D'} (u+v+2)dudv$ by the substitution $u=x-1, v=y-1$, $D'$ being $\{(u,v): u^{2}+v^{2} \leq 2\}$. By symmetry the integral of $u$ and $v$ over $D'$ is $0$. Hence the value is just $\iint_{D'} 2dudv=2(\pi) (2)=4\pi$.