Check to see if polar coordinates are correct This is an extension of a question I asked earlier, but with some slight changes to it. We have $x^2+y^2\leqslant2x+2y$ as a $2D$ region and we need to solve $\int\int{y\,dA}$
To see what this is I did the following:
$x^2+y^2-2x-2y$
$(x^2-2x)+(y^2-2y)$
$(x^2-2x+1)+(y^2-2y+1)$
$(x-1)^2+(y-1)^2 \leqslant 1^2$
So this has a center at (1,1) and a radius of 1 in the first and second quadrant of the disk. Therefore, my new limits of integration look like:
$$\int_{0}^{\pi}\int_{0}^{2\cos\theta+2\sin\theta} r^2\sin\theta \,dr\,d\theta$$
Is this correct? I believe it's $\pi$ and not $\frac{\pi}{2}$ for $d\theta$, but I want to make sure the answer so far is fine.
 A: No, it's not correct. In the work that you showed, three lines don't even have right-hand sides, and that's very suspicious. Even worse, that lead to an actual mistake: when you finally decided to write a right-hand side as "$\color{red}{\le1^2}$" — where did that $\color{red}{1^2}$ come from? It's actually wrong there. While your work on the left-hand side is correct, you have to work with the entire inequality to get a correct result. This shape is indeed a disk (filled circle) centered at $(1,1)$, but its radius is NOT $\color{red}{1}$.
You can also check your answer with some special points. It's easy to see that the origin $(0,0)$ satisfies the equation $x^2+y^2=2x+2y$, so it has to lie on the boundary of the given region; but it doesn't lie on the circumference of your circle centered at $(1,1)$ of radius $1$, so this is not the right circle.
As for the integral in polar coordinates that you need to obtain: your integrand $\color{green}{r^2\sin\theta}$ is correct, and your limits of integration for $r$, viz. $\color{green}{0}$ and $\color{green}{2\cos\theta+2\sin\theta}$, are also correct. But both limits of integration for $\theta$, viz. $\color{red}{0}$ and $\color{red}{\pi}$, are wrong because of the mistake explained above.
A: There's something thing wrong here, for one thing, the radius of the circle centered at $(1,1)$ has a radius of $\sqrt{2}$. I believe that I see the solution to this problem so that you can check your integral later. The integral you are calculating is related to the centroid of the circle, which is clearly at the center of the circle, i.e. $(1,1)$. Now, the centroid is given by
$$R_y=\frac{\int\!\!\!\int y~dA}{\int\!\!\!\int dA}=\frac{1}{A}\int\!\!\!\int y~dA$$
So, clearly
$$\int\!\!\!\int y~dA=R_yA=R_y\pi r^2=2\pi$$
since $R_y=1$ and $r=\sqrt{2}$.
