Polar coordinates in double integral of two circles 
Use polar coordinates to calculate the integral $\int\int_R(x²+y²)\,dx\,dy$ where $R$ is the region inside $x²-4x+y²=0$ and outside $x²-2x+y²=0$.

This is the graphic of the region: http://i.imgur.com/ejRA7gZ.png
I assumed with such information the point $(1,0)$ as the center of the region, but don't know how to set the upper limit of the integral of $r$.
$$2\int_0^\pi\int_1^?r^2*r\, dr\,d\theta$$
What I put in '?', or I'm doing it wrong?
 A: The outer circle is 
\begin{equation}
r=4\cos \theta 
\end{equation}
and the inner circle is 
\begin{equation}
r=2\cos \theta 
\end{equation}
so the integral is 
\begin{equation}
2\int _{0}^{\pi/2}\int _{2\cos \theta }^{4\cos \theta}r^{3}drd\theta  
\end{equation} 
Note: symmetry allows you to integrate from $0$ to $\pi /2$ and multiply the result by $2$
A: I think it should be like this:
$S = 2\int\limits_0^{\pi/2}  {\int\limits_{ 2\cos (\theta )}^{ 4\cos (\theta )} {{r^3}drd\theta } }$
A: Do the integral in the outer circle minus the integral in the inner circle! You can use translated polar coordinates for both of them. Let $C_1$ be the outer circle and $C_2$ be the inner circle, that is: $$C_1 : (x-2)^2+y^2 = 4 \\ C_2: (x-1)^2+y^2=1$$So: $$\iint_R x^2+y^2\,{\rm d}x\,{\rm d}y = \iint_{C_1}x^2+y^2\,{\rm d}x\,{\rm d}y - \iint_{C_2}x^2+y^2\,{\rm d}x\,{\rm d}y.$$For the first one make $x = 2+r\cos \theta$ and $y = r\sin \theta$, with $0 < \theta < 2\pi$ and $0 < r < 2$.
For the second one make $x = 1+r\cos \theta$ and $y = r\sin \theta$, with $0 < \theta < 2\pi$ and $0 < r < 1$.
