Polar coordinates in double integral of two circles

Question

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: https://i.stack.imgur.com/Y2E3C.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?

Answer 1

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 $2$

Answer 2

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 $0 < r < 1$.

Answer 3

I think it should be like this:

$S = 2\int\limits_0^{\pi/2} {\int\limits_{ 2\cos (\theta )}^{ 4\cos (\theta )} {{r^3}drd\theta } }$