solve the following integral in cartesian system . $$\int_0^{\frac{\pi}{3}}\int_{r=1}^{r=2\cos\theta}r\mathrm dr\,\mathrm d\theta$$
First we know that \begin{cases} x=r\cos\theta\\ y=r\sin\theta\\ r^2=x^2+y^2\\ \end{cases} and that $r$ is $1\leq r \leq 2\cos\theta$ so we get (after making it cartesian again)
\begin{cases}
x^2+y^2=2x\\
x^2+y^2=1\\
\end{cases}
and then I checked the intersection between the circles and got $2x=1$ So $x=\frac{1}{2}$ since $x=\frac{1}{2}$ we can substitute it to find $y$ so we get $y=\frac{\sqrt3}{2}$ and $y=-\frac{\sqrt3}{2}$
then I tried to find $\theta$ and got $r\cos\theta=0.5$ which is $\theta=\frac{\pi}{3}$ because $r=1$ and $r\sin\theta=\frac{\sqrt{3}}{2}$ and also got $\theta=\frac{\pi}{3}$ and lastly $rsin\theta=-\frac{\sqrt{3}}{2}$ and got $\theta=-\frac{\pi}{3}$ which is not in our bounds.
the graph I got is this 
and the graph in the bounds we want is:
I tried to separate the integral into two parts and got $$\int_0^{0.5}\int_{0}^{y=\sqrt{2x-x^2}}\sqrt{x^2+y^2}\mathrm dy\,\mathrm dx+\int_{0.5}^{1}\int_{0}^{y=\sqrt{1-x^2}}\sqrt{x^2+y^2}\mathrm dy\,\mathrm dx$$
But I got stuck , I am not able to solve this integral and I cannot find a different way to approach this question.
Appreciate any tips and help , and if someone can tell me if any of my approach is right?
The region from $r = 1$ to $r = 2\cos \theta$ is outide the circle centered at the origin and inside the circle centered at 1. It is the region that looks like a wave crashing on the origin.
The crest of the wave is at $(\frac 12, \frac {\sqrt 3}{2})$
To integrate in cartesian we will need to break up the region into two integrals.
The other thing we need to do is to consider the Jacobian.
$\displaystyle\int_{\frac 12}^1\int_{\sqrt {1-x^2}}^\sqrt{2x - x^2} \ dy\ dx + \int_1^2\int_0^\sqrt{2x - x^2} \ dy\ dx$