Evaluating $\iint_D\frac{y}{\sqrt{x^2+y^2}}\ln(x^2+y^2)\,dx\,dy$ using polar transformation 
Evaluate $$\iint_D \frac{y}{\sqrt{x^2+y^2}}\ln(x^2+y^2) \, dx \, dy$$, where $$D = \left\{(x,y):\frac{1}{4}\leq x^2+y^2\leq1,x^2+y^2\leq2x,y\geq0\right\}$$

I tried solving it by changing to polar coordinates but I can't find the interval for the polar angle.
At all, can someone explain me how to find the interval for the polar angle and the polar radius? Every time I solve problems with polar change I can't find the right intervals.
 A: The limits are difficult to calculate due to the $x^2+y^2\le 2x$ however without limits it is of the form:
$$I=\iint r\sin\theta\ln (r^2)drd\theta=2\iint r\sin\theta\ln|r|drd\theta$$
which is easy enough to solve once you have worked out the limits
A: Result
The integral can be calculated exactly to give
$$i = \frac{5}{72} (\log (8)-4) \simeq -0.133372\tag{1}$$
Derivation
Transforming to polar coordinates $x\to r \cos(\phi)$, $y\to r \sin(\phi)$, $dx dy = r dr d\phi$ the integral becomes
$$i = 2\int r\sin(\phi) \log(r)\,dr d\phi$$
The restrictions transform to polar coordinates as follows 
$$\frac{1}{4} \le x^2+y^2\le 1 \to \frac{1}{2} \le r \le 1\tag{2a}$$
$$x^2+y^2\le2x \to 0\le \phi\le \cos ^{-1}\left(\frac{r}{2}\right)\tag{2b}$$
so that we have
$$i = 2 \int_ {\frac{1}{2}}^1 r \log(r) \int_{0}^{\cos ^{-1}\left(\frac{r}{2}\right)} \sin(\phi) \,dr \,d\phi$$
or, doing the $\phi$-integral,
$$i = 2 \int_ {\frac{1}{2}}^1 r\left(1-\frac{r}{2} \right) \log(r)\,dr$$
The final $r$-integral gives $(1)$. QED.
