Evaluating $\iint_D\frac{y}{\sqrt{x^2+y^2}}\ln(x^2+y^2)\,dx\,dy$ using polar transformation

Question

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.

Answer 1

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.

Answer 2

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