Compute the following integral: $\int_D \ln(x^2+y^2)dxdy$ where $D$ is the disc $x^2+y^2 \le1$ in $R$ into polar coordinates.
What I have tried: $r^2\le1 \implies r\le1$ we also have $0 \le\theta\le2\pi$
This produces $$\int_0^{2\pi} d\theta\int_0^1\ln(r)\cdot r dr \implies 2\pi\int \ln(r)\cdot r dr$$
Taking the integral by parts to tackle the $dr$ integral we have $du = \frac{1}{r}$ and $v = \frac{r^2}{2}$
Which gives $$2\pi \left[\frac{\ln(r)r^2}{2}\Biggr|_0^1-\frac{1}{2}\int_0^1 rdr\right]\implies-\frac{2\pi}{4}$$
However, the right answer should be $-\pi$ any idea where I went wrong? Also, the actual question asks me to let "$D_\delta$ be the annulus $\delta^2\le x^2+y^2\le1$ as I'm supposed to show that this integral has a finite limit which is $-\pi$ although I'm unsure on how to proceed with this and would really appreciate the communities support.
Using annulus area $D_\delta=\{(x, y)|\delta^2\leq x^2+y^2\leq 1\}$, I'll solve that integral. Let $I_\delta$ as that value.
$$I_\delta=\int_{D_\delta}\ln(x^2+y^2)$$
Then, represent that annulus area as that... $D_\delta=\{(r\cos\theta,r\sin\theta)|\delta\leq r\leq1, 0\leq\theta\lt2\pi\}$ Then the integral $I_\delta$ goes like this. $$I_\delta=\int_0^{2\pi}d\theta\int_\delta^12r\ln rdr=4\pi\int_\delta^1r\ln rdr$$ ($\ln(x^2+y^2)=\ln r^2=2\ln r$) So... $$I_\delta=4\pi \biggl[{\cfrac{r^2(2\ln r-1)}{4}}\biggr]_\delta^1=-\pi(2\delta^2\ln\delta-\delta^2+1)$$ Then, Gave an limit. $$I=\lim \limits_{\delta \to 0+}{-\pi(2\delta^2\ln\delta-\delta^2+1)}=-\pi$$
That is the answer.