Answer
We prove:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln r{\rm{d}}A = - \frac{\pi }{2}$
Work Step by Step
We have the region ${\cal D}$, the unit disk ${x^2} + {y^2} \le 1$.
We have ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1,0 \le \theta \le 2\pi } \right\}$
Since $\ln r$ is not defined at $\left( {0,0} \right)$, we define ${{\cal D}_1}$ by
${{\cal D}_1} = \left\{ {\left( {r,\theta } \right)|a \le r \le 1,0 \le \theta \le 2\pi } \right\}$,
where $0 < a < 1$, and let $a \to 0$.
Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} \ln r{\rm{d}}A$ in polar coordinates:
(1) ${\ \ \ \ }$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} \ln r{\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = a}^1 r\ln r{\rm{d}}r{\rm{d}}\theta $
Consider the inner integral: $\mathop \smallint \limits_{r = a}^1 r\ln r{\rm{d}}r$.
Write $u = \ln r$ and $dv = rdr$. So, $du = \frac{1}{r}dr$ and $v = \frac{1}{2}{r^2}$. Using Integration by Parts Formula (Section 8.1),
$\smallint u{\rm{d}}v = uv - \smallint v{\rm{d}}u$
yields
$\mathop \smallint \limits_{r = a}^1 r\ln r{\rm{d}}r = \left( {\frac{1}{2}{r^2}\ln r} \right)|_a^1 - \frac{1}{2}\mathop \smallint \limits_{r = a}^1 r{\rm{d}}r$
$\mathop \smallint \limits_{r = a}^1 r\ln r{\rm{d}}r = - \frac{1}{2}{a^2}\ln a - \frac{1}{4}\left( {{r^2}|_a^1} \right)$
$\mathop \smallint \limits_{r = a}^1 r\ln r{\rm{d}}r = - \frac{1}{2}{a^2}\ln a - \frac{1}{4}\left( {1 - {a^2}} \right)$
Substituting this result back to (1) gives
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} \ln r{\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = a}^1 r\ln r{\rm{d}}r{\rm{d}}\theta $
$ = \left( { - \frac{1}{2}{a^2}\ln a - \frac{1}{4}\left( {1 - {a^2}} \right)} \right)\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta $
$ = \left( { - \frac{1}{2}{a^2}\ln a - \frac{1}{4}\left( {1 - {a^2}} \right)} \right)2\pi $
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} \ln r{\rm{d}}A = \left( { - \frac{1}{2}{a^2}\ln a - \frac{1}{4}\left( {1 - {a^2}} \right)} \right)2\pi $.
Now, we take the limit such that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln r{\rm{d}}A = \mathop {\lim }\limits_{a \to 0} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} \ln r{\rm{d}}A = \mathop {\lim }\limits_{a \to 0} \left( { - \frac{1}{2}{a^2}\ln a - \frac{1}{4}\left( {1 - {a^2}} \right)} \right)2\pi $
$ = - \pi \mathop {\lim }\limits_{a \to 0} {a^2}\ln a - \frac{\pi }{2}\mathop {\lim }\limits_{a \to 0} \left( {1 - {a^2}} \right)$
$ = - \pi \mathop {\lim }\limits_{a \to 0} \frac{{\ln a}}{{{a^{ - 2}}}} - \frac{\pi }{2}\mathop {\lim }\limits_{a \to 0} \left( {1 - {a^2}} \right)$
Using L'Hôpital's Rule:
$ = - \pi \mathop {\lim }\limits_{a \to 0} \frac{{1/a}}{{ - 2{a^{ - 3}}}} - \frac{\pi }{2}\mathop {\lim }\limits_{a \to 0} \left( {1 - {a^2}} \right)$
$ = \pi \mathop {\lim }\limits_{a \to 0} \frac{{{a^2}}}{2} - \frac{\pi }{2}\mathop {\lim }\limits_{a \to 0} \left( {1 - {a^2}} \right)$
$ = - \frac{\pi }{2}$
Hence, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \ln r{\rm{d}}A = - \frac{\pi }{2}$.
