Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {r^{ - a}}{\rm{d}}A$ converges if and only if $a < 2$.
Work Step by Step
We have the region ${\cal D}$, the unit disk ${x^2} + {y^2} \le 1$.
In polar coordinates, ${\cal D}$ is described by
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1,0 \le \theta \le 2\pi } \right\}$
Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {r^{ - a}}{\rm{d}}A$ in polar coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {r^{ - a}}{\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {{r^{ - a}}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\mathop \smallint \limits_{r = 0}^1 {r^{ - \left( {a - 1} \right)}}{\rm{d}}r} \right){\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 {r^{ - \left( {a - 1} \right)}}{\rm{d}}r} \right)$
$ = 2\pi \mathop \smallint \limits_{r = 0}^1 {r^{ - \left( {a - 1} \right)}}{\rm{d}}r$
Recall that the improper integral $\mathop \smallint \limits_0^1 {x^{ - a}}{\rm{d}}x$ converges if and only if $a < 1$. Therefore, the integral $\mathop \smallint \limits_{r = 0}^1 {r^{ - \left( {a - 1} \right)}}{\rm{d}}r$ converges if and only if $a - 1 < 1$ or $a < 2$.
Hence, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {r^{ - a}}{\rm{d}}A$ converges if and only if $a < 2$.
