Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left| {xy} \right|{\rm{d}}A = \frac{1}{2}$
Work Step by Step
We have $f\left( {x,y} \right) = \left| {xy} \right|$ and the region ${x^2} + {y^2} \le 1$.
In polar coordinates we have
$f\left( {r\cos \theta ,r\sin \theta } \right) = \left| {{r^2}\cos \theta \sin \theta } \right| = \frac{1}{2}{r^2}\left| {\sin 2\theta } \right|$
The description implies that the region is a unit disk. Thus, the region description in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1,0 \le \theta \le 2\pi } \right\}$
Using the Change of Variables Formula for ${\cal D}$, we evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {r^3}\left| {\sin 2\theta } \right|{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \left| {\sin 2\theta } \right|{\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r} \right)$
Notice that $\sin 2\theta $ is positive for $0 \le \theta \le \frac{\pi }{2}$ and negative for $\frac{\pi }{2} \le \theta \le \pi $; and so on. Since the integrand in the right-hand side is $\left| {\sin 2\theta } \right|$, the value is always positive. Thus, the periodicity occurs at each $\frac{\pi }{2}$. Thus, by symmetry we have
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \left| {\sin 2\theta } \right|{\rm{d}}\theta = {\bf{4}}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \sin 2\theta {\rm{d}}\theta $
$ = - 2\left( {\cos 2\theta |_0^{\pi /2}} \right)$
$ = - 2\left( { - 2} \right) = 4$
Thus,
$\frac{1}{2}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \left| {\sin 2\theta } \right|{\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^1 {r^3}{\rm{d}}r} \right) = \frac{1}{2}\left( 4 \right)\left( {\frac{1}{4}} \right) = \frac{1}{2}$
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left| {xy} \right|{\rm{d}}A = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {r^3}\left| {\sin 2\theta } \right|{\rm{d}}r{\rm{d}}\theta = \frac{1}{2}$
