Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\left( {{x^2} + {y^2}} \right)^{ - 3/2}}{\rm{d}}A = 2 - \frac{\pi }{2}$
Work Step by Step
We have $f\left( {x,y} \right) = {\left( {{x^2} + {y^2}} \right)^{ - 3/2}}$ and the region: ${x^2} + {y^2} \le 1$, $x + y \ge 1$.
In polar coordinates, $f\left( {r\cos \theta ,r\sin \theta } \right) = {r^{ - 3}}$.
The description of the region implies that it is bounded by the line $x+y=1$ and the circle of radius $1$.
First, we find the line $x+y=1$ in polar coordinates. Using $x = r\cos \theta $ and $y = r\sin \theta $, we get
$r\cos \theta + r\sin \theta = 1$
$r = \frac{1}{{\cos \theta + \sin \theta }}$
From the figure attached, we see that as $\theta$ varies from $\theta = 0$ to $\theta = \frac{\pi }{2}$, $r$ varies from $r = \frac{1}{{\cos \theta + \sin \theta }}$ to $r=1$. Thus, the region description:
${\cal D} = \left\{ {\left( {r,\theta } \right)|\frac{1}{{\cos \theta + \sin \theta }} \le r \le 1,0 \le \theta \le \frac{\pi }{2}} \right\}$
Using Eq. (4) of Theorem 1, 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 $
$ = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 1/\left( {\cos \theta + \sin \theta } \right)}^1 \left( {{r^{ - 3}}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = - \mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {{r^{ - 1}}|_{1/\left( {\cos \theta + \sin \theta } \right)}^1} \right){\rm{d}}\theta $
$ = - \mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {1 - \cos \theta - \sin \theta } \right){\rm{d}}\theta $
$ = - \left( {\theta - \sin \theta + \cos \theta } \right)|_0^{\pi /2}$
$ = - \left( {\frac{\pi }{2} - 1 - 1} \right) = 2 - \frac{\pi }{2}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\left( {{x^2} + {y^2}} \right)^{ - 3/2}}{\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 1/\left( {\cos \theta + \sin \theta } \right)}^1 \left( {{r^{ - 3}}} \right)r{\rm{d}}r{\rm{d}}\theta = 2 - \frac{\pi }{2}$
