Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x - y} \right){\rm{d}}A = 0$
Work Step by Step
We have $f\left( {x,y} \right) = x - y$ and the region: ${x^2} + {y^2} \le 1$, $x + y \ge 1$. Notice that the region is the same with the one in Exercise 18.
In polar coordinates, $f\left( {r\cos \theta ,r\sin \theta } \right) = r\left( {\cos \theta - \sin \theta } \right)$.
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( {\cos \theta - \sin \theta } \right){r^2}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {\cos \theta - \sin \theta } \right)\left( {{r^3}|_{1/\left( {\cos \theta + \sin \theta } \right)}^1} \right){\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {\cos \theta - \sin \theta } \right)\left( {1 - \frac{1}{{{{\left( {\cos \theta + \sin \theta } \right)}^3}}}} \right){\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {\cos \theta - \sin \theta } \right){\rm{d}}\theta - \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \frac{{\cos \theta - \sin \theta }}{{{{\left( {\cos \theta + \sin \theta } \right)}^3}}}{\rm{d}}\theta $
1. Consider the integral: $\frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {\cos \theta - \sin \theta } \right){\rm{d}}\theta $
$\frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {\cos \theta - \sin \theta } \right){\rm{d}}\theta = \frac{1}{3}\left( {\sin \theta + \cos \theta } \right)|_0^{\pi /2}$
$ = \frac{1}{3}\left( {1 - 1} \right) = 0$
2. Consider the integral: $\frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \frac{{\cos \theta - \sin \theta }}{{{{\left( {\cos \theta + \sin \theta } \right)}^3}}}{\rm{d}}\theta $
Write $u = \cos \theta + \sin \theta $. So, $du = \left( { - \sin \theta + \cos \theta } \right)d\theta $. Thus,
$\frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \frac{{\cos \theta - \sin \theta }}{{{{\left( {\cos \theta + \sin \theta } \right)}^3}}}{\rm{d}}\theta = \frac{1}{3}\mathop \smallint \limits_{u = 1}^1 {u^{ - 3}}{\rm{d}}u$
But the lower limit and upper limit of integration is the same, therefore
$\frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \frac{{\cos \theta - \sin \theta }}{{{{\left( {\cos \theta + \sin \theta } \right)}^3}}}{\rm{d}}\theta = \frac{1}{3}\mathop \smallint \limits_{u = 1}^1 {u^{ - 3}}{\rm{d}}u = 0$
Both integrals yield zero.
Thus,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x - y} \right){\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 1/\left( {\cos \theta + \sin \theta } \right)}^1 \left( {\cos \theta - \sin \theta } \right){r^2}{\rm{d}}r{\rm{d}}\theta = 0$
