Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A \simeq 13.2$
Work Step by Step
Write $f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $. In polar coordinates,
$f\left( {r\cos \theta ,r\sin \theta } \right) = r$
We have the domain ${\cal D}$ as is shown in Figure 20.
From Figure 20 and the figure attached, we see that the outer circle centered at the origin has radius $r=2$. The inner circle is an unit circle centered at $\left( {1,0} \right)$. Referring to Example 2, the inner circle has equation in polar coordinates $r = 2\cos \theta $.
We can divide ${\cal D}$ into two sub-domains ${{\cal D}_1}$ and ${{\cal D}_2}$. So, ${\cal D} = {{\cal D}_1}\bigcup {{\cal D}_2}$.
1. ${{\cal D}_1}$
In this sub-domain, the angle $\theta $ ranges from $ - \frac{\pi }{2}$ to $\frac{\pi }{2}$. The ray from the origin intersects ${{\cal D}_1}$ in the line segment where $r$ ranges from $r = 2\cos \theta $ to $r=2$.
Thus, the domain description of ${{\cal D}_1}$:
${{\cal D}_1} = \left\{ {\left( {r,\theta } \right)|2\cos \theta \le r \le 2, - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}} \right\}$
2. ${{\cal D}_2}$
In this sub-domain, the angle $\theta$ ranges from $\frac{\pi }{2}$ to $\frac{{3\pi }}{2}$. The ray from the origin intersects ${{\cal D}_2}$ in a single point at the circle $r=2$. So, $r$ ranges from $r=0$ to $r=2$.
Thus, the domain description of ${{\cal D}_2}$:
${{\cal D}_2} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2,\frac{\pi }{2} \le \theta \le \frac{{3\pi }}{2}} \right\}$
Using Eq. (4) of Theorem 1, and the linearity property of the double integral we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {r^2}{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} {r^2}{\rm{d}}r{\rm{d}}\theta + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} {r^2}{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 2\cos \theta }^2 {r^2}{\rm{d}}r{\rm{d}}\theta + \mathop \smallint \limits_{\theta = \pi /2}^{3\pi /2} \mathop \smallint \limits_{r = 0}^2 {r^2}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {{r^3}|_{2\cos \theta }^2} \right){\rm{d}}\theta + \frac{1}{3}\mathop \smallint \limits_{\theta = \pi /2}^{3\pi /2} \left( {{r^3}|_0^2} \right){\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {8 - 8{{\cos }^3}\theta } \right){\rm{d}}\theta + \frac{8}{3}\mathop \smallint \limits_{\theta = \pi /2}^{3\pi /2} {\rm{d}}\theta $
$ = \frac{8}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\rm{d}}\theta - \frac{8}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^3}\theta {\rm{d}}\theta + \frac{8}{3}\mathop \smallint \limits_{\theta = \pi /2}^{3\pi /2} {\rm{d}}\theta $
$ = \frac{{8\pi }}{3} - \frac{8}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^3}\theta {\rm{d}}\theta + \frac{{8\pi }}{3}$
(1) ${\ \ \ \ \ \ }$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A = \frac{{16\pi }}{3} - \frac{8}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^3}\theta {\rm{d}}\theta $
Consider the integral $\smallint {\cos ^3}\theta {\rm{d}}\theta $.
Using Eq. (6) of the Table of Trigonometric Integrals (Section 8.2), we obtain
$\smallint {\cos ^3}\theta {\rm{d}}\theta = \frac{{{{\cos }^2}\theta \sin \theta }}{3} + \frac{2}{3}\smallint \cos \theta {\rm{d}}\theta $
$\smallint {\cos ^3}\theta {\rm{d}}\theta = \frac{{{{\cos }^2}\theta \sin \theta }}{3} + \frac{2}{3}\sin \theta $
Thus, equation (1) becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A = \frac{{16\pi }}{3} - \frac{8}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^3}\theta {\rm{d}}\theta $
$ = \frac{{16\pi }}{3} - \frac{8}{3}\left( {\frac{{{{\cos }^2}\theta \sin \theta }}{3} + \frac{2}{3}\sin \theta } \right)|_{ - \pi /2}^{\pi /2}$
$ = \frac{{16\pi }}{3} - \frac{8}{3}\left( {\frac{2}{3} + \frac{2}{3}} \right)$
$ = \frac{{16\pi }}{3} - \frac{{32}}{9} \simeq 13.2$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A \simeq 13.2$.
