Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\sqrt {{x^2} + {y^2}} {\rm{d}}A = \frac{2}{{15}}$
Work Step by Step
Write $f\left( {x,y} \right) = x\sqrt {{x^2} + {y^2}} $. In polar coordinates,
$f\left( {r\cos \theta ,r\sin \theta } \right) = {r^2}\cos \theta $
We have the shaded region ${\cal D}$ enclosed by the lemniscate curve ${r^2} = \sin 2\theta $ as is shown in Figure 21. In this region, the angle $\theta$ ranges from $0$ to $\frac{\pi }{2}$. The ray from the origin intersects ${\cal D}$ in a single point at $r = {\left( {\sin 2\theta } \right)^{1/2}}$. Thus, the polar description of ${\cal D}$:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le {{\left( {\sin 2\theta } \right)}^{1/2}},0 \le \theta \le \frac{\pi }{2}} \right\}$
Using Eq. (4) of Theorem 1, we evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\sqrt {{x^2} + {y^2}} {\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {r^3}\cos \theta {\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^{{{\left( {\sin 2\theta } \right)}^{1/2}}} {r^3}\cos \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {{r^4}|_0^{{{\left( {\sin 2\theta } \right)}^{1/2}}}} \right)\cos \theta {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {{{\sin }^2}2\theta } \right)\cos \theta {\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{\pi /2} {\sin ^2}\theta {\cos ^3}\theta {\rm{d}}\theta $
Using Eq. (7) of the Table of Trigonometric Integrals (Section 8.2), the last integral becomes
$ = \frac{{{{\sin }^3}\theta {{\cos }^2}\theta }}{5}|_0^{\pi /2} + \frac{2}{5}\mathop \smallint \limits_{\theta = 0}^{\pi /2} {\sin ^2}\theta \cos \theta {\rm{d}}\theta $
The first term on the right-hand side is zero. Now, we evaluate the second term on the right-hand side:
$\frac{2}{5}\mathop \smallint \limits_{\theta = 0}^{\pi /2} {\sin ^2}\theta \cos \theta {\rm{d}}\theta $
Write $u = \sin \theta $. So, $du = \cos \theta d\theta $.
$\frac{2}{5}\mathop \smallint \limits_{\theta = 0}^{\pi /2} {\sin ^2}\theta \cos \theta {\rm{d}}\theta = \frac{2}{5}\mathop \smallint \limits_{u = 0}^1 {u^2}{\rm{d}}u = \frac{2}{5}\left( {\frac{1}{3}{u^3}|_0^1} \right) = \frac{2}{{15}}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\sqrt {{x^2} + {y^2}} {\rm{d}}A = \frac{2}{{15}}$.
