Answer
The total mass is $95.66$ g.
Work Step by Step
We have a mass density of $\delta \left( {x,y} \right) = \frac{{{x^2}}}{{{x^2} + {y^2}}}$ and the region as is shown in Figure 10. Let ${\cal D}$ be the shaded region.
In this exercise, we choose to evaluate the integral in polar coordinates. Using $x = r\cos \theta $ and $y = r\sin \theta $, we get
$\delta \left( {r\cos \theta ,r\sin \theta } \right) = \frac{{{r^2}{{\cos }^2}\theta }}{{{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta }} = {\cos ^2}\theta $
From Figure 10 we see that the region is part of a disk of radius $20$ centered at the origin.
The vertical line $x=10$ has polar equation $r\cos \theta = 10$. So,
$r = 10\sec \theta $
Thus, a ray of angle $\theta $ intersects ${\cal D}$ in the line segment where $r$ ranges from $r = 10\sec \theta $ to $r=20$. So, ${\cal D}$ has polar description:
${\cal D} = \left\{ {\left( {r,\theta } \right)|10\sec \theta \le r \le 20,0 \le \theta \le \frac{\pi }{3}} \right\}$
Using Eq. (1), the total mass is given by
${\rm{total{\ }mass}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{{{x^2}}}{{{x^2} + {y^2}}}{\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^{\pi /3} \mathop \smallint \limits_{r = 10\sec \theta }^{20} \left( {{{\cos }^2}\theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{\pi /3} {\cos ^2}\theta \left( {{r^2}|_{10\sec \theta }^{20}} \right){\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{\pi /3} \left( {400{{\cos }^2}\theta - 100} \right){\rm{d}}\theta $
$ = 200\mathop \smallint \limits_{\theta = 0}^{\pi /3} {\cos ^2}\theta {\rm{d}}\theta - 50\mathop \smallint \limits_{\theta = 0}^{\pi /3} {\rm{d}}\theta $
Using Eq. (4) of the Table of Trigonometric Integrals in Section 8.2:
$\smallint {\cos ^2}x{\rm{d}}x = \frac{x}{2} + \frac{{\sin 2x}}{4} + C$
we obtain
${\rm{total{\ }mass}} = 200\left( {\left( {\frac{\theta }{2} + \frac{{\sin 2\theta }}{4}} \right)|_0^{\pi /3}} \right) - \frac{{50}}{3}\pi $
$ = 200\left( {\frac{\pi }{6} + \frac{{\sqrt 3 }}{8}} \right) - \frac{{50}}{3}\pi = \frac{{50}}{3}\pi + 25\sqrt 3 \simeq 95.66$
Thus, the total mass is $95.66$ g.
