Answer
The total charge is $\frac{{31}}{{12}}\pi $ coulombs.
Work Step by Step
We have a region ${\cal D}$ as is shown in Figure 11, and a charge density of $\rho \left( {r,\theta } \right) = 3{r^{ - 4}}$ $C/c{m^2}$.
The ellipse has polar equation:
${r^2} = {\left( {\frac{1}{6}{{\sin }^2}\theta + \frac{1}{9}{{\cos }^2}\theta } \right)^{ - 1}}$
The unit circle ${x^2} + {y^2} = 1$ in polar coordinates is $r=1$. Thus, the description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|1 \le 0 \le {{\left( {\frac{1}{6}{{\sin }^2}\theta + \frac{1}{9}{{\cos }^2}\theta } \right)}^{ - 1/2}},0 \le \theta \le 2\pi } \right\}$
Using Eq. (1), the total charge is given by
${\rm{total{\ }charge}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \rho \left( {x,y} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 1}^{{{\left( {\frac{1}{6}{{\sin }^2}\theta + \frac{1}{9}{{\cos }^2}\theta } \right)}^{ - 1/2}}} \rho \left( {r,\theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 1}^{{{\left( {\frac{1}{6}{{\sin }^2}\theta + \frac{1}{9}{{\cos }^2}\theta } \right)}^{ - 1/2}}} 3{r^{ - 3}}{\rm{d}}r{\rm{d}}\theta $
$ = - \frac{3}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{r^{ - 2}}|_1^{{{\left( {\frac{1}{6}{{\sin }^2}\theta + \frac{1}{9}{{\cos }^2}\theta } \right)}^{ - 1/2}}}} \right){\rm{d}}\theta $
$ = - \frac{3}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{6}{{\sin }^2}\theta + \frac{1}{9}{{\cos }^2}\theta - 1} \right){\rm{d}}\theta $
Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$, ${\ \ \ }$ ${\cos ^2}x = \frac{1}{2}\left( {1 + \cos 2x} \right)$
we get
${\rm{total{\ }charge}} = - \frac{3}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{{12}}\left( {1 - \cos 2\theta } \right) + \frac{1}{{18}}\left( {1 + \cos 2\theta } \right) - 1} \right){\rm{d}}\theta $
$ = - \frac{3}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( { - \frac{{31}}{{36}} - \frac{1}{{36}}\cos 2\theta } \right){\rm{d}}\theta $
$ = \frac{1}{{24}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {31 + \cos 2\theta } \right){\rm{d}}\theta $
$ = \frac{1}{{24}}\left( {31\theta + \frac{1}{2}\sin 2\theta } \right)|_0^{2\pi }$
$ = \frac{1}{{24}}\left( {62\pi } \right) = \frac{{31}}{{12}}\pi $
Thus, the total charge is $\frac{{31}}{{12}}\pi $ coulombs.
