Answer
The total charge is $7.5 \times {10^{ - 7}}\pi $ coulombs.
Work Step by Step
We have a charge density $\delta \left( {x,y,z} \right) = \left( {3\cdot{{10}^{ - 8}}} \right){\left( {{x^2} + {y^2} + {z^2}} \right)^{1/2}}$ and the region of a solid ball ${x^2} + {y^2} + {z^2} \le 5$.
Using Eq. (1), the total charge is given by
${\rm{total{\ }charge}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \delta \left( {x,y,z} \right){\rm{d}}V$
We choose to evaluate the triple integral using spherical coordinates. First, we define the solid ball ${x^2} + {y^2} + {z^2} \le 5$ in spherical coordinates:
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|0 \le \rho \le \sqrt 5 ,0 \le \phi \le \pi ,0 \le \theta \le 2\pi } \right\}$
In spherical coordinates:
$f\left( {\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi } \right) = \left( {3\cdot{{10}^{ - 8}}} \right)\rho $
Thus,
${\rm{total{\ }charge}} = \left( {3\cdot{{10}^{ - 8}}} \right)\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \mathop \smallint \limits_{\rho = 0}^{\sqrt 5 } {\rho ^3}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{{3\cdot{{10}^{ - 8}}}}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \left( {{\rho ^4}|_0^{\sqrt 5 }} \right)\sin \phi {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{{75\cdot{{10}^{ - 8}}}}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\phi {\rm{d}}\theta $
$ = - \frac{{75\cdot{{10}^{ - 8}}}}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \phi |_0^\pi } \right){\rm{d}}\theta $
$ = - \frac{{75\cdot{{10}^{ - 8}}}}{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( { - 1 - 1} \right){\rm{d}}\theta $
$ = \frac{{75\cdot{{10}^{ - 8}}}}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = 75\cdot{10^{ - 8}}\pi $
So, the total charge is $7.5 \times {10^{ - 7}}\pi $ coulombs.
