Answer
We show that: $\bar d = \frac{6}{5}R$
Work Step by Step
Let $\left( {x,y,z} \right)$ be any point in the ball. Using the Law of Cosines in Section 1.4, the square distance from the North Pole $P = \left( {0,0,R} \right)$ to $\left( {x,y,z} \right)$ is given by
${R^2} + {\rho ^2} - 2\rho R\cos \phi $,
where $\phi $ is the angle between the position vectors $P$ and $\left( {x,y,z} \right)$; and ${\rho ^2} = {x^2} + {y^2} + {z^2}$.
So, the distance is
${d_P}\left( {x,y,z} \right) = \sqrt {{R^2} + {\rho ^2} - 2\rho R\cos \phi } $
By definition, the average value of the distance is given by (Eq. (5) in Section 16.3):
$\bar d = \frac{1}{{{\rm{Volume}}\left( {\cal W} \right)}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {d_P}\left( {x,y,z} \right){\rm{d}}V$
Since the volume of the ball of radius $R$ is $\frac{4}{3}\pi {R^3}$, the average distance in spherical coordinates is
$\bar d = \frac{1}{{\frac{4}{3}\pi {R^3}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\rho = 0}^R \mathop \smallint \limits_{\phi = 0}^\pi {\rho ^2}\sin \phi \sqrt {{R^2} + {\rho ^2} - 2\rho R\cos \phi } {\rm{d}}\phi {\rm{d}}\rho {\rm{d}}\theta $
$ = \frac{1}{{\frac{4}{3}\pi {R^3}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\rho = 0}^R {\rho ^2}\left( {\frac{1}{{3\rho R}}{{\left( {{R^2} + {\rho ^2} - 2\rho R\cos \phi } \right)}^{3/2}}|_0^\pi } \right){\rm{d}}\rho {\rm{d}}\theta $
$ = \frac{1}{{4\pi {R^4}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\rho = 0}^R \rho \left( {{{\left( {{R^2} + {\rho ^2} + 2\rho R} \right)}^{3/2}} - {{\left( {{R^2} + {\rho ^2} - 2\rho R} \right)}^{3/2}}} \right){\rm{d}}\rho {\rm{d}}\theta $
$ = \frac{1}{{4\pi {R^4}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\rho = 0}^R \rho \left( {{{\left( {R + \rho } \right)}^3} - {{\left( {R - \rho } \right)}^3}} \right){\rm{d}}\rho {\rm{d}}\theta $
$ = \frac{1}{{4\pi {R^4}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\rho = 0}^R \rho \left( {6{R^2}\rho + 2{\rho ^3}} \right){\rm{d}}\rho {\rm{d}}\theta $
$ = \frac{1}{{4\pi {R^4}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\rho = 0}^R \left( {6{R^2}{\rho ^2} + 2{\rho ^4}} \right){\rm{d}}\rho {\rm{d}}\theta $
$ = \frac{1}{{4\pi {R^4}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {2{R^2}{\rho ^3} + \frac{2}{5}{\rho ^5}} \right)|_0^R} \right){\rm{d}}\theta $
$ = \frac{1}{{4\pi {R^4}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {2{R^5} + \frac{2}{5}{R^5}} \right){\rm{d}}\theta $
$ = \frac{{3R}}{{5\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{6}{5}R$
So, $\bar d = \frac{6}{5}R$.
