How to find the mean distance between a point inside a ball and the surface of this ball?

Question

Say I have a sphere of radius $R$, I would like to know the average distance between a point $M$ inside the ball and the sphere of radius $R$.

Because of the spherical symmetry, we can say without loss of generality that my point $M$ inside the ball is defined by the coordinates: $$ (x=r,y=0,z=0) $$, and the points on the sphere are defined by the coordinates: $$(x=R\cos(\phi)\cos(\theta),y=R\cos(\phi)\sin(\theta),z=R\cos(\theta))$$ then the average distance is :

\begin{equation} d = \frac{1}{4 \pi} \int_0^{\pi} \int_0^{2\pi} \sqrt{ (R \cdot \cos(\phi) \cdot \cos(\theta)-r)^2 + (R \cdot \cos(\phi) \cdot \sin(\theta) )^2 + ( R \cdot \sin(\theta) )^2 } \cdot \sin(\theta) d\theta d\phi \end{equation}

\begin{equation} d = \frac{1}{4 \pi } \int_0^{\pi} \int_0^{2\pi} \sqrt{ R^2 + r^2 - 2 R r \cos(\theta)\cos(\phi) } \sin(\theta) d\theta d\phi \end{equation}

does someone know how to solve this integral?

I computed it numerically (the mean distance between a point inside a ball and the surface) and the result should be: $$ d = \frac{r^2}{3R}+R $$

Thank you

Answer 1

my point $M$ inside the ball is defined by the coordinates: $$ (x=r,y=0,z=0) $$

It is best to choose point $M$ on the z-axis to best take advantage of spherical symmetry. Hence,

$$M(0, 0, r)$$

and the points on the sphere are defined by the coordinates: $$(x=R\cos(\phi)\cos(\theta),y=R\cos(\phi)\sin(\theta),z=R\cos(\theta))$$

This is incorrect. The coordinate on the sphere should be

$$x=R\sin(\theta)\cos(\phi),y=R\sin(\theta)\sin(\phi),z=R\cos(\theta)$$

$$d = \frac{1}{4 \pi } \int_0^{\pi} \int_0^{2\pi} \sqrt{ R^2 + r^2 - 2 R r \cos(\theta)\cos(\phi) } \sin(\theta) d\theta d\phi $$

Your integral limits for $\theta$ and $\phi$ are reversed. As said above, if you set point $M$ on z-axis, then you will get the integral as

$$d = \frac{1}{4 \pi } \int_0^{2\pi} \int_0^{\pi} \sqrt{ R^2 + r^2 - 2 R r \cos(\theta) } \sin(\theta) d\theta d\phi$$

Do substitution, $u=\cos\theta$, then you get,

$$d = \frac{1}{2} \int_{-1}^1 \sqrt{ R^2 + r^2 - 2 R r u } ~du=\color{red}{\frac{r^2}{3R}+R}$$