Offset Sphere Expected Distance

$\large D^{2} = x^{2}+y^{2}+z^{2}$

What we want to find out is $\large \frac{1}{\frac{4\pi}{3}}\iiint_{P} D^{2} dxdydz$ where $P$ is the volume within the sphere $(x-1)^{2}+(y-1)^{2}+(z-1)^{2} =1$ .

Let us use a variable change of $x=1+u$ , $y=1+v$ and $z=1+w$ . The Jacobian would simply be $1$ .

we need to find the integral $\large \iiint_{Q} ((1+u)^{2}+(1+v)^{2}+(1+w)^{2}) dudvdw$ where $Q$ is the volume within the sphere $u^{2}+v^{2}+w^{2}=1$

Now shifting into polar coordinates where $u=r\sin(\theta)\cos(\phi)$ , $v=r\sin(\theta)\sin(\phi)$ and $w=r\cos(\theta)$ . The Jacobian of this transformation is $\lvert J \rvert = r^{2}\sin(\theta)$ We have:-

$\large \int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{1}\left(3+r^{2} + 2r(\sin(\theta)\cos(\phi)+\sin(\theta)\sin(\phi)+\cos(\theta))\right) r^{2}\sin(\theta) drd\phi d\theta$

Now $\large \int_{0}^{2\pi}\sin(\phi)d\phi =\int_{0}^{2\pi}\cos(\phi)d\phi = 0$

and $\int_{0}^{\pi}\sin(\theta)\cos(\theta)d\theta = 0$ .

Hence we have to find:-

$\large \int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{1} (3+r^{2}) r^{2}\sin(\theta) drd\phi d\theta$ (as the remaining integrals become 0).

We know $\large \int_{0}^{\pi}\sin(\theta)d\theta = 2$ and $\large \int_{0}^{2\pi}d\phi = 2\pi$

So our integral reduces to $\large 4\pi(\int_{0}^{1}(3r^{2}+r^{5})dr)$

So this integral is easy enough to evaluate and we will get the answer as $\large 2\pi(2+\frac{2}{5})=\frac{24\pi}{5}$ .

So the expected value of $D^{2}$ is $\displaystyle \frac{\frac{24\pi}{5}}{\frac{4\pi}{3}} = \frac{18}{5} = 3.6$

Shift the coordinate system so that the centre of the sphere is at the origin, and $D$ is the distance from the point $(-1,-1,-1)$ . Let $\mathbf{a} = (1\;1\;1)^T$ . Then, integrating over the sphere, $\begin{aligned} \iiint |\mathbf{r} - \mathbf{a}|^2\,dV & = \; \iiint \big(|\mathbf{r}|^2 - 2\mathbf{a}\cdot\mathbf{r} + |\mathbf{a}|^2\big)\, dV \; = \; \iiint |\mathbf{r}|^2\,dV - 2\mathbf{a}\cdot \iiint\mathbf{r}\,dV + |\mathbf{a}|^2 \iiint \,dV \\ & = \; \int_0^1 r^2 \times 4\pi r^2\,dr - 2\mathbf{a}\cdot\mathbf{0} + \tfrac43\pi|\mathbf{a}|^2 \; = \; \tfrac45\pi + \tfrac43\pi \times 3 = \tfrac{24}{5}\pi \end{aligned}$ since the origin is the centre of mass of the sphere. Dividing by $\tfrac43\pi$ tells us that $\mathbb{E}[D^2] \; = \; \tfrac{18}{5} \; = \; \boxed{3.6}$

More generally, consider a sphere of radius $R$ centered at $O = (0,0,0)$ , and let $P = (a,b,c)$ .

Parametrically, we can represent a point inside the sphere by $(r \cos \theta \sin \phi, r \sin \theta \sin \phi, r \cos \phi)$ , where $0 \le r \le R$ , $0 \le \theta \le 2 \pi$ , and $0 \le \phi \le \pi$ . If $D$ is the distance from a point inside the sphere to $P$ , then the expected value of $D^2$ is $\begin{aligned} &\frac{1}{\frac{4}{3} \pi R^3} \int_0^R \int_0^{2 \pi} \int_0^\pi [(r \cos \theta \sin \phi - a)^2 + (r \sin \theta \sin \phi - b)^2 + (r \cos \phi - c)^2] r^2 \sin \phi \ dr \ d \theta \ d \phi \\ &= \frac{1}{\frac{4}{3} \pi R^3} \int_0^R \int_0^{2 \pi} \int_0^\pi (r^2 \cos^2 \theta \sin^2 \phi - 2ar \cos \theta \sin \phi + a^2 + r^2 \sin^2 \theta \sin^2 \phi - 2br \sin \theta \sin \phi + b^2 + r^2 \cos^2 \phi - 2cr \cos \phi + c^2) r^2 \sin \phi \ dr \ d \theta \ d \phi \\ &= \frac{1}{\frac{4}{3} \pi R^3} \int_0^R \int_0^{2 \pi} \int_0^\pi (r^2 - 2ar \cos \theta \sin \phi - 2br \sin \theta \sin \phi - 2cr \cos \phi + a^2 + b^2 + c^2) r^2 \sin \phi \ dr \ d \theta \ d \phi \\ &= \frac{1}{\frac{4}{3} \pi R^3} \int_0^R \int_0^{2 \pi} \int_0^\pi r^4 \sin \phi - r^3 (2a \cos \theta \sin^2 \phi + 2b \sin \theta \sin^2 \phi + 2c \cos \phi \sin \phi) + r^2 (a^2 + b^2 + c^2) \sin \phi \ dr \ d \theta \ d \phi \\ &= \frac{1}{\frac{4}{3} \pi R^3} \int_0^R \int_0^\pi 2 \pi r^4 \sin \phi - 4 \pi cr^3 \cos \phi \sin \phi + 2 \pi r^2 (a^2 + b^2 + c^2) \sin \phi \ dr \ d \phi \\ &= \frac{1}{\frac{4}{3} \pi R^3} \int_0^R 4 \pi r^4 + 4 \pi r^2 (a^2 + b^2 + c^2) \ dr \\ &= \frac{1}{\frac{4}{3} \pi R^3} \left[ \frac{4 \pi R^5}{5} + \frac{4 \pi R^3 (a^2 + b^2 + c^2)}{3} \right] \\ &= \frac{3}{5} R^2 + a^2 + b^2 + c^2 = \frac{3}{5} R^2 + OP^2. \end{aligned}$