Expected distance in a ball

Solution. By rotational symmetry, condition on $P_1$ at distance $r$ from the origin; WLOG take $P_1 = (0, 0, r)$. The marginal density of $|P_1|$ is $3 r^2$ for $r \in [0, 1]$ (the Jacobian of the radial-coordinate change in the unit ball). Similarly, $|P_2|$ has density $3 s^2$ for $s \in [0, 1]$.

For fixed $r, s$, the angle $\phi$ between $P_1$ and $P_2$ has density $\tfrac12 \sin\phi$ on $[0, \pi]$ (standard spherical coordinates). Hence $$\mathbb{E}[|P_1 - P_2| \mid r, s] = \int_0^{\pi} \sqrt{r^2 + s^2 - 2 r s \cos\phi} \cdot \tfrac12 \sin\phi \, d\phi.$$ Substitute $u = \cos\phi$, so $du = -\sin\phi \, d\phi$: $$= \tfrac12 \int_{-1}^{1} \sqrt{r^2 + s^2 - 2 r s u} \, du.$$ The inner integral evaluates (by the standard antiderivative $\int \sqrt{a - bu} \, du$) to $$\frac{(r + s)^3 - |r - s|^3}{6 r s}.$$ For $r \geq s$, $(r + s)^3 - (r - s)^3 = 6 r^2 s + 2 s^3$, so the conditional expectation simplifies to $$\mathbb{E}[|P_1 - P_2| \mid r, s] = \frac{6 r^2 s + 2 s^3}{6 r s} = r + \frac{s^2}{3 r}.$$

Combine over $r, s$. By symmetry in $r, s$, we may write $$\mathbb{E}[|P_1 - P_2|] = 18 \int_0^1 \!\! \int_0^r \!\! \left( r + \tfrac{s^2}{3 r} \right) r^2 s^2 \, ds \, dr,$$ where the factor of $18$ comes from $2 \cdot 3 \cdot 3$ ($2$ from the $r \leftrightarrow s$ symmetry, and $3 r^2 \cdot 3 s^2$ from the two radial densities). Carry out the inner integral: $$\int_0^r \left( r^3 s^2 + \tfrac{r s^4}{3} \right) ds = \frac{r^6}{3} + \frac{r^6}{15} = \frac{2 r^6}{5},$$ the integrand being $\left(r + \tfrac{s^2}{3r}\right) r^2 s^2 = r^3 s^2 + \tfrac{r s^4}{3}$ (keeping the $r^2 s^2$ density factor). Then $$\mathbb{E}[|P_1 - P_2|] = 18 \cdot \tfrac{2}{5} \int_0^1 r^6 \, dr = \frac{36}{5} \cdot \frac{1}{7} = \frac{36}{35}.$$ So $$\boxed{\mathbb{E}[|P_1 - P_2|] = \frac{36}{35} \approx 1.029.} \qedhere$$