Particle in Sphere

Question

The position for a particle is random, with uniform distribution on the sphere that has its centre in origin and radius equal to 7. Calculate the expected value of the particle's distance from the origin.

Uniform in this sens should mean

$$ \begin{cases} f(x,y,z) = \frac{1}{\pi}, \ \ \sqrt{x^2 + y^2 + z^2} \leq 7 \\ 0, otherwise \end{cases} $$

But how do I calculate the expected value from here?

Answer 1

By the way, the density should be $f(x,y,z) = \frac{1}{\frac{4}{3} \pi \cdot 7^3}$ when $\sqrt{x^2+y^2+z^2} \le 7$.

If $R$ is the random distance, then the tail sum formula yields $$E[R] = \int_0^\infty P(R \ge r) \, dr.$$

What is $P(R \ge r)$?

$$P(R \ge r) = \begin{cases}0 & r > 7 \\ 1 - \frac{r^3}{7^3} & 0 \le r \le 7\end{cases}$$