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Suppose we have a semisphere of radius 1. We choose two random points inside it with a uniform distribution. That is, if we pick random points insed it, they will be uniformly distributed.

Picking two random points inside the semisphere

What is the average distance between two randomly selected points inside the semisphere?

Similarly, conside a quarter-sphere with radius 1.

Picking two random points inside the quarter-sphere

What is the average distance between two randomly selected points inside the quarter-sphere?

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  • $\begingroup$ Is your semisphere $\{(x,y,z) : z \ge 0, x^2 + y^2 = 1\}$ or $\{(x,y,z) : z \ge 0, x^2 + y^2 \le 1\}$? (It's not clear whether you are including the interior, or only considering the boundary of the ball.) $\endgroup$
    – angryavian
    Commented Jun 3 at 15:29
  • $\begingroup$ The interior of the ball $\endgroup$
    – Matheman242
    Commented Jun 3 at 15:36

1 Answer 1

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Hemisphere only. Only need to change limits of integration and leading weight of integral to solve for qarter sphere case.

$d^2=(r_1\cos\theta_1\sin\phi_1-r_2\cos\theta_2\sin\phi_2)^2+(r_1\sin\theta_1\sin \phi_1-r_2\sin\theta_2\sin\phi_2)^2+(r_1\cos\phi_1-r_2\cos\phi_2)^2$

$d^2=r_1^2\cos^2\theta_1\sin^2\phi_1+r_2^2\cos^2\phi_2\sin^2\phi_2^2-2r_1r_2\sin\phi_2\sin \phi_1\cos\theta_1\cos\theta_2+r_1^2\sin^2\theta_1\sin^2\phi_1+r_2^2\sin^2\theta_2\sin^2\phi_2-2r_1r_2\sin\theta_1\sin\theta_2\sin\phi_1\sin\phi_2+r_1^2\cos^2\phi_1+r_2^2\cos^2\phi_2-2r_1r_2\cos\phi_1\cos\phi_2$

$d^2=r_1^2+r_2^2-2r_1r_2\sin\phi_1\sin\phi_2\cos(\theta_1-\theta_2)-2r_1r_2\cos\phi_1\cos\phi_2$

$\frac{3}{2\pi R^3}\int_0^R\int_0^{2\pi}\int_0^{\pi/2}\sqrt{r_1^2+r_2^2-2r_1r_2\sin\phi_1\sin\phi_2\cos(\theta_1-\theta_2)-2r_1r_2\cos\phi_1\cos\phi_2} \ r_2^2\sin \phi_2 d\phi_2 d\theta_2dr_2$

WLOG $\theta_1=0.$

$\frac{3}{2\pi R^3}\int_0^R\int_0^{2\pi}\int_0^{\pi/2}\sqrt{r_1^2+r_2^2-2r_1r_2\sin\phi_1\sin\phi_2\cos(\theta_2)-2r_1r_2\cos\phi_1\cos\phi_2} \ r_2^2\sin \phi_2 d\phi_2 d\theta_2dr_2$

$r_1=\alpha R$. $r_2=\beta R. dr_1=Rd\alpha. dr_2=Rd\beta $

$F(\alpha, \phi_1)=\frac{3R}{2\pi }\int_0^1\int_0^{2\pi}\int_0^{\pi/2}\sqrt{\alpha^2+\beta^2-2\alpha\beta\sin\phi_1\sin\phi_2\cos\theta_2-2\alpha \beta\cos\phi_1\cos\phi_2}\beta^2\sin \phi_2d\phi_2d\theta_2d\beta$

$<d>=\frac{4}{\pi }\int_0^1\int_0^{\pi/2} F(\alpha,\phi_1) \alpha d\alpha d\phi_1$

I don't think the integrals have a closed form. Probably need to use elliptical integrals or numerical methods to evaluate.

Average distanced is $<d>=6CR/\pi^2$ where $C$ results from the integration.

Is interesting this is proportional to $1/\zeta(2)$, the probability that any two positive integers are relatively prime.

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