The expected distance between two points on a sphere and on a circle 
Two points are randomly selected on the circle. What is the expected
distance between them? And what will be the expected distance between
the two points on a sphere?

An interesting problem, I had several ideas: we can generate a uniform distribution in an $n$-dimensional cube described around a unit ball, remove points outside the ball from the sample, and obtain a uniform distribution of vectors in the ball. We normalize the vectors - we get on the sphere. And we use Monte Carlo.... Of course, there are a lot of iterations, and the accuracy is low, but for a rough estimate and for checking the exact calculations, it will do well.
I also reasoned like this: since the task does not change when rotating, I can take one point fixed. We get the expectation of the distance from a random point of the circle to a fixed one. This is an obvious integral:
$$ \frac{1}{2 \pi} \int_{-\pi}^{\pi} \sqrt{(1-\cos x)^{2}+\sin ^{2} x}\,dx=\frac{1}{\pi} \int_{-\pi}^{\pi}\left|\sin \frac{x}{2}\right|dx=\frac{2}{\pi} \int_{0}^{\pi} \sin \frac{x}{2}\,dx=\frac{4}{\pi}$$ - for the unit circle, of course.
$$ \begin{aligned} &\frac{1}{4 \pi} \int_{0}^{\pi} \int_{-\pi}^{\pi} \sin \theta \sqrt{(1-\cos \theta)^{2}+\sin ^{2} \theta \cos ^{2} \varphi+\sin ^{2} \theta \sin ^{2} \varphi}\, d \varphi d \theta= \\ &=\frac{1}{2} \int_{0}^{\pi} 2 \sin \frac{\theta}{2} \sin \theta\, d \theta=\frac{1}{2} \int_{0}^{\pi} \cos \frac{\theta}{2}-\cos \frac{3 \theta}{2} d \theta=\left.\left(\sin \frac{\theta}{2}-\frac{1}{3} \sin \frac{3 \theta}{2}\right)\right|_{0} ^{\pi}=\frac{4}{3} \end{aligned}.$$ -and this is for the sphere.
Parameterization of the sphere: $z=\cos(\theta)$, $x=\sin(\theta)\cos(φ)$, $y=\sin(\theta)\sin(φ)$. As a fixed point we take $(0,0,1)$. Jacobian $\sin(\theta)$. I took a slightly non-standard parameterization relative to theta, so that later I would not mess with things like $\sin (\pi/4-\theta/2)$.
Here are my thoughts. I ask you to double-check me and, if possible, write your own version.
 A: I do not have a lot of time now, so let me look at the circle case only. The assumption of one fixed point was a wise shortcut. I might exploit the symmetry to suggest another here. Let $\theta$ be the central angle subtended by the two points. It is only necessary to integrate from 0 to $\pi$. Let $r$ be the circle radius.
$\frac{2r}{\pi} \int_{0}^{\pi} \sin \frac{\theta}{2} d\theta = \frac{4r}{\pi}$
Edit:
Having more time now, I might look at the sphere. I am not exactly following your integral, but we are reaching the same bottom line, so that looks good. Before starting the sphere, I want to come clean on my circle. There was a typo in the integral, but remarkably it led to the same result. I have since corrected it.
Let $\varphi$ be the angle subtended by the fixed point and the variable point on the sphere. Again use radius $r$. The locus of the variable points defining that angle is a circle with radius $r\sin \varphi$, so its circumference is $2\pi r\sin \varphi$. Let it mark a strip of width $rd\varphi$, giving it area $2\pi r^2 \sin \varphi d\varphi$. The probability of the point falling on that strip is that area divided by the total surface area of the sphere.
probability = $\frac{2\pi r^2 \sin \varphi d\varphi}{4\pi r^2} = \frac{1}{2} \sin \varphi d\varphi$
length of chord = $2r\sin \frac{\varphi}{2}$
expectation = $r \int_{0}^{\pi} \sin \varphi \sin \frac{\varphi}{2} d\varphi$
= $2r \int_{0}^{\pi} \sin^2 \frac{\varphi}{2}\cos \frac{\varphi}{2} d\varphi$
= $\frac{4r}{3} [\sin^3 \frac{\varphi}{2}]_{0}^{\pi}$
= $\frac {4r}{3}$
I tend to use geometry as far as it will carry me. In this case, that served to simplify both integrals. Not everyone responds to that. See what you think.
