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Numerical integration suggests that $$\mathcal U=\int_0^\pi\int_0^\pi\int_0^\pi\arccos\left(\cos x\cdot\cos y+\sin x\cdot\sin y\cdot\cos z\right) dx dy dz\stackrel{\small\color{gray}?}=\frac{\pi^4}2\tag1$$ (note that the function being integrated represents the great-circle distance$^{[1]}$$\!^{[2]}$$\!^{[3]}$ in spherical coordinates).

How can we prove it?

Is it possible to find a closed form for this one? $$\mathcal W=\int_0^{\tfrac\pi2}\!\int_0^\pi\int_0^\pi\arccos\left(\cos x\cdot\cos y+\sin x\cdot\sin y\cdot\cos z\right) dx dy dz\tag2$$

Vladimir Reshetnikov
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2 Answers2

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$$\mathcal U=\int_0^\pi\int_0^\pi\int_0^\pi\arccos\left(\cos x\cos y+\sin x\sin y\cos z\right) dx dy dz$$ $$\overset{\large {x\to\pi -x \atop \large z\to\pi-z}}=\int_0^\pi\int_0^\pi\int_0^\pi\left(\pi-\arccos\left(\cos x\cos y+\sin x\sin y\cos z\right)\right) dx dy dz$$ $$\Rightarrow 2\mathcal U=\pi \int_0^\pi\int_0^\pi\int_0^\pi dxdydz\Rightarrow \mathcal U=\frac{\pi^4}{2}$$

The original integral can be evaluated by symmetry since $\cos(\pi-a)=-\cos a$ (it is crucial to have the $z$ bounds from $0$ to $\pi$ in order to achieve this). However, in the second integral the upper bound is $\frac{\pi}{2}$, therefore this symmetry doesn't work.

Zacky
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In spherical coordinates the distance between two points on the unit sphere is

$$x \cdot y = \cos \alpha$$

With

$$ x\ \to\ (\cos \phi \ \sin \theta, \sin \phi \ \sin \theta, \cos \theta), \ y\ \to\ (\cos \psi \ \sin \eta, \sin \psi \ \sin \eta, \cos \eta)$$

     \[Pi]/2 - ArcCos[Cos[\[Theta]] Cos[\[Phi] - \[Psi]] Sin[\[Eta]] + 
 Cos[\[Eta]] Sin[\[Theta]]] // TrigToExp // FullSimplify

$$ \text{arcsin}(\sin (\eta )\ \cos (\theta ) \ \cos (\phi -\psi )\ + \ \cos (\eta ) \ \sin (\theta ))$$

This function is odd by its $\cos$ terms under $(\eta.,\theta) \to (\pi-\eta,\pi-\theta)$, for any value of $\alpha$, its integral vanishes.

What is the meaning of integral over a cube of arc length of meridians? Seems to be a constant Haar measure over geodesic orbits. It seems to depend on the difference betweeen the two angles $\phi-\psi$ in the equatorial plane, but not really from axial rotational invariance.

Holding the difference fixed, it yields the area of the square of two meridian lengths $F=\frac{\pi^2}{4}$ multiplied by the constant $\frac{\pi}{2}$.

Roland F
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