How can we calculate $\int_{r\in \mathbb{S}^2}\ {\rm Area}\ \Delta p_0q_0r\ d{\rm Area}_r$

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Define a function $f : \mathbb{S}^2\times \mathbb{S}^2\times \mathbb{S}^2\rightarrow \mathbb{R}$ by $f(p,q,r)$ to be a area of the geodesic triangle $pqr$ in the unit sphere $\mathbb{S}^2$.

(Here the area of the triangle $pqr$ is $\angle p + \angle q +\angle r -\pi$).

Question 1 : Here how can we find $$ \int_{\mathbb{S}^2\times \mathbb{S}^2\times \mathbb{S}^2 } \ f(p,q,r) d{\rm Vol}(p,q,r) $$

Question 2 : Furthermore, how can we find the following $$\int_{\mathbb{S}^2 } \ f(p_0,q_0,r) d{\rm Vol}(r) $$

Question 3 : Consider the triangle $\Delta pqr$ where the order of $p,\ q,\ r$ is positively oriented. And define $$A = \frac{ q\times p }{| q\times p| },\ B = \frac{r\times q}{|r\times q|},\ C = \frac{p\times r}{|p\times r|} $$

Note that $\angle \ (A,B) =\pi-\angle p $ so that $$ \angle (A,B) + \angle (B,C)+\angle (A,C) = 2\pi - {\rm Area}\ \Delta pqr $$

Hence ${\rm perim}\ \Delta ABC$, the sum of all side lengths in $\Delta ABC$, is equal to $2\pi - {\rm Area}\ \Delta pqr$.

Hence we want to calculate $$ \int_{(A,B,C)\in \mathbb{S}^2 \times\mathbb{S}^2\times \mathbb{S}^2}\ {\rm perim}\ \Delta ABC \ d{\rm Vol}_{ (A,B,C) }$$

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The area of a spherical triangle $ABC$ on a unit sphere with respective angles $\alpha,\beta,\gamma$ in radians has an extremely nice formula:

$$\text{Area}(\triangle ABC) = \alpha+\beta+\gamma-\pi$$

There is a very cute proof of this formula using some clever geometric arguments given here among other places.

Because this formula is linear in the angles, we can just use linearity of expectation: compute the average value $x$ of an angle in a spherical triangle, and our answer will be $x+x+x-\pi$.

Without loss of generality, let $A$ be at the north pole, and fix $B$ to be on the Prime Meridian. Then the longitude of $C$ will be uniformly distributed on $[-180,180]$, so the angle $\alpha$ is uniformly distributed between $0$ and $\pi$; in particular, its expectation is $\pi/2$. So the expected area is simply

$$3\cdot\frac{\pi}{2}-\pi = \frac{\pi}2$$

or $1/8$ of the sphere.