The distribution of areas of a random triangle on the sphere - what are the second, third, etc. moments?

I found a cute geometric argument that $\mathbb{E}[X^2]=\frac{\pi^2}2$, and it seemed worth presenting here. (Many statements like "almost surely" and "if points are in general position" and "except for a set of measure $0$" have been omitted for ease of reading.)

Claim: If five points are randomly chosen on the sphere, their convex hull will be a triangle with probability $\frac5{16}$.

Proof: We start with a lemma.

Lemma: Given any three random points on the sphere, adding a random fourth point has a $50\%$ chance of producing a triangular convex hull.

Proof of lemma: Draw the three great circles connecting each pair of the three points, which subdivides the sphere into eight regions. Four of these regions have the property that a fourth point inside them will yield a triangular convex hull, and each such region is opposite a congruent region inside which a fourth point would not yield a triangular convex hull. So for any location $P$ of the fourth point, exactly one of $P$ and $-P$ would work. This completes the proof.

Now, back to the main theorem.

Given five points, consider ways of labeling one point $A$ and another $B$. Obviously, every five points have $20$ such labelings. Say that a collection of labeled points is good if, starting with the three unnamed points, we can add $A$ to get a triangular convex hull, and add $B$ onto the previous four to again produce a triangular convex hull.

For instance, in the following diagram, the left arrangement is good, but the right two are not (the second one has a quadrilateral convex hull when we add $A$, the third when we add $B$).

Now, fixing a labeling, what are the odds that starting with the unnamed points, adding $A$ gives a triangular hull and adding $B$ to those four points does as well? By our lemma, $\frac12\cdot\frac12=\frac14$. So every choice of five points produces an expected $\frac{20}4=5$ good labelings.

Conversely, given $5$ points whose convex hull is a triangle, how many good labelings does it have? Well, any choice of three points gives a valid starting triangle, and by the time we add the fifth point, we'll have a triangle once again, so the only way it could fail is if the first four points form a quadrilateral. How many ways can that happen?

It shouldn't be too hard to convince yourself that every $5$-point arrangement with a triangular convex hull looks like this, in terms of incidences and intersections (I've drawn lines as straight for convenience):

Inspecting this a bit, we can see that only one of the five points, when removed, leaves a quadrilateral convex hull (the top one, in the above diagram). So that has to be our choice of $B$ for things to fail, and then any of the other four could be $A$. So there are $20-4=16$ good labelings that produce a given configuration.

So, a random five-point arrangement produces $5$ good labelings on average, and a working arrangement produces $16$ good labelings. (Obviously, if a five-point arrangement doesn't have a triangular hull, it won't produce any good labelings.) So the fraction of five-point arrangements that work must be $\frac{5}{16}$.

---

Okay, proof complete. Now what?

Observe that the probability five points have a triangular convex hull is $10$ times the probability that they have a triangular convex hull and the three points on said triangle were the first three chosen (since any of the ten possible three-subsets could have just as easily been that triangle). So there is a $\frac1{32}$ chance that, if we pick three random points, the next two random points we choose will be inside that triangle.

But given a triangle, the odds that two random points lie inside it is just the square of the fraction of the area of the sphere that it takes up! So what we have shown is that $\mathbb{E}[\left(\frac{X}{4\pi}\right)^2]=\frac1{32}$, and hence $\mathbb{E}[X^2]=\frac{\pi^2}{2}$.

Note that this sort of argument can't extend to $6$ points, because the resulting probability of a triangular convex hull will be ${6\choose 3}\cdot \mathbb{E}[\left(\frac{X}{4\pi}\right)^3] = \frac{15}{32} - \frac{15\log(2)}{4 \pi^2}$, which is almost certainly irrational.

What breaks with $6$ points? I think the fundamental problem is that there is more than one point/line arrangement with positive probability, so the number of good labelings isn't constant. Consider the below two configurations:

In the left case, removing any of the outer triangle's vertices will yield a quadrilateral convex hull, but in the right case, any vertex can be removed while preserving triangularity.

I'll confess some frustration with this answer, because I can quote various sources on this but not verify them readily for myself.

The major result to quote is the probability density for the area $x\in [0,2\pi]$ of a spherical triangle:

$$f_X(x)=-\frac{(x^2−4\pi x+3\pi^2−6)\cos(x)−6(x−2\pi)\sin(x)−2(x^2−4\pi x+3\pi^2+3)}{16\pi \cos(x/2)^4}.$$

This formula apparently goes back to at least 1867, appearing as problem 2370 in Mathematical Questions and Solutions from the “Educational Times” : It was proposed by M.W. Crofton and solved the pseudonymous "Exhumatus". This is available via Google Books here (p. 21-23). Finch and Jones 2010 review Exhumatus's approach; see also the discussion of Case 2 random spherical triangles in Philip 2015. Interestingly, there doesn't seem to be a direct derivation of the above formula from the trivariate density for the three angles (which is known.)

With this density in hand, one may use Mathematica to verify that the PDF is properly normalized and that

$$\mathbb E[X]=\frac{\pi}{2},\quad \mathbb E[X^2]=\frac{\pi^2}{2},\quad \mathbb{E}[(X-\pi/2)^2]=\frac{\pi^2}{4}$$ in agreement with the OP. Mathematica seems to struggle with finding closed-forms for the higher moments but does obtain

\begin{array}{ll} \mathbb{E}[X^3]=\frac32 \pi^3-12\pi \ln 2\approx 20.3784, &\mathbb{E}[(X-\pi/2)^3]=\pi^3-12\pi \ln 2,\\ \mathbb{E}[X^4]=4 \pi^4-24 \pi^2 \ln 2-\zeta(3) \approx 95.6281, &\mathbb{E}[(X-\pi/2)^4]=\frac{25}{16}\pi^4-108\zeta(3)\\ \mathbb{E}[X^5]=10 \pi^5-120 \pi^3 \ln 2 \approx 481.167, &\mathbb{E}[(X-\pi/2)^5]=\frac{13}{4}\pi^5 -90\pi^3 \ln 2+270\zeta(3)\\ \end{array} and so forth. At this point computing the central moments becomes too laborious for my sessions of Wolfram Cloud, but the next few non-central moments are

\begin{align} \mathbb{E}[X^6]&=24 \pi ^6-480 \pi^4 \ln 2-180 \pi^2 \zeta(3)+13500 \zeta(5)\\&\approx 2527.33 ,\\ \mathbb{E}[X^7]&=56 \pi^7-1680 \pi^5 \ln 2+1260 \pi^3\zeta(3)+47250 \pi \zeta(5)\\&\approx 13664.1 ,\\ \mathbb{E}[X^8]&=128 \pi^8-5376 \pi^6 \ln 2+12096 \pi^4 \zeta(3)+264600 \pi^2 \zeta(5)-1666980 \zeta(7)\\&\approx 75422.4,\\ \mathbb{E}[X^9]&=288 \pi^9-16128 \pi^7 \ln 2+66528 \pi^5 \zeta(3)+1020600 \pi^3 \zeta(5)-10001880 \pi \zeta(7)\\&\approx 422860.,\\ \mathbb{E}[X^{10}]&=640 \pi^{10}-46080 \pi^8 \ln 2+293760 \pi^6\zeta(3)+3175200 \pi^4 \zeta(5)-67869900 \pi^2 \zeta(7)+260253000 \zeta(9) \\&\approx 2399821.43. \end{align} I'm not sure how much farther this can be pushed using Wolfram Cloud. The best route would be to determine some generating function like $\mathbb{E}[e^{tX}]$ or some other convenient $\mathbb{E}[f(tX)]$, but I haven't succeeded in obtaining such.

As a last point of agreement with the OP, the density does not vanish at $x=2\pi$: $$f_X(2\pi)=-\frac{(-\pi^2-6)-2(-\pi^2+3)}{16\pi}=\frac{3}{4\pi}-\frac{\pi}{16}\approx 0.0424. $$ This seems in good agreement with the histogram obtained by the OP. My own intuitive explanation for this is that the area is nearly maximized occurs when the vertices lie near some common great circle, and there are "many" ways to form such triangles. Thus there's no reason for the probability density to vanish as $x\to 2\pi$.