The distribution of areas of a random triangle on the sphere - what are the second, third, etc. moments? Suppose that we choose three points independently and uniformly at random on the surface of a unit sphere as the vertices of a triangle, and consider the area of this triangle. Call this random variable $X$.
The area of such a triangle is the sum of its angles minus $\pi$, so by linearity of expectation the expected value of $X$ is just $3q-\pi$, where $q$ is the expected value of one of the angles. But by symmetry we can fix one point to be at the north pole and the other to lie on the Prime Meridian, from which it is obvious that the angle distribution is uniform on $[0,\pi]$. Thus $\mathbb{E}[X]=\pi/2$, or one-eighth of the sphere's area.
However, because the angles are not independent (they cannot sum to less than $\pi$, for instance), we cannot use this sort of logic to easily infer the values of the second and third moments of this distribution.
After gathering some numerical data, it appears that $\mathbb{E}[X^2]=\frac{\pi^2}2$, but I am not sure how to prove this. (Note that this is equivalent to the statement that the standard deviation of $X$ is $\pi/2$, which may be easier to show?)
I also have $\mathbb{E}[X^3]\approx 20.36$, or that the third moment of $X$ is approximately $4.86$ (either can be inferred from the other, given lower-order moments - they should differ by $\pi^3/2$). I haven't found any particularly nice formulas that match either of these values, though I'm not sure about the last digit in either of these estimates.
In general, what is $\mathbb{E}[X^n]$ or $\mathbb{E}[(X-\frac{\pi}2)^n]$? If any of the values are open, has it been discussed in the literature? Is there a nice geometric argument for $\mathbb{E}[X^2]$?
Edit: Here is a histogram of the area distribution from $1000000$ samples. Interestingly, it seems not to decay to $0$ at the upper bound of $2\pi$.

 A: 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.
