TITLE: Is there a neat formula for the volume of a tetrahedron on $S^3$?
QUESTION [23 upvotes]: There is a nice formula for the area of a triangle on the 2-dimensional sphere;
If the triangle is the intersection of three half spheres, and has angles $\alpha$, $\beta$ and $\gamma$, and we normalize the area of the whole sphere to be $4\pi$ then the area of the triangle is 
$$
\alpha + \beta + \gamma - \pi.
$$
The proof is a cute application of inclusion-exclusion of three sets, and involves the fact
that the area we want to calculate appears on both sides of the equation, but with opposite signs.
However, when trying to copy the proof to the three dimensional sphere the parity goes the wrong way and you get 0=0.
Is there a simple formula for the volume of the intersection of four half-spheres of $S^3$ in terms of the 6 angles between the four bounding hyperplanes?

REPLY [4 votes]: Nice answer, Greg. I looked at the linked paper and was sufficiently intimidated.
I just want to point out, again, that for those (like me) who have a phobia of differential geometry, and hence don't want to use (generalized)Gauss-Bonnet, it is easy to see, using inclusion-exclusion, that the formula in even dimensions is a neat linear combination of
the formulas in lower dimensions.