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?



Is there a neat formula for the volume of a tetrahedron on $S^3$?
[On the volume of a hyperbolic and spherical tetrahedron](http://www.f.waseda.jp/murakami/papers/tetrahedronrev3.pdf), by Murakami and Yano. The volume is obtained as a linear combination of dilogarithms and squares of logarithms. The origin of their formula is really interesting: Asymptotics of quantum $6j$ symbols. (These asymptotics have also been studied by many other people: D. Thurston, [Roberts](http://xxx.soton.ac.uk/abs/math-ph/9812013), Woodward, Frohman, Kania-Bartoszynska, etc.)


Note that the 3-dimensional formula has to be much more complicated. The 2-dimensional formula comes from Euler characteristic and Gauss-Bonnet, but the Euler characteristic of the 3-sphere, or any odd-dimensional manifold, vanishes. In fact every characteristic class of a 3-sphere vanishes, because the tangent bundle is trivial. There can't be a purely linear treatment of volumes in isotropic spaces in odd dimensions. In even dimensions, there is always a purely linear extension from lower dimensions using generalized Gauss-Bonnet.


