My problem is as follows: given a function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ where $n$ is some integer of order 10 and $f$ is defined by a product of (non-separable) linear piecewise functions, find the surface area of the polytope projected on the unit $n-1$-sphere. That is, find the (hyper)solid angle of a convex polyhedral cone defined by $f$, which is itself defined by $f(\vec{x}) = 1$ inside a particular convex polyhedral cone, and $0$ elsewhere (generalizing the formulae to calculate the solid angle of a convex polyhedral cone is something I have not yet tried, but I imagine it is difficult). Thus, if I project $f$ onto the $n-1$ dimensional hypersphere and integrate over the surface of the hypersphere, I should get the value I desire. I could also integrate over the $n$-dimensional unit ball, divide by the volume of the $n$-dimensional unit ball, and multiply by the surface area of the $n-1$ dimensional hypersphere. This would give an equivalent solution because the convex cone is both right and centered at the origin.
My approach so far has been to calculate the volume of the intersection of the cone and the unit ball, if only because integrating directly over the hypersphere requires either a change of coordinates or a non-trivial substition, both of which complicate matters.
Unfortunately, I don't think I can abandon the function $f$ entirely and simply integrate $1$ over the appropriate regions, because the region is difficult to determine, even in the case of only three defining inequalities - changing to spherical coordinates ends up requiring integrating very nasty things.
How does one go about integrating a piecewise function that is not easily separable over either a ball or a sphere? Is there some straightforward way I am missing to integrate such functions, or am I stuck using numerics? I do have numerical results, and I even have an analytical result for this particular cone (using spherical geometry result), but for higher-dimensional cases, the function becomes very sparsely supported and numerics get more and more difficult.