Find a transformation from tetrahedron to cube in $R^3$ to calculate a triple integral?

Question

I would like to calculate the triple integral of a function $f(x,y,z)$ over a region given by a tetrahedron with vertices $(0,0,0)$, $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. I am trying to do this by transforming the region to the unit cube but I am having trouble finding a transform to do this. I think that a transformation from a tetrahedron with corners $(0,0,0)$, $(0,0,1)$, $(0,1,0)$ and $(1,0,0)$ to the unit cube is $x=uvw$, $y=uv(1-w)$, $z=u(1-v)$. How can I adapt this for the tetrahedron with the more general vertices?

Answer 1

The linear map $(x,y,z)\mapsto (x/a,y/b,z/c)$ transforms the tetrahedron with vertices you mention into the standard simplex with vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. You can compose it with your transform onto the unit cube.

That said, I am not convinced that the gain outweighs the trouble. The triple integral over an axis-aligned tetrahedron like the one you start with is not hard to set up directly.