How to "split up" triple integrals when changing order of integration? I'm trying to set up six different triple integrals of a tetrahedron and am having difficulty changing the order of integration, particularly finding the bounds for dzdxdy/dzdydx and dxdydz/dxdzdy. 
I am given the vertices of the tetrahedron, (8,7,1), (7,4,1), (2,5,1), and (5,5,5), and I have found the equations of the planes (see image below).

I was able to come up with the projection on the xy-plane (see image below), but don't know how to "split it up" so I can find the bounds/integrate. 
 
I know with certain problems you need to divide the shape into subsections so each part can be integrated separately and then add them, but I've scoured my book, notes, and the internet and am still not sure how to split this particular triangle up. Is there a method to it (e.g. equations you can solve for), or do you just split it up however you like so that integration is possible for each of the sections?
I'd appreciate any help regarding how I can move forward with this problem specifically, or just general information about how to "break up" integrals that are 2D projections. Thanks!
 A: If you can't change variables, then I think the best way to do it is to approximate it in slices of height $\Delta z$ parallel to the $xy-$plane,, and then let $\Delta z\to 0.$  Basically, we will integrate $xy$ in cross-sections parallel to the $xy-$plane, which will, of course be triangles.  To integrate over the triangle, first surround in by a rectangle with sides parallel to the $x-$ and $y-$axes.  Integrate over the rectangle, and subtract the integral over four right triangles, each of which has two sides parallel to the axes.  This will result in a function of $z$.  Integrate this function for $1\leq z\leq5.$
I'll try to get you started.  First we determine the intersection with the plane with $z-$coordinate $z$ by parameterizing the edges that meet at $P=(5,5,5).$  The edge joining P to $(7,4,1)$ is 
$\{(5+2t,5-t,5-4t)|0\leq t \leq 1\}.$  Eliminating $t,$ we get $$(x_1,y_1) = \left({15-z\over2},{15+z\over4}\right)\tag{1}$$ as one vertex of the triangle.  Following the same procedure for the other vertices, we get $$(x_2,y_2)=\left({35-3z\over4},{15-z\over2}\right)\tag{2}$$ and $$
(x_3,y_3)=\left({3z+5\over4},5\right)\tag{3}$$
When $z\leq5,$ we have $${3z+5\over4}\leq{15-2z\over2}\leq{35-3z\over4}$$ so we will take $${3z+5\over4}\leq x\leq{35-3z\over4}\tag{4}$$  Similarly, when $z\leq5,$ we have $$
{15+z\over4}\leq5\leq{15-z\over2},$$ so we will take $$
{15+z\over4}\leq y\leq{15-z\over2}\tag{5}$$  Inequalities $(4)$ and $(5)$ define the rectangle in the cross-sectional plane.  It's easy to integrate $kxy$ over this rectangle.  (Remember that $z$ is constant.)  
It's straightforward, but lengthy, to figure out which triangles you have to integrate over and subtract.  (Or at least it should be; I admit I don't have the patience to do it, but this answer should give you a good start.)  If you do this, you will have polynomial of low degree in $z$ to integrate from $z=1$ to $z=5.$ 
