Integrate area of function over a tetrahedron I actually attempted to enlist my professor help on this problem, but what he said was quick and I must not have written everything down because I cannot understand how this problem is supposed to be done.
$$
\text{Evaluate the triple integral  }\displaystyle \iiint_{E} xy \, dV \text{ where }E\text{ is the solid tetrahedron with vertices }(0,0,0), (8,0,0), (0,8,0), (0,0,9).
$$
We need to pick a side and use it as the base of the shape, so we pick the easy side that's already on the $xy$ plane. Using $(0,0,0), (8,0,0), (0,8,0)$. Now we pick our height, $z = 9 - \frac{9}{8}x - \frac{9}{8}y$ which we can now use as one edge of our integration, or at least that's what my professor said to do.
He ended up with an equation looking like this:
$$
\int_0^8 \int_0^{9-\frac{9}{8}x} \int_0^{9-\frac{9}{8}x-\frac{9}{8}y} xy\:dz\:dy\:dx
$$
This does not make sense to me. After the innermost integraion has been completed, we will end up with a $y$ factor we can't get rid in the outer integrals because we've already used up our $dy$.
How do we do this problem?

UPDATE:
I miss read the order of integration he put. The correct order was $dz dy dx$ That is more than likely the source of my confusion.

UPDATE:
This took a while to do by hand, just because expanding terms became very physically tedious to do. The ending answer I resulted with was $\frac{1458}{5}$. To make sure I did in-fact get the correct answer, I was able to compare it against wolframalpha and it matches the number I got by hand. So this makes me believe that there is still a problem with the way I've setup the problem. Is this not the conceptually correct way to do this?
 A: The possible $6$ orderings of integration are
$$\begin{align}
&\mathrm{z-y-x\implies\int_0^{8}\int_{0}^{8-x}\int_{0}^{9-9(x+y)/8}\text{ (stuff) }\;dzdydx}\\
&\mathrm{z-x-y\implies\int_0^{8}\int_{0}^{8-y}\int_{0}^{9-9(x+y)/8}\text{ (stuff) }\;dzdxdy}\\
&\mathrm{y-z-x\implies\int_{0}^{8}\int_{9}^{9-9x/8}\int_{0}^{8-x-8z/9}\text{ (stuff) }\;dydzdx}\\
&\mathrm{y-x-z\implies\int_{0}^{9}\int_{0}^{8-8z/9}\int_{0}^{8-x-8z/9}\text{ (stuff) }\;dydxdz}\\
&\mathrm{x-z-y\implies\int_{0}^{8}\int_{0}^{9-9y/8}\int_{0}^{8-y-8z/9}\text{ (stuff) }\;dxdzdy}\\
&\mathrm{x-y-z\implies\int_{0}^{9}\int_{0}^{8-8z-9}\int_{0}^{8-y-8z/9}\text{ (stuff) }\;dxdydz}\\
\end{align}$$
Replacing "stuff" with $\mathrm{xy}$ and integrating I get ${1536\over5}$.
A: To carry out the integral computations themselves is a bit treacherous in this problem, particularly because of that trinomial with a rational multiplier in the terms.    Something can be done to cut down on the amount of arithmetic (although not much can be done about all the algebra).
I think the most convenient arrangement for the integral is the one which Jack's wasted life lists fourth.  If we take "slices" through the tetrahedron parallel to the $ \ xy-$ plane, they are all right isosceles triangles bounded by the $ \ x-$ and $ \ y-$ axes and lines of the form $ \ x + y \ = \ c \ $ , with the constant running linearly from $ \ c = 8 \ $ in the $ \ xy-$ plane itself ( $ z = 0 $  ) to $ \ c = 0 \ $ at the apex of the tetrahedron  ( $ z = 9 $  ) .  The integral on each of these "horizontal" slices is then $ \int_0^c \int_0^{c - x} \ xy \ \ dy \ dx \ $ .  It is here that we introduce the dependence on the $ \ z-$ coordinate of the slice with $ \ c = \ 8 \ - \ \frac{8}{9} z \ $ .  Now integrating the horizontal slices from the $ \ xy-$ plane up to the apex leads us to the integrand presented,
$$ \int_0^9  \ \int_0^{8 \ - \ \frac{8}{9} z} \  \int_0^{8 \ - \ \frac{8}{9} z \ - \ x} \ xy \ \ dy \ dx \ dz \ \ . $$
So far, nothing new.
We can see that the trinomial is going to mean real work for us, since it will be getting raised to powers, but that factor of $ \ \frac{8}{9} \ $ will get rather annoying.  Because all of the boundaries in this figure can be represented by linear combinations of the coordinates, however, we can perform a "re-scaling" in the "vertical" direction, using $ \ \zeta \ = \ \frac{8}{9} z \ \ \Rightarrow \ \  dz \ = \ \frac{9}{8} \ d\zeta \ $ .  This transforms our integral to
$$ \longrightarrow \ \ \int_0^{\mathbf{8}}  \ \int_0^{8 \ - \ \zeta} \  \int_0^{8 \ - \ \zeta \ - \ x} \ xy \ \ dy \ dx \ ( \frac{9}{8}  \ d\zeta \ ) \ \ . $$
This moves all of the complication in the arithmetic, resulting from working with $ \ \frac{8}{9} \ $ , to a single multiplier of the entire integral.
We still have a lot of algebra to contend with. It might seem that a "horizontal-shift" transformation, perhaps like $ \ \xi \ = \ 8 \ - \ \zeta \ $ , could be useful to reduce that trinomial; this idea proves to be helpful as well.  The integral now transforms as
$$ \longrightarrow \ \ \frac{9}{8} \ \int_8^0  \ \int_0^{\xi} \  \int_0^{\xi \ - \ x} \ xy \ \ dy \ dx \ ( -  \ d\xi \ ) \ = \ \frac{9}{8} \ \int_0^8  \ \int_0^{\xi} \  \int_0^{\xi \ - \ x} \ xy \ \ dy \ dx \  d\xi  \ \ , $$
which leaves us with nothing worse than a binomial to square.  Once we consolidate the polynomial in $ \ \xi \ $ , we have (and here's where this transformation really "pays off")
$$ \frac{9}{16} \ \int_0^8 \ \int_0^{\xi} \ \left(   \xi^2  x \ - \ 2 \xi \ x^2 \ + \ x^3 \right) \ \ dx \ d\xi \ =  \frac{9}{16} \ \int_0^8 \ \left( \frac{1}{2} \xi^2 x^2 \ - \ \frac{2}{3} \xi \ x^3 + \ \frac{1}{4} x^4 \right) \vert_0^{\xi} \ \ d\xi \ $$
$$ = \ \frac{9}{16} \ \int_0^8 \ \left(  \frac{1}{2}  \ - \ \frac{2}{3} \ + \ \frac{1}{4} \right) \ \xi^4 \ \ d\xi  \ = \ \frac{9}{16} \ \int_0^8 \   \frac{1}{12}    \ \xi^4 \ \ d\xi $$
$$ = \ \frac{9}{16} \ \cdot \ \frac{1}{12}    \  (  \ \frac{1}{5}  \xi^5 \ ) \vert_0^8     $$
$$ = \ \frac{9}{16} \ \cdot \ \frac{1}{12}    \  \cdot \ \frac{1}{5} \cdot \ 8^5 \  = \ \frac{3 \cdot 3 \cdot 2^{15}}{2^6 \cdot 3 \cdot 5} \ = \ \frac{3 \cdot  2^9}{5} \ = \ \frac{1536}{5} \ \ , $$
confirming Jack's result.
