Help evaluating triple integral over tetrahedron I have a triple integral of $\iiint xyz\,dx\,dy\,dz$  over the volume of a tetrahedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. 
Normally I would just have limits 0 to 1 but that does not seem to work. How do I solve a problem like this?

To get an idea what to do in 3D, try to understand the 2D case first and try to generalize. When integrating over the triangle with vertices $(0,0)$, $(0,1)$ and $(1,0)$, it is often a good idea to first let $x$ go from zero to $1-y$ and then let $y$ go from zero to 1.
In your case, you can proceed analogously: let $x$ range in $[0,1-y-z]$, then $y$ in $[0,1-z]$ and finally $z$ in $[0,1]$.
That is,
$$
\int_{\text{tetrahedron}}f(x,y,z)dxdydz
=
\int_0^1\int_0^{1-z}\int_0^{1-y-z}f(x,y,z)dxdydz.
$$
In your case $f(x,y,z)=xyz$.
To make the iterated integral structure clear, you can write it as
$$
\int_0^1\left(\int_0^{1-z}\left(\int_0^{1-y-z}xyzdx\right)dy\right)dz
$$
and start by doing the innermost integral.
The first integral is $\int_0^{1-y-z}xyzdx=\frac12yz(1-y-z)^2$.
If you get this right, you are on the right track.
