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?
Best Answer
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.