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 the limits for x and y, you can use the triangle in the xy-plane with vertices (0,0), (1,0), (0,1), since this is the projection of the tetrahedron in the xy-plane. To get the limits for z, you need to find the equation of the plane passing through (1,0,0), (0,1,0), (0,0,1), since this gives the top surface of the tetrahedron. Sep 7, 2014 at 20:17
• How do I find the equation? Is it 1-x-y Sep 7, 2014 at 20:20
• If a plane has intercepts a, b, and c, (in the order x,y,z), then it has equation $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, so in this case you do get $z=1-x-y$. Sep 7, 2014 at 20:25

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.

• If its taking a while, am I doing it right? Or should it be simple? Sep 7, 2014 at 20:38
• @JacksonHart, it takes a while to integrate, because there are three integrals to do. Each of them is quite simple (polynomial), but the whole thing takes a while. No need to be alarmed. Sep 7, 2014 at 20:52
• It seems like it gets really complicated - (1-y-z)(1-y-z)yz and then continuing seems like it will be monotonous Sep 7, 2014 at 20:54
• @JacksonHart, the innermost integral is $\int_0^{1-y-z}xyzdx=\frac12yz(1-y-z)^2$. Do you agree on this one? You can then expand the polynomial and calculate the $y$-integral. If you started this way, I think you are doing fine. Sep 7, 2014 at 21:02

I agree entirely with the accepted answers; that is how the integral should be attempted analytically. However, if you wish to compute integrals of this form over the unit tetrahedron efficiently, or even just check your answers, it is worth noting that there is an explicit formula for the result as a function of the powers.

$$\int_0^1 \int_0^{1-z} \int_0^{1-y-z} x^a y^b z^c \, dx \, dy \, dz = \frac{1}{3!} \frac{a! \, b! \, c!}{(a + b + c)!} \frac{1}{3 + a + b + c \choose 3}$$

See This mathoverflow question for proof and an extension to other dimensions (triangles, 4-D tetrahedra, etc...).

Using the facts that the projection of the solid in the xy-plane is the triangle with vertices (0,0), (0,1), and (1,0), and that the top of the solid is the plane $x+y+z=1$, we can set up the integral as

$\displaystyle\int_{T} f(x,y,z) \;dV =\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} f(x,y,z) \;dz dy dx$.

• If you're still around and getting notifications for comments to this, how did you figure out the $x+y+z=1$ part? Nov 9, 2016 at 7:16
• @windy401, plane determinated by the points... Nov 18, 2016 at 15:15