Volume integrals over rhombic dodecahedron

Question

I need to know the value of the integrals $$ \dfrac{\int_{\Omega} f(\mathbf{x})\,d^3\!\mathbf{x}} {\int_{\Omega} 1\,d^3\!\mathbf{x}}, $$ where $\Omega$ is a rhombic dodecahedron of unit radius and $f(\mathbf{x}) = x^2,\,y^2,\,z^2,\,x^2y^2,\,x^2z^2,\,y^2z^2,\,x^4,\,y^4, $ and $z^4$. Is there a good way to calculate these?

Answer 1

Notice first of all that, by symmetry, the integrals of $x^2$, $y^2$ and $z^2$ are the same, so you need only compute one of them. The same goes for the other functions, so you need only compute, for instance, the cases $f(\mathbf{x})=z^2$, $f(\mathbf{x})=x^2y^2$ and $f(\mathbf{x})=z^4$. The case $f(\mathbf{x})=1$ leads to the volume of the solid, which is ${16\sqrt3\over9}a^3$, where $a$ is the length of an edge.

A convenient disposition of a rhombic dodecahedron can be set up as follows:

a) four rhombi with common vertex $(0,0,2)$, middle vertices $(\pm1,\pm1,1)$ and lower vertices $(\pm2,0,0)$, $(0,\pm2,0)$;

b) four rhombi with upper vertices $(\pm1,\pm1,1)$, middle vertices $(\pm2,0,0)$, $(0,\pm2,0)$ and lower vertices $(\pm1,\pm1,-1)$;

c) four rhombi with upper vertices $(\pm2,0,0)$, $(0,\pm2,0)$, middle vertices $(\pm1,\pm1,-1)$ and common vertex $(0,0,-2)$.

Notice that this has an edge length of $\sqrt3$, so the volume is $16$. The rhombus can be then divided into $12$ pyramids, each having a face as base and $(0,0,0)$ as vertex.

By symmetry, the contribution of a pyramid in groups a) and c) to any of the three integrals $\int_{\Omega}z^2\,d^3\!\mathbf{x}$, $\int_{\Omega}x^2y^2\,d^3\!\mathbf{x}$ and $\int_{\Omega}z^4\,d^3\!\mathbf{x}$ is the same, so you just need to compute one of them and multiply the result by $8$. The same goes for the pyramids in group b): just compute one integral and multiply by $4$.

EDIT.

For instance, in the case of the face with vertices $(0,0,2)$, $(1,1,1)$, $(0,2,0)$ and $(-1,1,1)$ (group a) above) the integral over the corresponding pyramid should be: $$ 2\int_0^1 dx\int_x^{2-x}dy\int_x^{2-y}\!dz\ f(\mathbf{x}), $$ while for the face with vertices $(2,0,0)$, $(1,1,1)$, $(0,2,0)$ and $(1,1,-1)$ (group b) above) the integral is $$ 4\int_0^1 dy\int_y^{2-y}dx\int_0^{y}\!dz\ f(\mathbf{x}). $$ You can check that for $f(\mathbf{x})=1$ both integrals evaluate to $4/3$, as it should be.

Of course you must in the end scale your results by the appropriate power of $\sqrt3$, to get the correct values for the edge$\ =1$ case.