To establish a rigorous mathematical foundation without physical overreach, we can derive the exact second-volume moment of a regular octahedron (the 3-dimensional cross-polytope) normalized by its circumradius.

1. Definition of the Octahedral Envelope

Let $R$ denote the circumradius of a regular octahedron—the Euclidean distance from the geometric center $(0,0,0)$ to any of its 6 vertices located at $(\pm R, 0, 0)$, $(0, \pm R, 0)$, and $(0, 0, \pm R)$.

In Cartesian coordinates, the solid regular octahedron $V$ is bounded by the inequality:

$$\vert{}x\vert{} + \vert{}y\vert{} + \vert{}z\vert{} \le R$$

Due to full octahedral symmetry ($O_h$), the integral over the entire domain can be evaluated by restricting our domain to the first octant where $x, y, z \ge 0$ ($x + y + z \le R$) and multiplying the result by $8$.

2. Step-by-Step Integration

Step A: Total Volume Verification

The volume integral for a single octant is:

$$V_{\text{octant}} = \int_0^R dz \int_0^{R-z} dy \int_0^{R-z-y} dx$$

1. Inner integral ($x$):$$\int_0^{R-z-y} dx = R – z – y$$
2. Middle integral ($y$):$$\int_0^{R-z} (R – z – y) \, dy = \left[ (R-z)y – \frac{y^2}{2} \right]_0^{R-z} = \frac{(R-z)^2}{2}$$
3. Outer integral ($z$):$$\int_0^R \frac{(R-z)^2}{2} \, dz = \frac{1}{2} \left[ -\frac{(R-z)^3}{3} \right]_0^R = \frac{R^3}{6}$$

Multiplying by the 8 octants yields the total volume of the octahedron:

$$V = 8 \times \frac{R^3}{6} = \frac{4}{3}R^3$$

Step B: The Second-Volume Moment

The second-volume moment is defined as the integral of the squared radial distance $r^2 = x^2 + y^2 + z^2$ over the volume $V$:

$$\int_V r^2 \, dV = \int_V (x^2 + y^2 + z^2) \, dV$$

By coordinate symmetry:

$$\int_V x^2 \, dV = \int_V y^2 \, dV = \int_V z^2 \, dV$$

Therefore:

$$\int_V r^2 \, dV = 3 \int_V z^2 \, dV$$

Evaluating $\int_{\text{octant}} z^2 \, dV$ for the first octant:

$$\int_0^R z^2 \, dz \int_0^{R-z} dy \int_0^{R-z-y} dx = \int_0^R z^2 \left( \frac{(R-z)^2}{2} \right) dz$$

Expanding the integrand:

$$\frac{1}{2} \int_0^R (R^2 z^2 – 2R z^3 + z^4) \, dz$$

Integrating term-by-term:

$$\frac{1}{2} \left[ R^2 \frac{z^3}{3} – 2R \frac{z^4}{4} + \frac{z^5}{5} \right]_0^R = \frac{1}{2} R^5 \left( \frac{1}{3} – \frac{1}{2} + \frac{1}{5} \right)$$

Combining fractions ($\frac{1}{3} – \frac{1}{2} + \frac{1}{5} = \frac{10 – 15 + 6}{30} = \frac{1}{30}$):

$$\int_{\text{octant}} z^2 \, dV = \frac{1}{2} R^5 \times \frac{1}{30} = \frac{1}{60} R^5$$

Scaling across all 8 octants and multiplying by 3 (for $x^2, y^2, z^2$):

$$\int_V r^2 \, dV = 3 \times 8 \times \frac{1}{60} R^5 = \frac{24}{60} R^5 = \frac{2}{5} R^5$$

3. The Dimensionless Ratio

Dividing the second-volume moment by $R^5$ eliminates all dimensional units, yielding the exact pure number:

$$\frac{1}{R^5} \int_V r^2 \, dV = \frac{2}{5}$$

4. Summary of Pure Geometric Properties

• Vertices: $6$
• Edges: $12$ (with edge length $a = R\sqrt{2}$)
• Faces: $8$ equilateral triangles
• Symmetry Group: $O_h$ (full octahedral symmetry, order 48)
• Normalized Second-Volume Moment: $\frac{2}{5}$