A convex compact body $K$ in 3-space has well-defined volume, surface area, and mean width. Do these quantities enable one to say anything about the "mean cross-sectional area"?

I put the phrase in quotes because there are at least two natural ways to define the term. We could average with respect to the measure on the affine Grassmannian (which is noncompact but that's okay since the set of planes intersecting $K$ is compact in Graff) or with respect to the linear Grassmannian (where within each family of parallel planes we average the cross-sectional area over all those planes that intersect $K$). Note that the former is a weighted version of the latter in which the weight is the width of $K$ in the direction perpendicular to the family of hyperplanes.

Is either of these two notions of mean cross-sectional area expressible in terms of volume, surface area, and mean width?


1) The mean cross section by the affine Grassmannian is just $$\frac{ \iint(\text{average area orthogonal to }v) (\text{weight in direction } v)dv} {\iint(\text{weight in direction } v)dv}$$ $$=\frac{ \iint\frac{\text{volume}}{\text{width in direction }v} (\text{width in direction } v)dv} {\iint(\text{width in direction } v)dv}$$ $$= \frac{\text{volume}}{\text{mean width}}$$

2) The mean cross section by the linear Grassmannian is independent of the volume, surface area, and mean width.

Consider the convex hull of $$S=\{(0,a,0),(b,0,0),(0,-c,0),(-d,0,0),(0,0,1)\}$$ for \begin{align} (a,b,c,d)&=(0.50000,1.00000,3.00000,2.50000)\\ (a',b',c',d')&=(2.31144,0.49261,2.00000,2.34887) \end{align}

These are pyramids whose bases are as shown: enter image description here

We can calculate the mean width by the boxed formula here, or directly.

We can calculate the linear-mean cross section as the average of $$\frac{\text{volume}}{\max(v\cdot S)-\min(v\cdot S)}$$ over all unit vectors $v$.

This gives \begin{array}{ccc} & (a,b,c,d) & (a',b',c',d')\\ \text{volume} & 2.042 & 2.042 \\ \text{surface area} & 14.564 & 14.564 \\ \text{mean width} & 2.766 & 2.766 \\ \text{linear-mean cross section} & 0.803 & 0.812\\ \end{array} so the linear-mean cross-sectional area is indeed independent.

  • $\begingroup$ @JosephO'Rourke, illustrating in 3-d is always hard for me -- I decided to replace the example entirely with something easier to visualize. $\endgroup$ – Matt F. Jul 3 at 17:38
  • $\begingroup$ 'Tis OK, Matt. Nice examples. $\endgroup$ – Joseph O'Rourke Jul 3 at 17:39

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