Average area of the shadow of a convex shape

Question

What is the average area of the shadow of a convex shape taken over all possible orientations?

If we take a sphere, its surface area is exactly 4 times the area of its shadow. How can it be generalised for any convex shape? I know there are a lot of books like "Introduction to Geometric Probability", but I would appreciate an almost-intuitive explanation.

Answer 1

The average size of the shadow cast by any 3D convex shape is $\frac{1}{4}$ times its surface area.

1. Because the shape is convex, you can compute the size of its shadow along a particular direction $\vec{u}$ by summing the shadows cast by its surface pieces:

   • Divide the surface of the shape into little area patches.
   • Compute the size of each patch's shadow by measuring the size of the dot product with $\vec{u}$.
   • Add up the sizes of each of the shadow-patches.
   • Note that this is actually twice the size of the shape's shadow, because we've double-counted: every patch has a counterpart on the "other side" of the shape, and both patches cast the same, overlapping shadow. So we must divide our total by two.

2. Because the shadow of a convex shape can be computed by summing the shadows of its individual surface patches, the average shadow can be computed in the same way:

   • Divide the surface of the shape into little area patches.
   • Compute the average shadow cast by the patch over all directions.
   • Add up the average shadows of each of the patches.
   • Divide the total by two.

3. There is a constant of proportionality $\lambda$ which connects the surface area of any convex object to the average size of its shadow.

   • The size of the shadow cast by a patch, averaged over all directions, should be proportional to the area of the patch. (After all, both the patch and the shadow have units of area, and magnifying the patch by some amount should magnify the shadow by the same amount.)

   • It follows that there is a constant of proportionality $\lambda$ such that if the area of the patch is $A$, the average area of its shadow is $\lambda \cdot A$.

   • When we compute the average shadow of the shape by adding up the individual contributions from each patch, this constant $\lambda$ will factor out of the sum. The remaining sum is just the surface area of the shape.

   • Hence the average shadow cast by a convex shape should be $\lambda$ times its surface area (divided by two).

4. Because the constant of proportionality $\lambda$ is the same for all convex shapes, we can use a known shape to solve for its value. A sphere of radius 1 has surface area $4\pi$, and casts a shadow of area $\pi$ in each direction. Hence its average shadow area over all directions is $\pi$, and the constant of proportionality must be $$\lambda \equiv \frac{\pi}{4\pi} = \frac{1}{4}.$$

So the average size of the shadow cast by any 3D convex shape is $\frac{1}{4}$ times its surface area.

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5. Bonus: By extension to $n$ dimensions, there will be a constant of proportionality $\lambda_n$ relating the outer surface of a convex volume to the average size of its shadow. We can use $n$-dimensional spheres, whose geometries are known, to compute those constants.

   The constant $\lambda_n$ is the volume of an $n-1$ ball (a disc, in our example), divided by the surface area of an $n-1$ sphere (a sphere, in our example). So, by standard formulas,

   $$V_n = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}$$ $$A_n = \frac{2\cdot \pi^{\frac{n+1}{2}}}{\Gamma(\frac{n+1}{2})}$$

   $$\lambda_{n+1} = \frac{V_{n}}{A_n} = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}\cdot \frac{\Gamma(\frac{n}{2}+2)}{2\cdot \pi^{\frac{n+1}{2}}} = \frac{1}{2\sqrt{\pi}}\frac{\Gamma(\frac{n}{2}+2)}{\Gamma(\frac{n}{2}+1)} $$

   Note that the $\Gamma()$ function simplifies if we consider odd and even cases separately. After manipulation, we find:

$$\lambda_{n+1} = \begin{cases}\frac{1}{2^{n+1}} {n \choose n/2} & n\text{ even}\\ \frac{1}{\pi}\frac{2^n}{n+1} {n \choose {\lfloor n/2\rfloor} }^{-1} & n\text{ odd} \end{cases}$$ And so $$\lambda_n = \frac{1}{2},\; \frac{1}{\pi},\; \frac{1}{4},\; \frac{2}{3\pi},\; \frac{3}{16},\; \frac{8}{15\pi},\; \ldots .$$