An area of a convex polyhedral surface homeomorphic to a sphere vs an area of its shadow

Question

Consider a convex polyhedral surface $P$ in $\mathbb{R}^3$. If $\pi_v : \mathbb{R}^3\rightarrow \{ x\in \mathbb{R}^3| x\perp v\}$ where $v\in \mathbb{S}^2$ is an orthogonal projection, then $$ \frac{1}{{\rm area}\ \mathbb{S}^2 }\int_{v\in \mathbb{S}^2} \ {\rm area}\ \pi_v(P)\ d{\rm Vol}_v = \frac{1}{4}\cdot {\rm area}\ (P) $$

This can be proved by using mutivariable calculus. But by using the fact that unit ball in $\mathbb{R}^2$ has an area, which is $\frac{1}{4}$ of area of $\mathbb{S}^2$, how can prove this ?

Answer 1

This is true not just for polyhedral surfaces, but for any convex body in $\mathbb{R}^3$.

To reformulate the problem in a much more intuitive sense, this is saying that if we project a convex set onto a plane oriented in a random direction, the expected area of the "shadow" cast onto the plane by a perpendicular beam of light is $1/4$ of the surface area of $S$. (Equivalently, we can think of holding $S$ in a random orientation and projecting light onto a fixed horizontal plane.)

To prove this, we use linearity of expectation. Consider some very small part of the surface of $S$. We say that its contribution to the shadow is $0$ if the beam of light does not touch it, and is equal to the area of its projection if the beam of light does reach it. (This way, we don't overcount the shadow area from both the top and bottom parts.) Then obviously the expected area of the shadow is just the integral of the expected contributions from infinitesimal regions of the surface of $S$.

But because $S$ is convex, it's impossible for a portion of the surface to be in the shadow of another part of the surface - it has no "overhangs". So the contribution of a small patch of area $\epsilon$ depends only on the direction of the beam, and not on the rest of $S$ - this means that the expected contribution is just linear in $\epsilon$, because all small patches of $S$ act the same.

So the expected shadow area of a convex set $S$ is $C\cdot\text{Area}(S)$, for some universal constant $C$ independent of $S$. Taking $S$ to be the sphere, it is obvious that the shadow area is always $\pi$ and the surface area is $4\pi$, so we must have $C=1/4$.

Answer 2

Assume that $F$ is a face in convex polyhedral surface $P$ homeomorphic to a sphere. If $v\in \mathbb{S}^2$, and $\pi_v$ is an orthogonal projection of direction $v$, then

\begin{align*} & {\rm Average\ area\ of\ the\ shadow\ of}\ P \\&= \frac{1}{2} \sum_{F\subset P} \frac{1}{{\rm area} \ \mathbb{S}^2}\int_{v\in \mathbb{S}^2}\ {\rm area}\ \pi_v (F)\ dv \\&= \frac{1}{2} \sum_{F\subset P} \frac{1}{{\rm area} \ \mathbb{S}^2} \int_{v\in \mathbb{S}^2}\ {\rm area}\ (F) |v\cdot (0,0,1)|\ dv \\&= \frac{1}{2} {\rm area}\ (P) \frac{1}{{\rm area}\ \mathbb{S}^2 }\int_v\ |v\cdot (0,0,1) |\ dv \\&= \frac{1}{4} {\rm Area} (P) \end{align*}