Expected value of cube projection

Question

This problem is from the book of V. I. Arnold.

Find the expected area of the projection of a unit cube onto the plane under isotropic random direction of projection.

The direct evaluation of it seems to me very hard and long. Below is Arnold's solution. Could you provide another?

Arnold's solution

The expected area of the projection doesn't depend on the shape of the projected figure (if it's convex), but depends only on its surface area. [This is more or less clear for me.] Therefore, the ratio between the average area of the projection of a cube and its genuine area is the same as the ratio between the area of the equatorial section of a sphere and the genuine area of the sphere. For a sphere of radius $1$ the equatorial area section has area $\pi$. The surface of a sphere of radius $1$ has area $4\pi$. So, the average area of the projection of the cube is a quarter of its genuine surface area, which is $6$. Therefore, the average projected area of the unit cube is $3/2$.

Answer 1

Having acceded to your request for a different solution, I now feel at liberty to spell out the elegant solution. If we look at a convex body from a uniformly random direction, by linearity of expectation the expected integral over the projected area is the integral over the expected projected area; that is, we can consider the expectation value of the projected area for each surface element individually and then integrate. The expected projected area of a surface element $\mathrm dS$ is obtained by averaging $\max(0,\cos\theta)$ over all directions:

$$ \mathrm dS\cdot\frac12\int_{-1}^1\mathrm d\cos\theta\max(0,\cos\theta)=\mathrm dS\cdot\frac12\int_0^1\mathrm d\cos\theta\cos\theta=\frac{\mathrm dS}4\;. $$

Thus, the expected projected area is simply one quarter of the surface area. Alternatively, we could have argued by linearity that the result would be some constant factor times $\mathrm dS$ and derived the factor $1/4$ from the sphere like Arnold does.

Answer 2

This is what you get when you present the elegant proof and ask for other proofs :-)

Consider a random rotation characterized by a unit axis $\vec u$ and a rotation angle $\theta$. According to Wikipedia, the density for $\theta$ to get random rotations uniform over $SO(3)$ is $(1-\cos\theta)/\pi$.

Now consider a cube face spanned by orthonormal edge vectors $\vec e_1$ and $\vec e_2$. Applying the rotation to $\vec e_i$ yields

$$ \cos\theta\, \vec e_i+\sin\theta\, \vec u\times\vec e_i+(1-\cos\theta)\vec u(\vec u\cdot\vec e_i)\;. $$

Forming the dot products with the $\vec e_i$ gives the coordinates of these rotated vectors in the plane spanned by $\vec e_1$ and $\vec e_2$:

$$ \pmatrix{\cos\theta+(1-\cos\theta )u_1^2\\\sin\theta u_3+(1-\cos\theta)u_1u_2}\;,\; \pmatrix{-\sin\theta u_3+(1-\cos\theta)u_1u_2\\\cos\theta+(1-\cos\theta )u_2^2}\;, $$

where the $u_i$ are the coordinates of $\vec u$ in the coordinate system formed by $\vec e_1$, $\vec e_2$ and $\vec e_1\times\vec e_2$.

The projection of the rotated cube face into the plane spanned by $\vec e_1$ and $\vec e_2$ is a parallelogram spanned by these two vectors, whose signed area is given by

$$ \cos^2\theta+\cos\theta(1-\cos\theta)(u_1^2+u_2^2)+(1-\cos\theta)^2u_1^2u_2^2+\sin^2 u_3^2-(1-\cos\theta)^2u_1^2u_2^2\\ =\cos\theta+u_3^2(1-\cos\theta)\;, $$

where I used $u_1^2+u_2^2+u_3^2=1$.

The sign of the area indicates whether it faces towards or away from the projection direction perpendicular to the original face. Integrating the signed quantity would yield $0$ by symmetry, so we can choose which signed component to integrate. The sign is negative in a triangular-shaped region and positive in the more complicated rest of the rectangle $0\le u_3\le1$, $0\le\theta\le\pi$, so let's integrate the negative component. Equating the area to $0$ yields $u_3^2=\cos\theta/(\cos\theta-1)$, which has a solution for $\theta\ge\pi/2$, so we need

$$ -\frac1\pi\int_{\pi/2}^\pi\mathrm d\theta\;(1-\cos\theta)\int_0^{\sqrt{\cos\theta/(\cos\theta-1)}}\mathrm du_3\left(\cos\theta+u_3^2(1-\cos\theta)\right)\;, $$

where the density for the $u_3$ integration is $1$ due to the fact that the surface of a horizontal slice of a sphere is proportional to the height of the slice. Performing the inner integration yields

$$ -\frac1\pi\int_{\pi/2}^\pi\mathrm d\theta\;(1-\cos\theta)\sqrt{\frac{\cos\theta}{\cos\theta-1}}\left(\cos\theta+\frac13\frac{\cos\theta}{\cos\theta-1}(1-\cos\theta)\right)\\ =\frac2{3\pi}\int_{\pi/2}^\pi\mathrm d\theta\sqrt{\cos^3\theta(\cos\theta-1)}\;. $$

The remaining integration is quite a mess, but Wolfram|Alpha evaluates the integral to $3\pi/8$, so the final result for one face is $1/4$. Since the cube has $6$ faces, the expected result $3/2$ follows.