Find the mathematical expectation of the area of the projection of a cube with edge of length onto a plane with an isotropically distributed random direction of projection.
(Proposed by Bhargav Gnv/FB/Math)
Let the cube be positioned in a 3d coordinate system with centre at the origin and edges parallel to axes,and with edge length 1. Let the direction of projection be given by a unit vector n = (cos α, cos β, cos γ) where α, β, γ are the angles of the vector to the coordinate axes. Also, for simplicity we confine ourselves to the first octant. Because of symmetry that doesn’t affect the result.
Now, from any direction exactly three non-parallel faces of the cube can be seen, including the ”degenerate” cases when some face is seen from its own plane. Therefore the area of the projection is cos α + cos β + cos γ. Isotropically distributed random direction of projection means that every position of the tip of vector n on the unit sphere around the origin is equally probable. Since the area element of the sphere is sin θ dθ dφ in polar coordinates the probability density f(θ) for the tip’s position is proportional to sin θ (no φ-dependence by azimuthal symmetry). By normalization f(θ) = (2/π)*sin θ. Thus the expectation value
<A> = ∫∫ (cos α + cos β + cos γ)(2/(π))*sin θ dθ dφ
where Σ = [0,π/2]x[0,π/2].
To calculate this we must express the direction cosines cos α, cos β, cos γ in polar angles θ, φ. Let’s say that α, β, γ are the angles with the x-, y-, and x-axis respectively. Then the tip of the vector n is at (sin θ cos φ, sin θ sin φ, cos θ). Using the scalar product of n with the unit vetors e_x, e_y, e_z, these components are equal to the direction cosines. Thus,
<A> = ∫∫ (sin θ cos φ + sin θ sin φ + cos θ)(2/π)*sin θ dθ dφ =
= (2/π)*∫∫ sin²θ (cos φ + sin φ) + cos θ sin θ dθ dφ =
= (2/π) * ∫ [θ=0,π/2] sin²θ dθ * ∫ [θ=0,π/2] cos φ + sin φ dφ
+ (2/π) * (π/2) * ∫ [θ=0,π/2] cos θ sin θ dθ =
= (2/π)*(π/4)*2 + 1*1/2*2 = 1 + 1 = 2.
With edge a we get of course <A> = 2a². Seems resonable, since three faces are seen contracted and each face has area a², so <A> must be something between a² and 3a².
Ooppss! ∫ [θ=0,π/2] cos θ sin θ dθ = 1/4*2 = 1/2 so <A> = 3a²/2.
Also, the estimate can be improved. The greatest projected area occurs in the direction (1/√3, 1/√3, 1/√3) and that area is 3a²/√3 = a²√3. So we must have a² < <A> < a²√3, which satisfied by <A> = 3a²/2. – In general the expaction value of a non-constant random variable must be strictly between the smallest and the greatest value.