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Problem: Find the mathematical expectation of the area of the projection of a cube with edge of length 1 onto a plane with an isotropically distributed random direction of projection.

Source: Arnold's Trivium.

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up vote 3 down vote accepted

The area of the projection is the area that you see when looking at the cube. You see three faces, and the area under which a face appears is its actual area, $1$, times the absolute value of one of the coordinates of the uniformly random unit direction vector. Those coordinates are uniformly distributed over $[-1,1]$ (see Generate a random direction within a cone), so the mean absolute value is $1/2$, so by linearity of expectation the expected projected area is $3/2$.

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@Potato: The area under which a face appears is its actual area times the cosine of the angle between its normal and the viewing direction. We can consider the cube fixed and axis-parallel and the viewing direction uniformly random; then the normals are the canonical basis vectors and the cosines are the coordinates of the viewing direction. – joriki Oct 12 '12 at 20:35
If I'm following you, this would work only in three dimensions, not in two or $\ge$ four. – Michael Hardy Oct 12 '12 at 21:43
@Michael: You're following me :-) In two dimensions, the answer is twice the average value $2/\pi$ of the absolute value of the cosine, $4/\pi\approx1.27$. – joriki Oct 12 '12 at 21:44

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