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I have a planar rectangular shape, of dimensions $N$ by $M$, positioned in 3-space above a two-dimensional surface. Provided a large number of random 3-space rotational orientations of the shape, what is the average surface area $A$ of a projection of the shape on the two-dimensional surface?

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It is equivalent to keep the shape fixed and randomly rotate the plane. The orientation of the plane is completely determined by its normal. So you want to pick the normal of the plane, say $\hat n$, uniformly over the unit sphere. Then the area of the projection is equal to the area of the original shape times $\left|\hat n\cdot\hat z\right|$, where $\hat z$ is the fixed normal of the shape. Using spherical coordinates with $\hat z$ as the zenith, the expected value of $\left|\hat n\cdot\hat z\right|$ as $\hat n$ varies over the unit sphere is $$\mathrm E[\hat n\cdot\hat z] = \frac{\int_{-\pi}^\pi\int_0^\pi\left|\cos\theta\right|\cdot\sin\theta\,\mathrm d\theta\,\mathrm d\phi}{\int_{-\pi}^\pi\int_0^\pi1\cdot\sin\theta\,\mathrm d\theta\,\mathrm d\phi} = \frac{2\pi\int_0^\pi\left|\cos\theta\right|\cdot\sin\theta\,\mathrm d\theta}{4\pi} = \frac12.$$ So the expected area of the projection is half the area of the shape itself.

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What happened to the rotation around $\hat{n}$? This rotation can also change the projection area. –  nbubis Aug 18 '12 at 8:44
@nbubis, rotating the plane about its own normal does not change the plane, so I'm not sure what you mean. –  Rahul Aug 18 '12 at 18:17
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Any "surface element" ${\rm d}\omega$, whether it belongs to a rectangle or to a more general surface, has an average projected area of ${1\over2}{\rm d}\omega$. To arrive at the constant ${1\over2}$ note that the full area of a sphere is $4\pi$, whereas the average projected area is $2\pi$.

Therefore the average projected area of your rectangle is ${1\over2}MN$.

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is there a more rigorous proof for that? –  nbubis Aug 18 '12 at 9:48
@nbubis: See Rahul Narain's answer. Of course you could start with a parametric representation $(u,v)\mapsto{\bf x}(u,v)\in{\mathbb R}^3$ and "do the works". It all boils down to averaging $\cos\theta$ over $S^2$. –  Christian Blatter Aug 18 '12 at 10:25
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You cannot answer this question without first specifying what probability function describes the orientation of the rectangle. You can take a look at a similar problem.

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Come on, there is a very natural uniform distribution over the space of orientations. –  Rahul Aug 18 '12 at 6:58
Not really - the rotation is defined by three Euler angles which are far from uniform in angular area. What would you define as the natural space of orientations? –  nbubis Aug 18 '12 at 7:05
The uniform measure on the unit quaternions. Or the Haar measure on $SO(3)$. (Are the two equivalent? I hope so.) –  Rahul Aug 18 '12 at 7:09
I've asked a new question about this to make sure. –  Rahul Aug 18 '12 at 19:20
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