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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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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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    I've asked a new [question](http://math.stackexchange.com/q/184086/856) about this to make sure.2012-08-18
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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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    @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 $S^2$.2012-08-18
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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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    @nbubis, rotating the plane about its own normal does not change the plane, so I'm not sure what you mean.2012-08-18