The average surface area of a projection of a randomly rotated planar rectangular shape on a two-dimensional surface 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?  
 A: 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 ${1\over2}MN$.
A: 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.
A: 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. 
