Average projected area of an ellipsoid Consider an ellipsoid of semi-axes a, b, c (possibly prolate, b=c). I am interested in the "shadow" of this solid onto a distant plane, in a given direction d=(k,l,m) orthogonal to that plane. By shadow I mean the projected area onto the plane: each point on the surface of the ellipsoid is translated in the same direction until it intersects with a plane normal to it; the shadow is defined by the envelope of the intersection points. First question: Is the projected curve an ellipse? (and what is its equation in terms of a,b,c and the direction vector d)? Second question: What is the mean area of this shadow when averaged over all orientations of the ellipsoid (or, equivalently, the plane of projection) I'm guessing this problem has been solved in the past; any references would be very welcome. Thanks, baptiste

When a surface element $d\omega$ at a point $P$ in three-space is projected orthogonally into all possible directions then the average area of the image is $\rho\thinspace d\omega$ where $\rho$ is a universal constant. To determine the value of $\rho$ we argue as follows: The unit sphere $S^2$ has a surface area of $4\pi$, and the area of its projection is $2\pi$ in all directions (note that the "shadow" is doubly covered); whence also the average area of the projection is $2\pi$. So $\rho$ must be ${1\over2}$. What we have found out is valid for any surface element $d\omega$ of your ellipsoid $E$. It follows that the average area of the shadow of $E$ is equal to $\omega(E)/4$, where $\omega(E)$ is the total surface area of $E$ and we have taken into account that the shadow is covered twice. The value so obtained can be expressed as an elliptic integral in terms of $a$, $b$, $c$.