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I would like to compute the area of a high-dimensional hyper-ellipsoid that is projected (x/norm(x,2)) onto the unit hyper-sphere of the same dimensions. Any pointers would be appreciated.

Even a pointer to an approximate solution would be great. I know how to do this for 2d but am stuck with higher dimensions.

user484101
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  • Which kind of projection : Orthogonal, oblique, central ? Is the sphere $S_2$ in $R^3$ ? – Jean Marie Sep 24 '17 at 22:02
  • I edited the question to be a bit more precise. – user484101 Sep 25 '17 at 03:16
  • Thus a central projection. You should treat the case $\mathbb{R^3}$ first – Jean Marie Sep 25 '17 at 12:32
  • Take a look at (https://math.stackexchange.com/q/1187285) – Jean Marie Sep 25 '17 at 16:21
  • Thanks for the pointer, the solution in that link uses a polar parametrization and works for 3 dimensions. My problem is in 100s of dimensions though. Perhaps I can think along those lines and figure out a solution. – user484101 Sep 25 '17 at 21:26
  • I have had a different idea that works in 3D: it is to consider the cone with vertex in 0 and tangent to the ellipsoid. I know there is an easy manner to get it (but just at the moment I cannot retrieve it...). Substituting the cone to the ellipsoid, you have the contour of the area (which is easier to manipulate through a certain parameterization). But in nD, something puzzles me: what to you consider as a "surface" in the ball in nD: you will have to deal with hypervolumes... – Jean Marie Sep 25 '17 at 21:41
  • Are you looking for something that could be called a "generalized solid angle" under which a scene (here an ellipsoid) is seen from the origin ? Have you already met this concept of "solid angle" ? – Jean Marie Sep 27 '17 at 22:14
  • I have heard of the concept of solid angle, but need to study it. Is there a particular source (book, paper etc.) for reading up on this apart from Googling? – user484101 Sep 28 '17 at 12:32
  • Take a look for example at (http://drdrbill.com/downloads/optics/photometry/Solid_Angle.pdf) – Jean Marie Sep 28 '17 at 20:18
  • Thanks for the pointer, you have been very helpful. – user484101 Oct 02 '17 at 21:01

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