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    $\begingroup$ I'm not sure what you're asking. But if you want the area of the xy-projection, you could find the area of the convex hull of the projected vertices. Or you could project the mesh itself and take 0.5 it's area since a convex polyhedron has an 'up' side and a 'down' side. SeedRandom[1]; reg = RandomPolyhedron[{"ConvexHull", 20}]; convex hull area: Area[ConvexHullMesh[reg[[1, All, 1 ;; 2]]]] (* 0.477824 *) projection area: 0.5Area[MeshRegion[reg[[1, All, 1 ;; 2]], Polygon[reg[[2]]]]] (* 0.477824 *) $\endgroup$ Commented Oct 12, 2023 at 0:38
  • $\begingroup$ @GregHurst You gave a great comment! While, yes your solution finds the area after a particular rotation there is a drawback. It is numeric, so it doesn't give a general formula for finding the exact area after a known roll-pitch-yaw. It isn't computationally optimal and it doesn't help to find interesting mathematical relationships. $\endgroup$ Commented Oct 12, 2023 at 15:24
  • $\begingroup$ @GregHurst Here is the partial cuboid formula. It is not general, because it only works when $\gamma = 0$. It would be great if all of the angles can be between $[-2\pi, 2\pi]$ which would mean the general solution is a piecewise function. And it would be great to be able to do it for any convex polyhedron. I feel like there would be a way for Mathematica to solve this fairly easily instead of focusing so much on the Mathematics behind it. $\endgroup$ Commented Oct 12, 2023 at 15:27