![]() I'm not exactly sure how one would do this rigorously without essentially re-inventing calculus, but perhaps the somewhat intuitive argument here is sufficient for your needs. In dimensions eight and 24 (the latter dimension in collaboration with other researchers), she has proved that two highly symmetrical arrangements pack spheres. By analogy with determining the area of a circle as the limit of a sequence of triangles of ever more finely divided base, the hypervolume of the hypersphere should be one fourth the common altitude of the hyperpyramids, $R$, times the combined base volume, $2\pi^2R^3$, or In a pair of papers posted online this month, a Ukrainian mathematician has solved two high-dimensional versions of the centuries-old sphere packing problem. One can then apply this formula to the hypersphere, whose (hypersurface) volume you have determined to be $2\pi^2R^3$. By appropriate scaling and skewing, one can demonstrate the same formula for any hyperpyramid. This demonstrates that a hyperpyramid has a volume of one-fourth the base volume times the altitude. This is by analogy of dividing a square into two congruent right triangles from any vertex to the two sides that don't contain that vertex, or of dividing a cube into three congruent pyramids from any vertex to the three faces that don't contain that vertex. One way to do what I think you might want is to observe that you can dissect a hypercube into four congruent parts by drawing hyperpyramids from any vertex $V$ to the four hyperfaces that do not contain $V$. ![]() Thus the volume of the half-hypersphere isīasic approach. Taking the square root on both sides we get: . Rotations in 4-dimensional Euclidean space In mathematics, the group of rotations about a fixed point in four-dimensional Euclidean space is denoted SO (4). I have already proved (without integration) that the volume of a hypersphere in $R^=\circ$$ If we square both sides of the above equation it becomes: s2 c2t2, which can also be written as: s2 - c2t2 0.
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