Timeline for How to exactly calculate the volume?
Current License: CC BY-SA 4.0
11 events
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| Feb 10, 2023 at 6:39 | history | edited | Henrik Schumacher | CC BY-SA 4.0 |
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| Feb 9, 2023 at 16:55 | comment | added | user64494 | +1. Thank you. Your answer is solid, as usually. Let us wait some time for an exact answer. | |
| Feb 9, 2023 at 10:44 | history | edited | Henrik Schumacher | CC BY-SA 4.0 |
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| Feb 9, 2023 at 10:37 | history | edited | Henrik Schumacher | CC BY-SA 4.0 |
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| Feb 9, 2023 at 10:27 | history | edited | Henrik Schumacher | CC BY-SA 4.0 |
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| Feb 9, 2023 at 10:20 | comment | added | user64494 |
Show[{P, RegionPlot3D[(2 x/3 + y/3 + z/3)^2 + (2 y/3 + x/3 - z/3)^2 + (2 z/3 + x/3 - y/3)^2 <= 1 && (2 x/3 - y/3 - z/3)^2 + (2 y/3 - x/3 - z/3)^2 + (2 z/3 - x/3 - y/3)^2 <= 1 && (2 x/3 + y/3 - z/3)^2 + (2 y/3 + x/3 + z/3)^2 + (2 z/3 - x/3 + y/3)^2 <= 1 && (2 x/3 - y/3 + z/3)^2 + (2 y/3 - x/3 + z/3)^2 + (2 z/3 + x/3 + y/3)^2 <= 1 && x^2 + y^2 <= 1, {x, -3/2, 3/2}, {y, -3/2, 3/2}, {z, -3/2, 3/2}, PlotPoints -> 50]}, Opacity -> 0.5] demonstrates that some pieces of M1 are outside of reg.
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| Feb 9, 2023 at 10:18 | comment | added | user64494 |
Thank you for your edit. I can't reproduce your M on my weak comp. I can execute M1 = BoundaryDiscretizeRegion[reg, MaxCellMeasure -> (1 -> 0.2)]. Then P = Region[M1, PlotRange -> {{-3/2, 3/2}, {-3/2, 3/2}, {-3/2, 3/2}}];.
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| Feb 9, 2023 at 9:30 | history | edited | Henrik Schumacher | CC BY-SA 4.0 |
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| Feb 9, 2023 at 8:52 | comment | added | user64494 |
Volume[DiscretizeRegion[reg]] results in 0.323708.
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| Feb 9, 2023 at 8:45 | comment | added | user64494 |
Thank you for your work. Unfortunately, DiscretizeRegion[reg, MaxCellMeasure -> Infinity] produces "DiscretizeRegion::drf: DiscretizeRegion was unable to discretize the region ImplicitRegion[<<2>>]." and the result of DiscretizeRegion[reg] is not good. There is a possibility Mathematica uses numeric integration to calculate Volume[reg].
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| Feb 9, 2023 at 8:22 | history | answered | Henrik Schumacher | CC BY-SA 4.0 |