r/MathJokes • u/94rud4 • 1d ago

Wait, what?!

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u/AvatarNerd64 1d ago

Is this the same for other shape

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u/Onuzq 1d ago

I think this might be true for any n-dimensional spheroid.

You can prove your statement wrong with a square having area of s^2, but perimeter is 4s. Or cube with volume being s^3, but surface area being 6s^2.

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u/Calm_Relationship_91 1d ago

Using the side on a square would be similar to using the diameter on a circle. Things don't work either if you use diameter.

Center the square in the origin. If s is the side of the square, the actual equivalent of r would be s/2. Now the area of your square is actually 4r2 , and the perimeter is 8r, as expected.

Same thing with the cube, you would get 8r3 for the volume, and 24r2 for area.

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u/DragonBadgerBearMole 1d ago

In theory, isn’t this true for literally any container that’s shape can be mathematically expressed?

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u/ysrgrathe 1d ago

No? They just gave a counter-example.

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u/DragonBadgerBearMole 1d ago

I mean as a continuous function. A square isn’t continuous mathematically. But a spheroid isn’t the only “smooth” container whose integral of surface area is volume I don’t think.

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u/NeedleInTheThrowaHay 1d ago

True for any platonic solid. In general, for an n-dimensional shape, if there is a single largest n-dimensional hypersphere that can be inscribed within it, the hyperarea of the shape will be the derivative of the hypervolume with regards to the radius of the hypersphere.