• LifeInMultipleChoice@lemmy.world
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      14 hours ago

      I read it as surface area, thus being the amount of space on the sphere itself.

      A=4πr2 is the formula if I remember correctly, so I just figure only radius can be altered to match infinity.

      Maybe someone will tell me I’m missing something

      • thebestaquaman@lemmy.world
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        12 hours ago

        I believe the surface area of an n-dimensional hypersphere is (n - 1) pi r^{n - 1}. In that case (I may have some factors wrong here, just going off memory), an infinite-dimensional hypersphere has infinite surface area as long as it has non-zero radius.

        • loppy@fedia.io
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          7 hours ago

          It does indeed scale with r^(n-1), but your factors are not close at all. It involves the gamma function, which in this case can be expanded into various factorials and also a factor of sqrt(pi) when n is odd. According to Wikipedia, the expression is 2pi(n/2)r(n-1)/Gamma(n/2).

          • thebestaquaman@lemmy.world
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            2 hours ago

            Edit: ignore the below, I forgot my pi-factor in the gamma function for half-integers…

            Edit 2: Since you’re right, my missing gamma-factor completely changes this. An infinite-dimensional hypersphere will have zero surface area for any (finite?) radius.

            Original dum-dum:

            While I’m completely open that my factor is likely wrong here, the expression you provided is definitely wrong in the 3D case (I’m assuming the r superscript on the pi was a typo), since it doesn’t give n = 3 => A = 4 pi r^2.