this post was submitted on 25 Dec 2025
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No Stupid Questions

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[–] Zwuzelmaus@feddit.org 8 points 1 day ago

If the radius is infinite

If the radius is >0 and the definition of sphere is used in infinite-dimensional space.

[–] loppy@fedia.io 5 points 1 day ago

Planes can be thought of as spheres of infinite radius, and have infinite surface area. This point of view is very natural in conformal geometry.

[–] LifeInMultipleChoice@lemmy.world 12 points 1 day ago* (last edited 1 day ago) (1 children)

I would assume if and only if the radius is infinite

[–] sopularity_fax@sopuli.xyz 2 points 1 day ago (2 children)
[–] Knock_Knock_Lemmy_In@lemmy.world 2 points 16 hours ago

Math is weird. You can have infinite circumference but finite radius.

[–] LifeInMultipleChoice@lemmy.world 9 points 1 day ago* (last edited 1 day ago) (1 children)

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 5 points 1 day ago (1 children)

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 1 points 1 day ago (1 children)

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 1 points 1 day ago* (last edited 1 day 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.

[–] ch00f@lemmy.world 5 points 1 day ago (1 children)
[–] mumblerfish@lemmy.world 3 points 1 day ago (1 children)

one can further prove that the sphere S**n−1 can be partitioned into as many pieces as there are real numbers (that is, {\displaystyle 2^{\aleph \_{0}}}pieces)

Would the answer to OP be some argument along the lines of defining the surface area of the ball as the sum of the partitioned balls surface areas then?

[–] ch00f@lemmy.world 1 points 1 day ago

IDK what OP is even going on about. This just seemed relevant.

[–] Kolanaki@pawb.social -5 points 1 day ago* (last edited 1 day ago) (3 children)

Under the definition of what a circle is...?

It's a polygon with infinite sides. The circumstance would be "being a circle." If it has less than infinite sides, it's not a circle.

[–] TonyTonyChopper@mander.xyz 6 points 1 day ago

Ok but the sum of the infinite sides is a finite value

[–] sbeak@sopuli.xyz 1 points 1 day ago (1 children)

sphere != circle

Sphere is like a ball (3D), while circles are like those round coasters for drinks (flat)

[–] Kolanaki@pawb.social 1 points 17 hours ago (1 children)

Okay, and your point is? I did not describe a sphere.

[–] sbeak@sopuli.xyz 1 points 3 hours ago (1 children)

I'm pretty sure the question described a sphere

[–] Kolanaki@pawb.social 2 points 3 hours ago (1 children)

It said circle originally. OP edited the question after my initial reply.

[–] sbeak@sopuli.xyz 1 points 2 hours ago

Fair enough.

[–] TonyTonyChopper@mander.xyz 1 points 1 day ago (1 children)

Ok but the sum of the lengths of these infinite sides is a finite value

[–] Kolanaki@pawb.social -1 points 1 day ago

If OP just means the size of the entire inside of the circle, then it would need an infinite radius, too.