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Is there any pathological shape that has a finite surface area but an infinite volume, sort of like the opposite of a Gabriel's horn?

Rodrigo de Azevedo
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kevin
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    One could argue that all of $\Bbb R^3$ is an example. Infinite volume, zero surface area. If you want finite surface area, delete a small sphere. – Ross Millikan Mar 11 '14 at 22:18
  • @RossMillikan sorry, but why not an infinite surface area? How can we define the surface area to get zero? – Ian Mateus Mar 11 '14 at 22:24
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    I am taking advantage of the fact that there is no surface to $\Bbb R^3$. You are correct that if we take a larger and larger sphere the surface grows without bound. But there isn't even one point on the surface of $\Bbb R^3$ – Ross Millikan Mar 11 '14 at 22:28
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    Are you open to non-Euclidean spaces? – user76284 Sep 06 '21 at 06:04
  • Any region with infinite volume is unbounded, meaning you'll have to be more precise about what you mean by its surface. – eyeballfrog Mar 28 '23 at 13:40

4 Answers4

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Aside from my comment about all of $\Bbb R^3$, there is none. A sphere has the smallest surface for a given volume. One pathological case would be fractals that have well defined volume but no well defined surface area. In most cases you would like to say they have infinite area.

Ross Millikan
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Nope!

See the "Converse" section of http://en.wikipedia.org/wiki/Gabriel's_Horn .

Maroon
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kevin
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    this is only for surfaces of revolution. That does not take into account all possibilities. – clem Mar 11 '14 at 21:55
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Imagine a sphere outline in an infinite void. If the area within the sphere outline is empty space, and the space outside is solid, it is a 3D shape of infinite volume, and since it continues infinitely, there is no outer edge of the shape to apply surface area to, meaning the surface area is a finite value, on the same spherical plane as the outline.

tomas german
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    How does this differ from Ross Millikan's answer? – robjohn Feb 11 '18 at 18:22
  • @robjohn I do not see any similarity between the answers. – Anixx Dec 28 '20 at 22:42
  • @Anixx: Note that my comment was before Ross Millikan's posted answer. I was referring to his comment. – robjohn Dec 28 '20 at 22:48
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    To elaborate, years later: let $B^3$ be the open ball of radius 1. The set $\mathbb{R}^3-B^3$ has boundary $S^2$, the sphere of radius 1. The volume of $S^2$ is finite, but the volume of $\mathbb{R}^3-B^3$ is infinite. Hence this is a shape whose volume is infinite, but whose boundary has finite volume. – mwalth Oct 05 '21 at 17:40
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By calculus of variations it is established that a surface area of magnitude $A$ can enclose no more than this maximum volume among all possible shapes in 3-space:

$$ \dfrac{A^{\frac32}}{6 \sqrt{\pi}} $$ when it forms a sphere. So this is a finite limit, cannot go to infinity volume.

For example the 2D Koch curve/snow flakes have infinite perinmeter enclosing a finite area; so also some 3D fractal volumes have infinite surface area can enclose a finite volume.

Narasimham
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