What is the volume and surface area of a 1-D sphere?

[lostidentity]

I'm trying to find the volume and surface area of a 1-D dimensional sphere, i.e. retaining only the radial dependence.

I know that the volume element for a 3-D sphere would be

$$dV = r^2\sin\theta{d}\theta{d}\phi{d}r$$

If it's one-dimensional would it just be $$dV = r^2{d}r$$? Or would it just be $$dr$$?

With regards to the surface area vector in 3-D it is

$$d\boldsymbol{A} = r^2\sin\theta{d}\theta{d}\phi\hat{\boldsymbol{e}_r}$$

so in 1-D would it be

$$d\boldsymbol{A} = r^2\hat{\boldsymbol{e}_r}$$ or would it just be $$\hat{\boldsymbol{e}_r}$$?


Essentially what I'm trying to do is a Finite Volume Method for a 1-D sphere and I want to find the surface area vectors, and volume for my Finite Volume Cells.

 

[tiny-tim]

hi lostidentity!

perhaps I'm misunderstanding what you're saying, but if a 3D sphere is an ordinary sphere, then isn't a 2D sphere an ordinary circle, and a 1D sphere a pair of points at positions ± r from the origin?

 

[HallsofIvy]

hi lostidentity!

perhaps I'm misunderstanding what you're saying, but if a 3D sphere is an ordinary sphere, then isn't a 2D sphere an ordinary circle, and a 1D sphere a pair of points at positions ± r from the origin?

I think lostidentity is using "sphere" to mean "ball" and "2D sphere" to mean "disk".

If so by "1D sphere" he must an interval which has two points as "surface". In that case, the "volume element" of a "1D sphere" would be just "dx" and there is no "surface area element" since the surface is not continuous.