Exploring the problem of an n-dimensional sphere

[Shmi]

I recently came across the problem of finding the volume and surface area of a sphere of n-dimensions.

Two and three dimensions seemed to work out, but oddly enough it's one dimension that seems the strangest conceptually. How do you distinguish volume and surface area of a line?

 

[Dickfore]

Volume of a line is the measure of a line, namely its length $2 L$.

 

[Shmi]

Is it non-sensical or have I ill-defined what I'm looking for in searching for surface area?

 

[Dickfore]

The surface $A_{n}(r)$ are is related to volume $V_{n}(r)$ through the formula:

$$V_{n}(r) = \int_{0}^{r}{A_{}(t) \, dt} \Rightarrow A_{n}(r) = \frac{d V_{n}(r)}{d r}$$

Usung this rule, we get:

$$A_{1}(r) = \frac{d}{d r}\left(2 r\right) = 2$$

Is this what you get?

 

[Shmi]

$$V_{n}(r) = \int_{0}^{r}{A_{}(t) \, dt} \Rightarrow A_{n}(r) = \frac{d V_{n}(r)}{d r}$$


Maybe this is really elementary or really complex, but why does that work? Is volume defined this way, regardless of dimensionality? Or is there a deeper reason?

 

[Dickfore]

The integral tells you that if you divide up the sphere into infinitesimally thin spherical shells of thickness $dt$, then the volume of each is simply the surface area $A_{n}(t)$ times the thickness. This can be regarded as a definition of volume. Then, you sum up all the volumes, but summation (an innumerably infinite number) of infinitesimal quantities is integration, so we arrive at the given forumla. The implication follows from the Fundamental Theorem of Calculus.

 

[Shmi]

Thanks!