We are all familiar with the 3-dimensional notions of volume and area. For example, the volume of a 3-dimensional sphere is given by $V_3 = 4\pi R^3/3$ and its area by $S_3 = 4\pi R^2$, where $R$ is its radius.
My question concerns the one-dimensional equivalent notion of area. I understand that a 1-dimensional hypersphere is specified by two points, $\{-R, R\}$ and thus its 1-dimensional volume would be $V_1 = 2R$. But what would be the area $S_1$ here?
On
I came across this recently in the context of Green's formula stating that $$ \int_{U}\Delta u \,dx= \int_{\partial U}\frac{\partial u}{\partial \nu} dS(x) $$ where $U$ is a region of $\mathbb{R}^{n}$, $\partial U$ is its boundary, $\Delta u$ is the Laplacian of $u$ and $\frac{\partial u}{\partial \nu}$ is the "outward normal derivative" of $u$ at the boundary.
In one dimension this would become $$ \int_{a}^{b} \frac{d^{2}u}{d x^{2}}\,dx =\left(\frac{d u}{d x}(b)-\frac{d u}{d x}(a)\right)k$$ with $k$ the "area" at each of the two end points. This is just the fundamental theorem of calculus—so long as we have $k=1$.
The Wikipedia article about the n-sphere tells that:
"The 0-sphere (unitary radius) consists of its two end-points, {−1,1}. So, $S_0 = 2$ "
intending that in this case the Hausdorff measure which keeps congruency in the relation between the measure of the volume and the surface: see them in the cited Wikipedia article.
So also the surface of sphere {-R,R} equals $2$. In the limit for $R \to 0$ it remains equal to $2$.
-- note -- the surface of a 1-ball is denoted as $S_0$, the dimension of the sphere.