Let $I$ be a $n$-dimensional interval (i.e. Cartesian product of closed intervals).
For instance, when $n=3$, a typical interval looks like $[a_1,b_1]\times[a_2,b_2]\times[a_3, b_3]$.
When $n=3$, I am familiar with the usual notion of volume and surface area, what I am not sure is for other $n$ (both lower and higher).
My attempt:
When $n=2$, volume is "surface area of the rectangle" and surface area is "total length of the 4 sides of the perimeter"?
When $n=1$, volume is "length of the interval", and surface area is 1?
How do we generalise this notion nicely to higher dimensional intervals? (The resources online mostly talk about higher dimensional spheres).
Thanks.
First: for $n=1$, the "surface" is the set of the two end points. Its measure is $0$.
For $n\ge 2$, the "surface" is the sum of the measures of the "faces". A "face" is obtained by choosing a dimension $k$, $1\le k\le n$ and a "side", say "left" or "right". A "face" in the $k$-th dimension is a set $$[a_1,b_1]\times\cdots[a_{k-1},b_{k-1}]\times\{c_k\}\times[a_{k+1},b_{k+1}]\times\cdots\times[a_n,b_n]$$ where $c_k=a_k$ if you chose the "left" side and $c_k=b_k$ if you chose the "right" side.
The measure of a face in $k$-th dimension is $$\prod_{1\le j\le n,j\ne k}(b_j-a_j)$$
The total "surface" is $$2\sum_{k=1}^n \prod_{1\le j\le n,j\ne k}(b_j-a_j)$$
You can be interested in polytopes.