Intuitively, suppose I have a bounded “smooth” surface in $\mathbb{R}^3$, say a part of the graph of a function $f: \mathbb{R}^2 \to \mathbb{R}$. Suppose I want to approximate the volume of this surface by containing it within boxes in $\mathbb{R}^3$.
Why intuitively can we make the sum of the volumes of the boxes go to zero (and hence, intuitively, the surface has zero volume)?
Note: I am purposely being imprecise here, since I am just trying to get some geometric intuition. By a “smooth” surface I mean intuitively a surface that is smooth throughout, without any sharp peaks or wild behavior.
Also, sure I can "visualize geometrically" how the boxes containing the surface can be made to have small volume, but I am wondering if there is an elementary argument like the one below to make this argument more rigorous, in an elementary-analytic-geometry sense.
For the case of a bounded “smooth” arc (e.g. a spiral) in $\mathbb{R}^3$, I can geometrically argue as follows:
- Let $L$ be the length of the arc. Subdivide the arc into $n$ sub-arcs of equal length $L/n$. Geometrically, an arc of length $L/n$ can be contained in a cube of side length $L/n$. So, do this for each of our $n$ sub-arcs. The sum of the volumes of these $n$ cubes is $n \cdot (L/n)^3 = L/n^2$. This tends to zero as $n$ tends to infinity. So the volume of the arc is zero.
I tried to extend this geometric argument to the case of a surface, but met with some difficulties. I tried to first contain the surface in a box of sides $l, w, h$, and then tried to subdivide the base of the box into $n^2$ parts each of length $l/n$ and height $w/n$, and then tried to argue that $n^2$ boxes each of small height could cover the surface. But I couldn’t really complete the intuitive argument.