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When motivating the definitions of line and surface integrals, one usually defines the length and area elements \begin{align*} ds &:= \| \vec{r}^{\, '}(t) \| dt, \\ dA &:= \| \vec{\Sigma}_{u} \times \vec{\Sigma}_{v} \| du dv. \end{align*} The "justification" given is that both expressions approximate the actual real length and area elements. So, one gives geometric justifications (lines and parallelograms). However, there are many ways to approximate the actual real length and area elements. One must generally be careful when working with such "infitesimals", e.g. Is value of $\pi = 4$?.

Question : So, I would like to know if there is a more rigorous way of justifying that the definitions of line and surface integrals are really what we want. For example, why not take $$ ds := 0.9999 \cdot\| \vec{r}^{\, '}(t) \| dt $$ or approximate the length element by a slightly curved segment ?

Amateur
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  • What are your "actual real length and area"? – Christian Blatter Sep 04 '14 at 15:39
  • @ChristianBlatter Let's say I have a curve $C$ parameterized by $\vec{r}(t)$ for $t \in [a,b]$ for example. Then for $c \in [a,b)$ we approximate $| \vec{r}(c+\epsilon) - \vec{r}(c) |$ by the length of a line segment. The "actual length" would be the arc length of the curve segment when $t \in [c, c+\epsilon]$. – Amateur Sep 04 '14 at 15:44

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The simplest case is that of curves. When $$\gamma:\quad[a,b]\to{\mathbb R}^n, \qquad t\mapsto{\bf f}(t)$$ is a curve in ${\mathbb R}^n$ parametrized by a continuous map ${\bf f}$ then for any subdivision $${\cal P}:\quad a=t_0<t_1<t_2<\ldots<t_N=b$$ of the parameter interval $[a,b]$ we can compute the elementary euclidean length of the polygonal path through the points ${\bf x}_k:={\bf f}(t_k)$ $\>(0\leq k\leq N)$. This length is given by $$L_{\cal P}(\gamma)=\sum_{k=1}^N|{\bf x}_k-{\bf x}_{k-1}|=\sum_{k=1}^N|{\bf f}(t_k)-{\bf f}(t_{k-1}|\ .$$ It is then natural to define the "real" length of $\gamma$ by putting $$L(\gamma)=\sup_{\cal P}L_{\cal P}(\gamma)\ .$$ The sup here intuitively corresponds to our geometric idea, and is the simplest kinf of "limit" under the given circumstances.

It is then a theorem requiring about two pages for proof that when ${\bf f}$ is in fact continuously differentiable then this length $L(\gamma)$ can be computed as an integral: $$L(\gamma)=\int_a^b|{\bf f}'(t)|\>dt\ .\tag{1}$$ A "differential" form of equation $(1)$ is the formula $$ds=|{\bf f}'(t)|\>dt\ ,$$ it reminds one of the fact that $L(\gamma)$ should not depend on the chosen parametrization. I won't go into this here.

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These elements satisfy two conditions:

(1) When changing coordinates (i.e. reparametrizing the curve or surface), both of these quantities transform in such a way that the values of integral quantities computed using these elements does not depend on the choice of coordinates (parametrization) used.

(2) They give the usual formulas for lengths of lines and areas of rectangles (with any parametrization, though the standard ones will certainly do, provided (1) holds).

Note that your proposed definition for arc length satisfies (1) but not (2); indeed "lengths" are independent of your choice of parametrization but your line segments will not have the "right" length.

Travis Willse
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  • Indeed. But, I might as well ask to get corresponding results for all known lengths and areas of curves and surfaces. Does the fact that the definitions satisfy your (2) (and (1)) automatically guarantees it will match all known results ? – Amateur Sep 04 '14 at 15:54
  • What do you mean by "corresponding results for all known lengths and areas"? – Travis Willse Sep 04 '14 at 16:03
  • I mean that using the definitions we should get that the arc length of a circle corresponds (is equal to) with the known $2 \pi r$, for example. – Amateur Sep 04 '14 at 16:07
  • What is your method of deciding what the length of a curve is? – Travis Willse Sep 04 '14 at 16:13
  • Hmmm yeah my method must be equivalent to using the definition of the line integral, never mind. – Amateur Sep 04 '14 at 16:20