Why is curve length an integral (an area/volume)?

Question

Why is curve length

$$L(C)=\int_a^b |r'(t)| dt$$

(where $r(t)$ is a parametrization of the curve.)

an integral (an area/volume)?

Answer 1

An intuitive, completely unrigorous definition is that the integrand(function being integrated) is the derivative of arc length (or a tiny differential (shudders) bit of arc length). Taking the integral sums up these bits of arc lengths over whatever times you specify.

Answer 2

As has already been said, integration is just generalized summation. This summation can be used to calculate area or volume, but it can also be used to calculate other things. In this case, to sum up infinitesimal lengths to get the full curve length.

To see why we get a length here, it may be useful to consider dimensional analysis. The parameter $t$ is measured in some unit, perhaps time (though it could be anything). $r$ is a vector in space, so it's components are all measured by length. Therefore the derivative $$\frac {dr}{dt}$$ (and its norm) would have units of $\frac {\text{length}}{\text{time}}$. When integrated, we multiply by $dt$, which has units of time. So our result is $$\frac {\text{length}}{\text{time}}(\text{time}) = \text{length}$$

And so we should expect to get a length from it.

Contrast this with your standard area-under-the-curve integral, say of $y = f(x)$. Since $f(x)$ is a height, it has units of length. And since $x$ is a horizontal distance, it too is in units of length. Thus $f(x) dx$ has units of $\text{length}\times\text{length}$, which is area.