Line Integral: why does it require |r'(t)|

calculusline-integrals

I'm a confused by the following definition of the line integral from Wiki:

For some scalar field $f : U \subset R^n \to R$ the line integral
along some piecewise curve $C \subset U$ is defined as: $\int_C f(r) \ ds = \int_a^b f(r(t)) | r'(t) | dt$

where $r : [a,b] \to C$ is an arbitrary bijective parametrization
of the curve $C$ such that $r(a)$ and $r(b)$ give endpoints of $C$.

QUESTIONS

On the LHS $r$ is used. Is it still defined as a bijective parametrization?

On the RHS why is the Euclidean norm $|r'(t)|$ required in the equation? Intuitively, it seems the line integral could just be calculated $\int_a^b f(r(t)) dt$?

Best Answer

A stupid example of why you need the $|r'(t)|$ factor.

Suppose you have your parametrization of $x$ and $y$ in terms of some parameter t. We could replace $t$ with $2t$ and it would still describe the same curve. You would be moving along the curve at twice the speed, but it is the same curve. As a result of moving along that curve at twice the speed, the interval of integration will be cut in half though. The $|r'(t)|$ will correct for such shenanigans.

$\int_a^b f(r(t))|r'(t)|\ dt = \int_{\frac {a}{2}}^{\frac {b}{2}} f(r(2t))|r'(2t)|\ dt$

We want to analyze the integrand such that ever piece of the curve of equal length has equal contribution regardless of the "speed" we are moving across those lengths.

The $|r'(t)|$ factor gives us a consistent integral for a path regardless of the parameterization of that path.