Line Integral Notation: What is r in f(r)? I've had a tough time trying to interpret line integrals such as this one, which may be found on Wikipedia's Line Integral article:
$$\int_C f(\mathbf r) \,ds$$
More specifically, I'm trying to understand what $\mathbf r$ is in this integral. I understand that:

*

*$C$ is a curve and $ds$ is the differential arc length function of $C$, which depends on the parameterization variable.

*$f$ is a scalar field, which takes in a vector and returns a real number.

But, what is $\mathbf r$? Wikipedia says that $\mathbf r$ is a parameterization of $C$. Does this mean $\mathbf r$ is a vector-valued function and not a vector? Wikipedia certainly treats it like a function of the parameterization variable. If so, how is the expression $f(\mathbf r)$ valid if $f$ takes in a vector? This is like trying to fit a square peg into a round hole.
Then, I see this equivalent expression, which makes sense because we are getting the value of $\mathbf r$ at $t$:
$$\int_a^b f(\mathbf r(t))|\mathbf r'(t)| \,dt$$
However, this makes matters worse because now I'm more convinced that $\mathbf r$ is a function. How is the expression $f(\mathbf r)$ in the first integral possible? Is this just another way to write a composite function like $f \circ \mathbf r$?
As I gain more knowledge in mathematics, I'm discovering that I appear to be a perfectionist regarding syntax, and things like this make me go crazy. I have a background in object oriented programming, so when I see notation like this and fail to find logical consistency, I freak out a little bit. This is my first post here so I'm excited to see what people will say.
 A: If you want to be strict with the notation, then you simply write $\int_Cf\,ds$, and state appropriate regularity conditions on $f$ and $C$, and state that $ds$ denotes the arclength measure along $C$. Note that the symbol $dl$ might often be used in place of $ds$. Then, given a parametrization $\mathbf{r}:[a,b]\to\Bbb{R}^n$ with image $C$, one defines
\begin{align}
\int_Cf\,ds&:=\int_a^bf(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt.
\end{align}
The LHS is just a symbol, and the RHS is the meaning of the LHS.
As for why we write $f(\mathbf{r})$, well, this sort of thing is the usual abuse of language where people refer to functions not by the name of the function, but rather by the function values (and here often one overloads the symbol $\mathbf{r}$ to mean an element of $\Bbb{R}^n$ and also a mapping $[a,b]\to\Bbb{R}^n$). For instance, we might often speak of "the function $f(x)=x^2$", when in reality, what we should say is "the function $f:A\subset\Bbb{R}\to\Bbb{R}$, defined on some set $A$, having the 'rule' $f(x)=x^2$".
A: A parameterization of a curve $C$ $\mathbf r:[a,b]\to\Bbb R^n$, is a vector function, for which a $t\in [a,b]$ gives the vector $\mathbf r(t)$ of a point of the curve and conversely for a point of the curve there is a parameter $t\in [a,b]$ which gives the position vector $\mathbf r(t)$ of the point.
A: Is $r$ a vector-valued function?  Yes, it has a one-dimensional input and a multi-dimensional output.
An example is in order.
Suppose, the contour we want to evaluate is a circle.  Then we can parameterize the circle as:
$x = \cos t\\ 
y = \sin t$
or  $r(t) = (\cos t, \sin t)$
$f(r) = f(\cos t, \sin t)$
$r'(t) = (-\sin t, \cos t)$
$\|r'(t)\| = (-\sin t)^2 + (\cos t)^2 = 1$
Which means $ds = dt$
If our contour was a square with coordinates $(0,0),(1,0),(1,1),(0,1)$ we can parameterize each line segment as $r(t)=(t,0), r(t)=(1,t),r=(1-t, 1),r=(0,1-t)$ respectively.
I hope this helps.
