I am doing vector calculus and am having trouble gaining intuition/understanding about the difference between a line integral with respect to ds, and line integrals with respect to dx.
$$\int_C f(x,y)\ ds$$
$$\int_C f(x,y)\ dx$$ or $$\int_C f(x,y)\ dy$$
Notably, in the consequence that:
$$\int_C f(x,y)\ ds=\int_{-C} f(x,y)\ ds$$
But:
$$\int_C f(x,y)\ dx=-\int_{-C} f(x,y)\ dx$$
Would anyone have pointers to better understand these nuances? Thanks!
These are very good questions, and in trying to understand the differences it helps to have a little insight into this topic from the slightly more formal point of view of integrating forms on curves.
For ease of reference, call these integrals (1) and (2) respectively.
$\int_C f(x,y)\ ds$
$\int_C f(x,y)\ dx$
To be sure, (1) and (2) are not the same integral but differently parametrized. They are truly distinct.
Assume throughout that $C$ is a smooth or piecewise smooth curve in $\mathbb{R}^2$ and that $f$ is continuous on some open set $D$ in $\mathbb{R}^2$ that contains $C$. That is, the usual basic assumptions for line integrals.
Also, this entire discussion would be valid in $\mathbb{R}^n$ for $n>2$ but we can stick to $n=2$ without any real loss of generality.
(1) is a line integral of a scalar field (real-valued function) along a curve $C$ with respect to arc-length $ds$ and is calculated as:
$\int_C f(x,y)\ ds = \int_a ^b f(x(t),y(t))\sqrt{x'(t)^2+y'(t)^2}\space dt$
where $x(t)$ and $y(t)$ parametrize $C$ for $a\leq t\leq b$.
For (2), rather than try to view it as a special case of (1), which it really is not, it's better to view it as a special case of a line integral
$\int_C f(x,y)\space dx+g(x,y)\space dy$
which is often written as
$\int_C P\space dx+Q\space dy$
where $Q=g(x,y)=0\space$ in this case.
In this case, $\vec F=(P,Q)$ is a vector field on $D$ and we are integrating it along $C$ (vs. integrating a scalar function in (1)).
In this case the line integral is evaluated as
$\int_C P\space dx+Q\space dy = \int_a^b \vec F (\vec r(t))\cdot \vec r'(t)\space dt$
where $\vec r(t) = (x(t),y(t))$ for $a\leq t\leq b$ and when $Q=g(x,y)=0$ we just get
$\int_a^b f(x(t),y(t))\space x'(t)\space dt $
and comparing to the evaluation of (1) this is not a special case.
As to your observation on signs when reversing orientation of $C$:
Integral (1) is an unoriented line integral. If $f(x,y)=1$ this calculates the the arc length of $C$ and we would like this to be the same positive value regardless of which way $C$ is traversed. Clearly if we parametrize as above and then integrate from $b$ to $a$ the integral will reverse sign however if the unoriented integral is required to be on an interval $a\leq t\leq b$ it will not even under an orientation-reversing parametrization.
Integral (2) on the other hand as the integral of a vector field (its associated 1-form) on a curve (1-chain) is naturally an oriented integral and as defined will naturally reverse sign upon reversing the orientation.