I am studying vector analysis (Mathematical Methods by Hassani). There is a passage in the book that talks about how parameterisation ensures that a line integral will have the correct sign.
What are the conditions that make a non-parameterised line integral have a different sign than a parameterised one? I tried it for other examples but I couldnt produce the conflict.
Regarding the example given, its weird to me (altho not wrong?) that $d\vec{r} = -\hat{e}_x dx$. My instinct is to have a "positive" $d\vec{r}$ and let the sign resolve itself depending on the direction of integration.
Excerpt:
Problem:
Figure:
Note:
It seems that the Author is aimed to make the reader aware about the risk to proceed by coordinates, in that case assuming
clearly along path $(iv)$
$$\vec A \cdot d\vec r=-A_xdx$$
and it leads to the wrong result
$$\int_{(a,a)}^{(0,0)} \vec A \cdot d\vec r=\int_{a}^{0} -A_xdx=\int_{0}^{a} A_xdx$$
The mistake here is in the limits for the integral, indeed as the parametrization shows, we have that for $x\in [0,a]$
$$r_x=a-x \implies dr_x=-dx$$
and
$x=a \iff r_x=0$
$x=0 \iff r_x=a$
therefore the correct step should be
$$\int_{(a,a)}^{(0,0)} \vec A \cdot d\vec r=\int_{a}^{0} A_x \cdot dr_x=\int_{0}^{a} -A_xdx=-\int_{0}^{a} A_xdx$$