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.


Problem: problem

Figure: figure

Note: text


It seems that the Author is aimed to make the reader aware about the risk to proceed by coordinates, in that case assuming

  • $\vec A=(A_x,A_y,A_z)$ with $A_x>0$

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 $x\in [0,a]$

$$r_x=a-x \implies dr_x=-dx$$


  • $x=a \iff r_x=0$

  • $x=0 \iff r_x=a$

therefore the correct step should be

$$\int_{(a,a)}^{(0,0)} \vec A \cdot d\vec r=\int_{a}^{0} A_x \cdot dr_x=\int_{0}^{a} -A_xdx=-\int_{0}^{a} A_xdx$$


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