In multivariable calculus the line integrals was parameterized and denoted: \begin{gather*} \int_l \mathbf{A} \cdot \, d\mathbf{r} = \int_\gamma \mathbf{A}(\mathbf{r}(t)) \cdot \, \frac{d\mathbf{r}(t)}{dt}dt \tag{1}\\ \text{where} \quad \mathbf{r}(t)=x(t)\mathbf{\hat i}+y(t)\mathbf{\hat j}+z(t)\mathbf{\hat k} \end{gather*} However in electromagnetism the line integrals are confusing.

Exemple: Suppose a charge is moving from point $b(\rho_b,\phi_b,z_b)$ to point $a(\rho_a,\phi_a,z_a)$ along the direction of $\rho$ (cylindrical coordinates): \begin{gather*} -\int_b^a \frac{Q\phi_L}{2\pi\epsilon_0 \rho}\mathbf{\hat \rho} \cdot (\mathbf{\hat \rho}\, d\rho+\mathbf{\hat \phi} \, \rho d\phi + \hat z \, dz)=\\ =-\int_b^a \frac{Q\phi_L}{2\pi\epsilon_0 \rho}\mathbf{\hat \rho} \cdot \mathbf{\hat \rho} \, d\rho = -\int_b^a \frac{Q\phi_L}{2\pi\epsilon_0 \rho} \, d\rho \end{gather*} This notation confuses me. In math the procedure was; parametrize the curve, take the derivative of it, dot it with the parameterized field, as in (1). How can I use this approach in the exemple?

How can I derive the notation in the second integral from the notation in (1)?

Can I calculate the exemple using $\mathbf{r}(t)$ in cartesian coordinates?

  • $\begingroup$ Very briefly, $\rho$ is the parameter. Usually you can use coordinates as parameters. $\endgroup$
    – Javier
    Commented Aug 13, 2016 at 17:11

1 Answer 1


That is exactly what you did here. You'r parametrization is - $$\vec{r} = (t,\phi_a,z_a),\;t\in[\rho_a,\rho_ b]$$ so - $$\vec{r}'=(1,0,0)$$ since $\phi_a=\phi_b, z_a=z_b$.

  • $\begingroup$ Thanks! I edited my post, I missed one thing. In my formula, $\mathbf{r}(t)$ is cartesian coordinates, i.e. $\mathbf{r}(t)=x(t)\mathbf{\hat i}+y(t)\mathbf{\hat j}+z(t)\mathbf{\hat k}$. Can I calculate the exemple using cartesian coordinates? Before electromagnetism I have only used this approach. $\endgroup$
    – JDoeDoe
    Commented Aug 14, 2016 at 7:59
  • $\begingroup$ Your first formula works for any set of coordinates, it does not require the cartesian coordinates specifically. If you want to calculate your example in cartesian coordinates you first have to change variables and then calculate the integral. $\endgroup$
    – proton
    Commented Aug 14, 2016 at 8:40

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