Difference between uniform and non-uniform conductor Am I correct to say that a uniform conductor is cylindrical, while a non-uniform conductor can have any geometry as long as it is open at both ends.
The textbook does not seem to define these 'uniform' or 'non-uniform' conductors.
 A: The integration is performed on a path, usually a section of a circuit
$\mathbf{B}(\mathbf{r}) = - \dfrac{\mu}{4 \pi} \displaystyle \int_{\mathbf{r}' \in\ell}  i \dfrac{\mathbf{r} - \mathbf{r}' }{|\mathbf{r}-\mathbf{r}'|^3} \times d \mathbf{r}'$.
So far, this is a equation of physics. When you need to evaluate this integral, you usually need to describe the path $\ell$ in a parametric way, i.e. the points of the path $\ell$ in space are function of a parameter $s$, namely $\mathbf{r}'(s)$. It's convenient to use a regular parametrization, so that each point of the path is associated with one value of the parameter $s$.
The choice of the parametrization induces the extreme of integration as well, being $\mathbf{r}_0 = \mathbf{r}(s_0)$ the starting point of the path, and $\mathbf{r}_1 = \mathbf{r}(s_1)$ the other extreme point, so that you can recast the integral as
$\mathbf{B}(\mathbf{r}) =  - \dfrac{\mu}{4 \pi}\displaystyle \int_{s_0}^{s_1}  i \dfrac{\mathbf{r} - \mathbf{r}'(s) }{|\mathbf{r}-\mathbf{r}'(s)|^3} \times \dfrac{d \mathbf{r}'(s)}{ds} ds$
On the other hand, you can write
$\mathbf{B}(\mathbf{r}) = \displaystyle \int_{\mathbf{r}' \in \ell} d\mathbf{B}(\mathbf{r}, \mathbf{r}')$, being
$d\mathbf{B}(\mathbf{r}, \mathbf{r}') = -  \dfrac{\mu}{4 \pi}  i \dfrac{\mathbf{r} - \mathbf{r}' }{|\mathbf{r}-\mathbf{r}'|^3} \times d \mathbf{r}'$
