$$\vec{B} = \frac{cos(\alpha)}{p^2}e_p + (\frac{sin(\alpha)}{p^2}+\frac{1}{ap})e_\alpha + 0e_z$$
Calculate the line integral $$\int_C \vec{B} \cdot d\vec{r}$$, where C is parameterized by
$$ \vec{r} = (bcos(s),bsin(s),b)$$$$ b>0$$ $$s: 0\rightarrow2\pi$$
I have tried writing $d\vec{r} = \frac{d\vec{r}}{ds}ds$, and rewriting $\vec{B}$ in only s, but I really don't know how to solve it.
Note that
$$ \frac{{\rm d}\mathbf{r} }{{\rm d}s} = -b\sin s\;\mathbf{e}_x + b\cos s\;\mathbf{e}_y = b \mathbf{e}_\alpha \tag{1} $$
where $\mathbf{e}_\alpha$ is the unit tangential vector in cylindrical coordinates. And $\mathbf{B}$ can be written as
$$ \mathbf{B} = \frac{\cos \alpha}{\rho^2} \mathbf{e}_\rho + \left(\frac{\sin \alpha}{\rho^2} + \frac{1}{a \rho} \right) \mathbf{e}_\alpha \tag{2} $$
From Eqns (1) and (2):
$$ {\rm d}\mathbf{r}\cdot\mathbf{B} = {\rm d}s\frac{{\rm d}\mathbf{r} }{{\rm d}s}\cdot\mathbf{B} = {\rm d}s\; b \left.\left(\frac{\sin \alpha}{\rho^2} + \frac{1}{a \rho} \right)\right|_{\rho = b, \alpha = s} = {\rm d}s\; \left(\frac{\sin s}{b} + \frac{1}{a} \right) $$
We then arrive to
$$ \int_C {\rm d}\mathbf{r}\cdot \mathbf{B} = \int_0^{2\pi}{\rm d}s\; \left(\frac{\sin s}{b} + \frac{1}{a} \right) = \frac{2\pi}{a} $$