A line integral 
If $\mathbf{B}(\mathbf{x})=\rho^{-1}\mathbf{e}_{\phi}$ in cylindrical polars, find:
$$\int_{C}\mathbf{B}\cdot\mathrm{d}\mathbf{x}$$
where $C$ is the circle $z=0,\rho=1,\;0\leq \phi\leq 2\pi$. Also find $\nabla\times\mathbf{B}$

My try:
$$\int_{C}\mathbf{B}\cdot\mathrm{d}\mathbf{x}=\int_C (\rho^{-1}\mathbf{e}_{\phi})\cdot (\mathbf{e}_{\rho}d\rho+\rho\mathbf{e}_{\phi}d\phi+\mathbf{e}_{z}dz)=\int_C \mathrm{d}\phi=2\pi$$
This seems way too easy. Is it wrong? For the second part I get $\nabla\times\mathbf{B}=-\rho^{-3}\mathbf{e}_z$ but a friend of mine says it should be $0$...
 A: This answer addresses part 2 which asks for the curl of $\mathbf B$. Your friend is correct that the curl is identically zero.
The components of the curl of a vector field $\mathbf{B}(\mathbf{x})=\mathbf{e}_{\rho}B_{\rho}+\mathbf{e}_{\phi}B_{\phi}+\mathbf{e}_{z}B_{z}$ in cylindrical coordinates are:
$$\nabla\times\mathbf{B} = \mathbf{e}_{\rho}\left(\frac{1}{\rho}\frac{\partial B_z}{\partial\phi}-\frac{\partial B_{\phi}}{\partial z}\right) + \mathbf{e}_{\phi}\left(\frac{\partial B_{\rho}}{\partial z}-\frac{\partial B_z}{\partial \rho}\right) + \mathbf{e}_{z}\frac{1}{\rho}\left(\frac{\partial}{\partial \rho}\left(\rho B_{\phi}\right)-\frac{\partial B_{\rho}}{\partial \phi}\right)$$
For the function $\mathbf{B}(\mathbf{x})=B_{\phi}(\rho)\mathbf{e}_{\phi}=\rho^{-1}\mathbf{e}_{\phi}$, the $\rho$ and $z$ components are zero: $B_{\rho}=B_z=0$. And since the $\phi$ component $B_{\phi}=\rho^{-1}$ is independent of $\phi$ and $z$, $\frac{\partial B_{\phi}}{\partial z}=0$. Thus, the formula for the curl reduces to:
$$\nabla\times\mathbf{B} = \mathbf{e}_{\rho}\left(-\frac{\partial B_{\phi}}{\partial z}\right) + \mathbf{e}_{z}\frac{1}{\rho}\left(\frac{\partial}{\partial \rho}\left(\rho B_{\phi}\right)\right) \\
 = \mathbf{e}_{z}\frac{1}{\rho}\left(\frac{\partial}{\partial \rho}\left(\rho B_{\phi}\right)\right)\\
 = \mathbf{e}_{z}\frac{1}{\rho}\left(\frac{\partial}{\partial \rho}\left(\rho \rho^{-1}\right)\right) = \mathbf{e}_{z}\frac{1}{\rho}\left(\frac{\partial}{\partial \rho}\left(1\right)\right) = \mathbf{e}_{z}\frac{1}{\rho}\left(0\right) = \mathbf0.$$
