Line integral of conservative field in polar coordinates I am solving the vector equation:
$$\vec \nabla P(r,\phi) = \vec f(r,\phi)$$
where $\vec f$ is conservative, in polar coordinates.
Am I allowed to the following?
$$\partial_r P= f_r$$
$$\partial_\phi P=r f_\phi$$
Or, if I am to solve the line integral, using the following fact 
$$P(r,\phi)-P(r_0,\phi_0)=\int_\mathcal{C} \vec f(r,\phi)\cdot \vec dl$$
with $\mathcal{C}$ any path from $(r_0,\phi_0)$ to $(r,\phi)$, I could go along the direction $\hat r$ and write:
$$P=\int_{r_0}^r  f_r(r,\phi) dr$$
or not?

The gradient operator in polar coordinates is given by
$$\nabla =\hat r \frac{\partial }{\partial r}+\hat \theta \frac{1}{r}\frac{\partial }{\partial \theta}$$
Since the gradient of a scalar field is a conservative, then the line integral between two points is not affected by the path taken between points.  Therefore, one can write
$$\begin{align}
P(\vec r_2)-P(\vec r_1)&=\int_{\vec r_1}^{\vec r_2} \nabla P \cdot d\vec \ell\tag 1
\end{align}$$
The points $\vec r_1$ and $\vec r_2$ are not in general collinear.  In other words, the unit vectors $\hat r_1 =\hat x \cos \theta_1+\hat y \cos \theta_1$ and $\hat r_2 =\hat x \cos \theta_2+\hat y \cos \theta_2$ with $\theta_1\ne \theta_2$ point in different directions.  
Therefore, in general the line integral in $(1)$ is not equivalent to a path integral along a constant $\theta$ direction.
