Problem: Determine the values A and B for which the vector field \begin{align*} F = Ax \ln(z) \hat{i} + By^2 z \hat{j} + (\frac{x^2}{z} + y^3) \hat{k} \end{align*} is conservative. If $C$ is the straight line from $(1,1,1)$ to $(2,1,2)$, find \begin{align*} \int_C 2x \ \ln(z) \ dx + 2y^2 z dy + y^3 dz \end{align*}
Attempt at solution: For the field to be conservative, we demand that \begin{align*} \frac{\partial F_1}{\partial y} = 0 = \frac{\partial F_2}{\partial x}, \frac{\partial F_1}{\partial z} = \frac{Ax}{z} = \frac{2x}{z}, \frac{\partial F_2}{\partial z} = By^2 = 3y^2. \end{align*} Hence $A = 2$ and $B =3$. Now, to evaluate the integral above I first need to find a potential function $\phi$ for my vectorfield, then apply a vector field extension.
So let $\frac{\partial \phi}{\partial x} = 2x \ln(z)$. Then $\phi(x,y,z) = x^2 \ln(z) + C(y,z)$. Taking the partial of this with respect to $y$ gives $C'(y,z)$. But we also have that $\frac{\partial \phi}{\partial y} = 3y^2 z$. Hence $C'(y,z) = 3y^2 z$. Integrating this with respect to $y$ gives $y^3 z+ D(z)$. Thus our potential function so far becomes $\phi(x,y,z) = x^2 \ln(z) + y^3 z + D(z)$. I'm not sure if I'm done now, or if I should take the partial of this with respect to $z$. But then I don't know how to solve for $D'(z)$.
And how to evaluate the above integral with the help of this potential function?