Let $C$ be curve along surfaces $z=\ln(1+x)$ and $y=x$ from $(0,0,0)$ to $(1,1,\ln(2))$. Calculate the work done by vector field
$$\mathbf{F}(x,y,z)=(2x\sin(\pi y)-e^z)\mathbf{i}+(\pi x^2 \cos (\pi y)-3e^z)\mathbf{j}-xe^z \mathbf{k}$$
along the curve $C$.
I got this hint: $$\mathbf{F}(x,y,z)=\overbrace{(2x\sin(\pi y)-e^z)\mathbf{i}+(\pi x^2 \cos \pi y)\mathbf{j} -xe^z\mathbf{k}}^{=\mathbf{H}}-\overbrace{3e^z\mathbf{j}}^{=\mathbf{G}}$$ And that I should find the potential function $\phi$ and then $\nabla\phi=\mathbf{G}$. Also that $\int_C \nabla \mathbf{G} \cdot d\mathbf{s} =\phi(1,1,\ln(2))-\phi(0,0,0)$. But i'm still a bit confused at where to start..
EDIT. copied the question wrong. There is $e^z$ not $e^x$ in all terms.
Let's check if $\mathbf{F}$ is conservative:
$$\begin{align} \nabla\times\mathbf{F}&=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}} \\ 2x\sin(\pi y)-e^z & \pi x^2 \cos (\pi y)-3e^z & -xe^z \end{vmatrix} \\ &=\mathbf{i}(0+3e^z)-\mathbf{j}(-e^z+e^z)+\mathbf{k}(2\pi x\cos(\pi y)-2\pi x\cos(\pi y)) \\&=3e^z\mathbf{i} \end{align}$$
It's not?? If I split $\mathbf{F}$ to $\mathbf{H}$ and $\mathbf{G}$, then $\mathbf{H}$ is conservative, but what to do with $\mathbf{G}$?
First, check to see if it has a potential function at all: a vector field $\boldsymbol F$ has a potential function if and only if $\nabla \times \boldsymbol F = 0$.
If it does have a potential, find it as follows:
(1) First integrate the $x$ component with respect to $x$, $$ \phi = \int F_x(x,y,z) \, dx = \phi_1 (x,y,z) + C_1(y,z),$$
since the constant of integration can be any function of $y$ and $z$.
(2) Then take the $y$-derivative of this, $$ \frac{d \phi}{dy} = \frac{d\phi_1}{dy} + \frac{d C_1}{dy}.$$
Since this derivative must be the $y$-component of $\boldsymbol F$, we have the equation
$$ \frac{d C_1}{dy} = F_y - \frac{d\phi_1}{dy}.$$
Solving this gives $C_1$, up to an integration constant- any function of $z$.
(3) Repeat step (2) with $F_z$ and the $z$ derivative to find the $z$ dependence, leaving only an arbitrary constant left--this doesn't matter once you take the difference at two points.
(4) Check the potential function by computing the gradient $\nabla \phi$--it should be $\boldsymbol F$.