How to show that a line integral is independent of the path of integration Show that the  following line integral is path-independent 
$$
\int_C (\ln y) e^{-x} dx - \dfrac{e^{-x}}{y}dy + zdz
$$
 A: It's enough to find a potential function $f$ such that:
\begin{align*}
f_x &= e^{-x}\ln y \\
f_y &= \frac{-e^{-x}}{y} \\
f_z &= z
\end{align*}
Integrating the first equation with respect to $x$, we obtain:
$$
f(x, y, z) = -e^{-x}\ln y + g(y, z)
$$
where $g$ is some function of $y$ and $z$. By taking the partial derivative with respect to $y$, combining with the second equation, and integrating with respect to $y$, we obtain:
$$
\frac{-e^{-x}}{y} + g_y(y, z) = f_y = \frac{-e^{-x}}{y} \iff g_y(y, z) = 0 \iff g(y, z) = h(z)
$$
where $h$ is some function of $z$, so that:
$$
f(x, y, z) = -e^{-x}\ln y + h(z)
$$
By again taking the partial derivative with respect to $z$, combining with the third equation, and integrating with respect to $z$, we obtain:
$$
h'(z) = f_z = z \iff h(z) = \frac{z^2}{2} + C
$$
where $C$ is some constant. So we conclude that the desired potential function is:
$$
f(x, y, z) = -e^{-x}\ln y + \frac{z^2}{2} + C
$$
A: Consider the vector field $$\vec{F} = \left((\ln y) e^{-x},-\dfrac{e^{-x}}{y},  z \right)$$
If you can show that $\nabla \times \vec{F} = \vec{0}$, then by Stokes' theorem, the closed path integral will be zero.
