How to show that a line integral is independent of the path of integration

Question

Show that the following line integral is path-independent

$$ \int_C (\ln y) e^{-x} dx - \dfrac{e^{-x}}{y}dy + zdz $$

Answer 1

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.

Answer 2

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: $$ f(x, y, z) = -e^{-x}\ln y + \frac{z^2}{2} + C $$