Path-independence example in three dimensions

In three dimensions, we can write a vector field as

Assume that F is defined everywhere in R³, except for possibly a finite number of points. Then, F is path-independent if and only if the curl of F is zero, i.e., if

$$\nabla \left(\vphantom{\frac{\partial F_3}{\partial y} - \frac{\partial ... ... x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} }\right. {\frac{{\partial F_3}}{{\partial y}}} {\frac{{\partial F_2}}{{\partial z}}} {\frac{{\partial F_1}}{{\partial z}}} {\frac{{\partial F_3}}{{\partial x}}} {\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial F_1}}{{\partial y}}} \left.\vphantom{\frac{\partial F_3}{\partial y} - \frac{\partial ... ... x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} }\right)$$

We can write this condition as

$${\frac{{\partial F_1}}{{\partial y}}} {\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial F_1}}{{\partial z}}} {\frac{{\partial F_3}}{{\partial x}}} {\frac{{\partial F_2}}{{\partial z}}} {\frac{{\partial F_3}}{{\partial y}}}$$

It's the same thing.

Here are two examples.

Example 1: $\int_{C}^{}$(x² - ze^y)dx + (y³ - xze^y)dy + (z⁴ - xe^y)dz.
Is the integral path-independent?

This corresponds to F(x, y, z) = (x² - ze^y, y³ - xze^y, z⁴ - xe^y). The vector field F defined on R³. So we can use the above condition.

$${\frac{{\partial F_1}}{{\partial y}}} {\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial F_1}}{{\partial z}}} {\frac{{\partial F_3}}{{\partial x}}}$$

$${\frac{{\partial F_2}}{{\partial z}}} {\frac{{\partial F_3}}{{\partial y}}}$$

The curl is zero, so the integral is path-independent.

Example 2: $\int_{C}^{}$(x² - xe^y)dx + (y³ - xze^y)dy + (z⁴ - xe^y)dz.
Is the integral path-independent?

This corresponds to F(x, y, z) = (x² - xe^y, y³ - xze^y, z⁴ - xe^y).

$${\frac{{\partial F_1}}{{\partial y}}} \neq {\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial F_1}}{{\partial z}}} \neq {\frac{{\partial F_3}}{{\partial x}}}$$

$${\frac{{\partial F_2}}{{\partial z}}} {\frac{{\partial F_3}}{{\partial y}}}$$

The curl is not zero, so the integral is path-dependent.

Note: since path-independence doesn't depend on $${\frac{{\partial F_1}}{{\partial x}}}$$, $${\frac{{\partial F_2}}{{\partial y}}}$$, $${\frac{{\partial F_z}}{{\partial z}}}$$, you can easily conclude from Example 1 that

is also path-independent.