Path-independence example in three dimensions

In three dimensions, we can write a vector field as

F(x,y,z) = (F1(x,y,z),F2(x,y,z),F3(x,y,z)).

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

∇× F = ( ) ∂F3- ∂F2- ∂F1- ∂F3- ∂F2- ∂F1- ∂y - ∂z , ∂z - ∂x , ∂x - ∂y = (0, 0, 0)
We can write this condition as
∂F1 ---- ∂y = ∂F2 ---- ∂x, ∂F1 ---- ∂z = ∂F3 ---- ∂x, and ∂F2 ---- ∂z = ∂F3 ---- ∂y.
It’s the same thing.

Here are two examples.

Example 1: ∫ C(x2 - zey)dx + (y3 - xzey)dy + (z4 - xey)dz.
Is the integral path-independent?

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

∂F1 ---- ∂y = ∂F2 ---- ∂x = -zey, ∂F1 ---- ∂z = ∂F3 ---- ∂x = -ey
∂F2- ∂z = ∂F3- ∂y = -xey
The curl is zero, so the integral is path-independent.

Example 2: ∫ C(x2 - xey)dx + (y3 - xzey)dy + (z4 - xey)dz.
Is the integral path-independent?

This corresponds to F(x,y,z) = (x2 - xey,y3 - xzey,z4 - xey).

∂F1- ∂y = -xey≠∂F2- ∂x = -zey, ∂F1- ∂z = 0≠∂F3- ∂x = -ey
∂F2 ---- ∂z = ∂F3 ---- ∂y = -xey
The curl is not zero, so the integral is path-dependent.

Note: since path-independence doesn’t depend on ∂F1- ∂x, ∂F2- ∂y, ∂Fz- ∂z, you can easily conclude from Example 1 that

F(x,y,z) = (x2 - zey + x1000,y3 - xzey - cos y372,z4 - xey + ez+99)
is also path-independent.