Path independence

A line integral is path independent when its value depends only on the two endpoints. On a connected domain, this holds for the line integral of every exact differential.

For a vector-field line integral, this condition is equivalent to the field being a conservative vector field.

Path independence requires the same line integral for every admissible path between fixed endpoints, not merely a few selected paths. For example, the smooth field $\mathbf{F}=(y-z,0,0)$ vanishes on the entire plane $y=z$. Every path in that plane from $(0,0,0)$ to $(1,1,1)$ therefore has zero integral. Yet the path through $(0,1,0)$ and $(1,1,0)$ has integral one: only the horizontal middle segment contributes. The curl is $(0,-1,-1)$, confirming that the field is not conservative. Special paths can hide a nonconservative component.

Equality of line integrals on two selected paths is a necessary test for a conservative vector field, but it is not sufficient. The paths may both miss the component responsible for nonzero curl.

For an explicit example, take $W(x,y,z)=(z-y,x-z,y-x{)}^T$. Its curl is the nonzero constant $2(1,1,1{)}^T$, so it is not a gradient field on ${\mathbb{R}}^3$. Nevertheless, on the plane $x=y$, the field equals $(z-x,x-z,0{)}^T$. Every tangent vector to that plane has equal first and second components, giving zero dot product with $W$. Thus every line integral of $W$ along a path confined to that plane is zero, regardless of endpoints or the path's shape.

In particular, a broken straight path and a genuinely curved path in that plane can have matching line integrals while revealing nothing about global path independence. To establish a conservative vector field, construct a vector-field potential, or verify the appropriate global criterion such as zero curl on a simply connected domain.