Agreement of two path integrals does not imply conservativeness

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.