= 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.
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