Conservative Fields, Potentials and Path Independence

$\mathbf{F}=(-y,x)$ — circulating, path dependent.

Cross-partials $-1$ and $1$: unequal, so no potential. From $(0,0)$ to $(1,1)$ the diagonal gives $0$, the route through $(1,0)$ gives $1$, the route through $(0,1)$ gives $-1$. Around the unit square counterclockwise, $2$.

The swirl $\partial Q/\partial x-\partial P/\partial y=2$ is constant and positive, matching a uniform counterclockwise rotation.

$\mathbf{F}=(2xy,x^2)$ — a gradient, path independent.

Cross-partials both $2x$: equal, and the plane is simply connected, so a potential exists. Integrating $P$ in $x$ gives $x^{2}y+g(y)$; matching $f_y$ against $Q=x^2$ forces $g^{\prime }=0$. The potential is $f=x^{2}y$, verified by $f_x=2xy$ and $f_y=x^2$.

All three paths from $(0,0)$ to $(1,1)$ give $1$, equal to $f(1,1)-f(0,0)$. Every closed path gives $0$.

$\mathbf{F}=({\displaystyle \frac{-y}{x^2+y^2}},{\displaystyle \frac{x}{x^2+y^2}})$ — passes the test, not conservative.

Both cross-partials equal ${\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}$ at every point of the domain. The plane minus the origin. Yet on the unit circle $\mathbf{r}(t)=(\cos t,\sin t)$ the integrand is

so the integral is ${\int }_0^{2\pi }1dt=2\pi$. A conservative field gives $0$ on every closed path, so this one is not conservative, despite passing the test everywhere it is defined.

On a simply connected subregion avoiding the origin, such as the right half-plane, the same formula is conservative and does have a potential there. Conservativeness depends on the region as well as the formula.

$\mathbf{F}=(-y,x)$ used as an area meter.

The swirl is the constant $2$, so half the closed integral gives area. On the triangle $(0,0)$, $(4,0)$, $(0,3)$: the side along the $x$-axis and the side along the $y$-axis each contribute nothing, and the hypotenuse parametrised as $(4-4t,3t)$ gives integrand $(4-4t)(3)-(3t)(-4)=12$, hence ${\int }_0^{1}12dt=12$. Half of that is $6$, matching $\frac{1}{2}\times 4\times 3$.

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The first two are the basic dichotomy: circulation means the route matters, a gradient means it does not, and one pair of derivatives distinguishes them. The third shows the test's converse failing on a region with a hole, which is the only way the implication can fail. The fourth shows the same circulating field from the first example put to work, its uniform swirl is a nuisance for path independence and exactly what makes it an area meter.

Two fields, same endpoints, same three paths.

The first row decides everything below it, at the cost of two derivatives. When the cross-partials disagree, no amount of integration will produce a consistent answer; when they agree on a simply connected region, one subtraction replaces every integral in the column.

A test that fails against one that passes and proves nothing.

For $(-y,x)$ the cross-partials are $-1$ and $1$: unequal, and the field is definitively not conservative. For the punctured-plane field they are equal everywhere in the domain, and the field is still not conservative, as its $2\pi$ loop integral shows.

So a failed test is conclusive and a passed test is not. The asymmetry is the practical content of the theorem: use the test to rule out, and use it to rule in only after checking the region.

The same formula on two regions.

$(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$ on the punctured plane: not conservative, loop integral $2\pi$.

The identical formula on the right half-plane : conservative, with a potential, every loop integral $0$.

Nothing about the formula changed. Conservativeness is a property of a field together with a region, and reporting it without naming the region is reporting half a claim.

A closed path with a singularity inside against one without.

Green's theorem requires continuous partials on the enclosed region, not merely on the curve. The punctured-plane field is perfectly well behaved on the unit circle itself and has a singularity at the centre, so the theorem does not apply, which is how a field with zero swirl everywhere it is defined can have a nonzero loop integral. A field with no enclosed singularity cannot do this.

Two orientations of one curve.

Traversing the unit square counterclockwise gives $2$ for $(-y,x)$; clockwise gives $-2$. The curve is the same set of points, and the integral is a property of the oriented curve. A stated value without a stated direction is ambiguous by a sign.

An interval against a route.

${\int }_a^{b}fdx$ has one path from $a$ to $b$, so path dependence cannot arise and the question never comes up. Adding a second dimension creates the choice, and with it a genuine possibility that the answer differs. The fundamental theorem for line integrals is the statement that gradient fields behave like the one-dimensional case, and the point of this unit is that most fields do not.

That $\partial P/\partial y=\partial Q/\partial x$ is what it means for a vector field to be conservative, so checking it settles the question wherever the field is defined. The test is necessary and becomes sufficient only on a simply connected domain. One with no holes. The standard counterexample is $\mathbf{F}=({\displaystyle \frac{-y}{x^2+y^2}},{\displaystyle \frac{x}{x^2+y^2}})$ on the plane with the origin removed. Its cross-partials agree at every point where it is defined: both $\partial P/\partial y$ and $\partial Q/\partial x$ equal ${\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}$. Yet integrating around the unit circle $(\cos t,\sin t)$ gives ${\int }_0^{2\pi }({\sin }^{2}t+{\cos }^{2}t)dt={\int }_0^{2\pi }1dt=2\pi$, not zero, and a conservative field has zero integral around every closed path, since the endpoints coincide. So the field is not conservative on the punctured plane, despite passing the test everywhere. What the hole costs is the ability to shrink a loop to a point without leaving the domain, which is exactly what the proof of sufficiency requires. On any simply connected subregion avoiding the origin, a half-plane, say, the same field is conservative and does have a potential there. The lesson is that conservativeness is a property of a field together with a region, not of a formula.