The fundamental theorem for line integrals turns work into endpoint values, but only for gradient fields. How can you tell whether a field is a gradient field without finding the potential? Partial derivatives give a quick test, with one important condition on the shape of the region.
Key idea
Let $\mathbf{F} = \langle P, Q \rangle$ have continuous partial derivatives on an open, simply connected region $D$ (connected, with no holes). Then
$$\mathbf{F} \text{ is conservative on } D \iff P_y = Q_x \text{ everywhere in } D.$$
In space: $\mathbf{F} = \langle P, Q, R \rangle$ on a simply connected region (such as all of space) is conservative if and only if $\nabla \times \mathbf{F} = \mathbf{0}$.
Why it works
Necessary: if $\mathbf{F} = \nabla f$, then $P_y = f_{xy} = f_{yx} = Q_x$ (Clairaut).
Sufficient, when there are no holes: take any simple closed curve $C$ in $D$. Because $D$ has no holes, the whole region $R$ inside $C$ lies in $D$, so Green's theorem gives $\oint_C \mathbf{F} \cdot d\mathbf{r} = \pm\iint_R (Q_x - P_y)\,dA = 0$. Zero around every loop means path independence, and a path-independent field is a gradient.
Where holes break it: if the inside of $C$ contains a point where $\mathbf{F}$ is not defined, Green's theorem cannot be applied to that region, and the circulation need not be ${}0$.
Sufficient, when there are no holes: take any simple closed curve $C$ in $D$. Because $D$ has no holes, the whole region $R$ inside $C$ lies in $D$, so Green's theorem gives $\oint_C \mathbf{F} \cdot d\mathbf{r} = \pm\iint_R (Q_x - P_y)\,dA = 0$. Zero around every loop means path independence, and a path-independent field is a gradient.
Where holes break it: if the inside of $C$ contains a point where $\mathbf{F}$ is not defined, Green's theorem cannot be applied to that region, and the circulation need not be ${}0$.
Worked examples
Worked example: A conservative field
Is $\mathbf{F} = \langle 2xy + y^2, x^2 + 2xy \rangle$ conservative on the plane?
- $P_y = 2x + 2y$ and $Q_x = 2x + 2y$.
- They agree everywhere, and the plane has no holes, so yes. (A potential is $x^2y + xy^2$.)
Answer: yes
Worked example: A field that fails
Is $\mathbf{F} = \langle y, 2x \rangle$ conservative? Confirm with a loop.
- $P_y = 1$, $Q_x = 2$: not equal, so no.
- Around the unit circle: $$\int_0^{2\pi} (-\sin^2 t + 2\cos^2 t)\,dt = -\pi + 2\pi = \pi \ne 0.$$
Answer: no; circulation $\pi$
Worked example: In space
Which of $\mathbf{F} = \langle yz, xz, xy \rangle$ and $\mathbf{G} = \langle y, x, x \rangle$ is conservative?
- $\nabla \times \mathbf{F} = \langle x - x, y - y, z - z \rangle = \mathbf{0}$: conservative ($\mathbf{F} = \nabla(xyz)$).
- $$\nabla \times \mathbf{G} = \langle 0 - 0, 0 - 1, 1 - 1 \rangle = \langle 0, -1, 0 \rangle \ne \mathbf{0}\text{:}$$ not conservative.
Answer: $\mathbf{F}$ only
Common mistake
Forgetting the region. The field $\left\langle \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right\rangle$ passes the test $P_y = Q_x$ at every point where it is defined, yet its circulation around the unit circle is ${}2\pi$, so it is not conservative on the plane without the origin. The test only guarantees a potential on a region with no holes. On the right half-plane $x > 0$, which has no hole, it is conservative: there it is $\nabla \arctan(y/x)$.