Determining from its graph whether a vector field is conservative

One cannot always make this determination visually, but one can apply some ad hoc tests that guarantee one or the other. For example (here we assume the given vector field $\bf F$ is $C^1$):

• If one can find, for example, an smooth, oriented loop $\gamma$ such that at every point of $\gamma$ the unit tangent vector makes a zero, acute, or right angle, and at least at one point where the vector field is nonzero makes a zero or acute angle with the vector field, then the vector field cannot be conservative: This condition guarantees that $\int_{\gamma} {\bf F} \cdot d{\bf s} > 0$, but for a conservative vector field that integral is zero for all loops $\gamma$.
• If the vector field is invariant under rotation about some point, then it is conservative: By translating we may take the distinguished point to be the origin, and by construction $\bf F$ has potential $f\left(\sqrt{x^2 + y^2}\right)$, where $f(r) := \int_a^r {\bf F}(x, 0) \cdot d{\bf x}$, where $d{\bf x}$ is the infinitesimal vector pointing in the positive $x$-direction and $(a, 0)$ is some point in the domain of $\bf F$ on the positive $x$--half-axis. (Strictly speaking, this construction assumes that the domain of $\bf F$ is connected.)

If it is conservative then $\vec F = \nabla \phi $ for some potential $\phi$, using a very useful identity, $\nabla \times \vec F = \nabla \times \nabla\phi = 0$, this mean that if the field is conservative, it won't curl around any point it will be straight lines, something that looks like electric or gravitationnal field !