Fundamental theorem of line integrals, what if the curl DOESN'T equal 0?

Question

I am getting really frustrated here and have had many arguments with people. Is it true or not, that if you compute the curl, and you get something other than 0, THEN you can NOT use the fundamental theorem? My understanding is that, logically, conservative F implies curl=0, therefore, curl=/=0 implies F not conservative, (if not q then not p) therefore you can not use the theorem. Thank you to anyone who clarifies this.

Answer 1

Your logic is mostly correct, but with a caveat.

If $\vec{F}$ is conservative, then there exists a scalar function $f$ such that $\vec{F} = \nabla f$ by definition. Then $\nabla \times \vec{F} = \nabla \times (\nabla f) = \vec{0}$ by the symmetry of second derivatives. However, note that this requires the mixed partials of $f$ to be continuous on the domain of interest! It is possible to construct pathological vector fields that are conservative ($\vec{F} = \nabla f$), but for which the curl is non-zero or even non-existent.

For an explicit example, consider the scalar function

$$f(x,y) = \begin{cases} \frac{x y (x^2 - y^2)}{x^2 + y^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0)\end{cases}.$$

It can be seen that the vector field $\vec{F} = (P,Q) = \nabla f$ exists and is continuous everywhere. However, $Q_x(0,0) \neq P_y(0,0)$, i.e. the curl of this vector field is not identically zero.

As such, your proposition that $\vec{F}=\nabla f$ $\iff$ $\vec{F}$ conservative $\implies \nabla \times \vec{F} = \vec{0}$ isn't strictly true, but can be amended by requiring $f$ to be $C^2$ (i.e., second partial derivatives exist and are continuous).

You may also be interested in the following theorem:

If $\nabla \times \vec{F} = \vec{0}$ and the domain of $\vec{F}$ is simply connected, then $\vec{F}$ is conservative.

Note that this theorem implicitly assumes that the curl of $\vec{F}$ exists, i.e., the component functions can be differentiated. Taking the contrapositive of this theorem gives that

If $\vec{F}$ is not conservative, then either $\nabla \times \vec{F} \neq \vec{0}$ or the domain is not simply connected.

Wikipedia gives an example that highlights the importance of the domain being simply connected.