Visualizing the exactness of two-dimensional differentials

Question

One of the very first notions one encounters when introduced to total differentials $\text{d}f$ is the notion of exactness, that is the property of $f(x,y)$ that in $$\text{d}f=\frac{\partial f(x,y)}{\partial x}\text{d}x + \frac{\partial f(x,y)}{\partial y}\text{d}y$$ we have $$ \tag{$\spadesuit$} \frac{\partial}{\partial y}\left(\frac{\partial f(x,y)}{\partial x}\right)=\frac{\partial}{\partial x}\left(\frac{\partial f(x,y)}{\partial y}\right). $$ Now we can interpret in the above $\text{d}f$ as a scalar product of the gradient $\vec{\nabla}f$ with $\text{d}\vec{r}=(\text{d}x,\text{d}y)$ and so the integrability condition $(\spadesuit)$ reads as an equality of the directional derivatives of the gradient along the $x$ and $y$ axes.

This means if we were to plot the gradient of $f$ as a vector field, it should be symmetric under reflections along the diagonal $x=y$. These reflections are displayed by red lines in the following.

As an example, take $\text{d}f=\cos(x)\sin(y)\text{d}x-\sin(x)\cos(y)\text{d}y$, which is not exact:

Compare with $\text{d}f=\sin(x)\cos(y)\text{d}x+\cos(x)\sin(y)\text{d}y$, which is exact:

Now, assuming the above interpretation is correct, is there a nice way to see in the vector field plot the following two things:

• Whether $\oint_C \text{d}f$ will be $=0$ in the exact and $\neq0$ in the inexact case? So, is it possible to gain a hint as to the exactness of a given $\text{d}f$ from the visualization alone?${}^{[1]}$
• How an integrating factor $\mu(x,y)$ should look like, so that ($\spadesuit$) holds for $\mu\cdot \text{d}f$ if it doesn't for $\text{d}f$?

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_{${}^{[1]}:$ Apart from those cases where $\oint_C \text{d}f \neq 0$ even for exact differentials, where because of terms like $\frac{1}{x^2+y^2}$ we have to exclude certain points, which lead to singularities that change the closed loop integral when integrated around them.}

Answer 1

Your statement

Now we can interpret in the above $\text{d}f$ as a scalar product of the gradient $\vec{\nabla}f$ with $\text{d}\vec{r}=(\text{d}x,\text{d}y)$ and so the integrability condition $(\spadesuit)$ reads as an equality of the directional derivatives of the gradient along the $x$ and $y$ axes.

and the associated picture with the red diagonal line, suggests that the exactness of a differential is dependent only on its values at or near the diagonal, which is completely untrue(*). I think you need to rethink that interpretation.

(*) Brief explanation: suppose that $\omega$ is everywhere inexact. Multiply $\omega$ by $f$, where $f$ is zero on a unit-width strip along the diagonal $x = y$, and $1$ outside a two-unit-wide strip along the diagonal. Then $f\omega$ will have zero gradient along the diagonal, hence appear "exact" by your criterion, but outside that wide strip, $f\omega = \omega$, and hence is inexact.