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Let $\mathbf{F} (x, y) = \frac{- y}{x^2 + y^2} \mathrm{d} x + \frac{x}{x^2 + y^2} \mathrm{d} y$ be defined on the region $R = \mathbb{R}^2 \setminus \{ (0, 0) \}$. The vector field is closed but not exact. The way I know to show this is by taking $\oint_C \mathrm{F} \cdot \mathrm{d} \mathbf{r}$ along a circle $C$ around the origin, which would come to $0$ were the field conservative, but doesn't.

I have in mind a different tactic for demonstrating that $\mathbf{F}$ isn't conservative, without using contour integrals, and I want to know whether my method will work, and how to fill in the details. My idea is to try to find the function $\phi (x, y)$ for which $\mathbf{F} = \nabla \phi$ and hope a contradiction falls out. Here's what I've got so far.

If $\phi$ fits the bill, then \begin{align*} \phi_x & = \frac{- y}{x^2 + y^2} \\ \Rightarrow \int \phi_x \mathrm{d} x & = \int \frac{- y}{x^2 + y^2} \mathrm{d} x \\ & = - y \int \frac{\mathrm{d} x}{x^2 + y^2} \\ & = - y \frac{1}{y} \arctan{\frac{x}{y}} + p(y) \\ & = - \arctan{\frac{x}{y}} + p(y) , \end{align*} where $p(y)$ is some differentiable function in $y$. But $\arctan\frac{x}{y}$ isn't defined on the $x$-axis, a contradiction.

Is this sufficient? Is there some nuance I'm missing that needs to be pointed out, or is this it?

Thanks!

EDIT: Someone pointed out that it's possible in principle that $p(y)$ could "make up" for $\arctan \frac{x}{y}$ being undefined on the $y$-axis, so I decided to continue. If $\phi(x, y) = - \arctan \frac{x}{y} + p(y)$, then \begin{align*} \phi_y & = \frac{\partial}{\partial y} \left( - \arctan \frac{x}{y} + p(y) \right) \\ = \frac{x}{x^2 + y^2} & = \frac{x}{x^2 + y^2} + p'(y) \\ \Rightarrow 0 & = p'(y) . \end{align*} This means that $\phi(x, y) = - \arctan \frac{x}{y} + p$, for some constant $p$, which has no continuous extension to all of $R$.

AJY
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    It's not sufficient, you can also write the function in the form $-\operatorname{arccot} \frac{y}{x} + q(y)$, and then the problem doesn't occur on the $x$-axis. You need to show that you can't glue the local primitives together to obtain a global primitive. And that's not easier than computing the integral over the unit circle. – Daniel Fischer Jun 05 '20 at 18:00

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$p(y)$ may be some other function that is undefined on the $x$-axis, and it may sufficiently cancel the arctangent term you are focussing on so that the result has limits on the $x$-axis. (It doesn't in this case, but you would need to do the analysis on $\phi_y$ to know for sure.)

You might be better off using that the curl of a gradient of a scalar function is zero, so $$ \nabla \times\mathbf{F} = \nabla \times (\nabla \phi) = 0 $$ and taking the curl of the given $\mathbf{F}$ is pretty easy. (The basic idea: the $xy$ and $yx$ second derivatives of $\phi$ are equal but appear with opposite signs.)

Eric Towers
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