Prove that $\hat{\mathbf{F}}$ is a Conservative Field

Question

If $\hat{\mathbf{F}}$ is a vectorial field defined by:

$\hat{\mathbf{F}}(x, y) = \frac{1}{y} \hat{\mathbf{i}} - \frac{x}{y²} \hat{\mathbf{j}} $

Prove that $\hat{\mathbf{F}}$ is conservative.

My issue with this problem is the fact that the domain of $\hat{\mathbf{F}}$ doesn't appear to be path-connected and, therefore, not simply connected. Because of this detail, I can't determine this statement by showing that the curl equals zero. Can somebody help me, please?

Answer 1

A conservative vector field is by definition the gradient of a function $f$.

To find the function from the gradient, $F$xy=$F$yx. Which are both equal to $\frac{-1}{y^2}$. This shows that our field is the gradient of a function (the function can be found by integrating each component of the field).

Note that the only property that a field is a conservative is that it the gradient of a function. This means that you do not have to worry about the domain or continuity.

Answer 2

Well, it's the gradient of $\displaystyle f(x,y)=\frac{x}{y}$.

The function and its gradient do not exist on $y=0$, but the only requirement for conservative is that it satisfies being the gradient of another function.