Quickest way to determine if a vector field is conservative? Say I have some vector field given by $$\vec{F} (x,y,z)=(zy+\sin x)\hat \imath+(zx-2y)\hat\jmath+(yx-z)\hat k$$ and I need to verify that $\vec F$ is a conservative vector field. What would be the most convenient way to do this?
Indeed I managed to show that this is a vector field by simply finding an $f$ such that $\nabla f=\vec{F}$. But I'm not sure if there is a nicer/faster way of doing this.
Side question I found $$f(x, y, z) = xyz-y^2-\frac{z^2}{2}-\cos x,$$ so would I be correct in saying that any $f$ that shows $\vec{F}$ is conservative is of the form $$f(x, y, z) = xyz-y^2-\frac{z^2}{2}-\cos x+\varphi$$ for $\varphi \in \mathbb{R}$?
 A: A vector field $\bf G$ defined on all of $\Bbb R^3$ (or any simply connected subset thereof) is conservative iff its curl is zero $$\text{curl } {\bf G} = 0 ;$$ we call such a vector field irrotational. This is easier than finding an explicit potential $\varphi$ of $\bf G$ inasmuch as differentiation is easier than integration. (NB that simple connectedness of the domain of $\bf G$ really is essential here: It's not too hard to write down an irrotational vector field that is not the gradient of any function.)
The answer to your second question is yes: Given two potentials $g$ and $h$ for a vector field $\Bbb G$ on some open subset $U \subseteq \Bbb R^n$, we have
$$\nabla (h - g) = \nabla h - \nabla g = {\bf G} - {\bf G} = {\bf 0};$$
in components, this says that the partial derivatives of $h - g$ are $0$, and hence $h - g$ is constant on the connected components of $U$. In particular, if $U$ is connected, then for any potential $g$ of $\bf G$, every other potential of $\bf G$ can be written as
$$g(x, y, z) + c$$
for some constant $c$.
A: A vector field $\textbf{A}$ on a simply connected region is conservative if and only if $\nabla \times \textbf{A} = \textbf{0}$.
Check out https://en.wikipedia.org/wiki/Conservative_vector_field
https://en.wikipedia.org/wiki/Conservative_vector_field#Irrotational_vector_fields
