Determining whether a vector field is conservative For a vector field $\vec{F}(x,y,z) = \langle F_1(x,y,z), F_2(x,y,z), F_3(x,y,z) \rangle$ in $\mathbb{R}^3$, how can I use mixed second-order partial derivatives of each of the components to determine whether it is conservative? Which partial derivatives should I compare?
 A: You should check that
$$ \frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}, \\
\frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x}, \\
\frac{\partial F_2}{\partial z} = \frac{\partial F_3}{\partial y}. $$
An easy way to remember this is to call your variables $x^1,x^2,x^3$ (instead of $x,y,z$). Then you need to check that
$$ \frac{\partial F_i}{\partial x^j} = \frac{\partial F_j}{\partial x^i} $$
for all $i \neq j$. This works in all dimensions.
A: Assume the derivatives of the components of $\vec F$ are continuous functions.  If $\vec F = \left<P(x,y,z),Q(x,y,z),R(x,y,z)\right>$ were the gradient of a function $f(x,y,z)$, then 
$$
    \frac{\partial f}{\partial x} = P \text{,}\qquad
    \frac{\partial f}{\partial y} = Q \text{,}\qquad\text{and}\qquad
    \frac{\partial f}{\partial z} = R 
$$
Taking the derivative of both sides of the first equation with respect to $y$, and the second equation with respect to $x$, we get
$$
    \frac{\partial^2 f}{\partial y\,\partial x} = \frac{\partial P}{\partial y}
\qquad\text{and}\qquad
    \frac{\partial^2 f}{\partial x\,\partial y} = \frac{\partial Q}{\partial x}
$$
By Clairaut's theorem the left-hand sides of the two equations are equal.  Therefore
$$
    \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}
$$
Similarly, it can be shown that
$$
    \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}
\qquad\text{and}\qquad
    \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y}
$$
These are all necessary conditions for $\vec F$ to be conservative.  They can be written in a form that doesn't assume only three variables, or, in the case of three variables, in a vector form.  If these equations do not hold, $\vec F$ cannot be conservative.
The big question is when these conditions are sufficient as well.  The answer is that as long as the domain of $\vec F$ is connected and simply connected, they are sufficient.  This follows from Stokes's Theorem.
A: Hint: show that $$\nabla\times\vec F(x,y,z) = \vec 0$$
Then, it easily follows that $\vec F $ is conservative.
A: If $\nabla G = <G_x,G_y,G_z> \ = F = \ <F_1,F_2,F_3>$ for some $G(x,y,z)$, then $F$ is conservative.
$(F_1)_y = (G_x)_y = G_{xy} = G_{yx} = (G_y)_x = (F_2)_x$.
So, $\displaystyle \frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$ must hold.
The other two equations are similar, and given in other answers.
A: The criterion is 
$$
\DeclareMathOperator{curl}{curl}
\curl F = \epsilon_{ijk} e_i \partial_j F_k = 0
$$
where we sum over same indices, and $\epsilon_{ijk}$ is the sign of the permutation $(ijk)$ or zero if index values repeat.
