Q:

Is $y^2 i − z^2 j + x^2 k$ conservative vector field?

I want to prove if $y^2 i − z^2 j + x^2k$ is a  conservative vector field. Is it sufficient to say that it is 'simply connected', and show that the curl is $0$? Or do I need to prove it by attempting to find a potential function?

A:

Our vector field is
$$\mathbf{F}(x,y,z)=y^2\hat{\mathbf{i}}-z^2\hat{\mathbf{j}}+x^2\hat{\mathbf{k}}$$
It is conservative if $\nabla \times\mathbf{F}=0~~\forall (x,y,z)\in\mathbb{R}^3$.
So you need to check if
$$\nabla \times \mathbf{F}=\det\left(\begin{bmatrix}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\
\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\
y^{2} & -z^{2} & x^{2}
\end{bmatrix}\right)=0$$
Which is in fact not the case.

