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

Question

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?

Answer 1

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.

Answer 2

This is a simple example where by trying to find a potencial function you would obtain the answer. An alternative could be the following- any open ball in the neighbourhood of $(0,0,0)$ is a convex set and then if it was a conservative field it would satisfy $\frac{\partial F_{i}}{\partial x_{j}}=\frac{\partial F_{j}}{\partial x_{i}} $ for $i,j=1,2,3$ which is not the case.