# Curl and Vector Fields

I am having real difficulty knowing how to approach this question, so any help or pointers would be appreciated.

Consider the vector field:

$$\vec{G} = -3xz^2\vec{i} + z^3\vec{k}$$

FInd a vector field $\vec{F}$, such that:

$$\vec{G} = \nabla \times \vec{F}$$

Hint: Look for a vector field in the form $\vec{F} = F\vec{j}$

I am assuming I am looking for a way to exploit the fact that the $\vec{j}$ term in the original vector field is zero?

Many thanks to anyone who can help.

• Approaching what question? – anon Nov 5 '14 at 1:20
• Sorry - I was editing the equations. The post should make sense now. – Proioxis Nov 5 '14 at 1:26
• Have you even tried using the hint? – anon Nov 5 '14 at 1:30
• Is $G$ stated correctly? It's not divergence-free and so cannot be written as the curl of another vector field. – Semiclassical Nov 5 '14 at 1:34
• Sorry - I realised I missed out a z term (I was trying to get to grips with the notation). I have corrected it now. – Proioxis Nov 5 '14 at 1:37

Definition of the curl is $$-3xz^2\mathbf{i} + z^3\mathbf{k} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ F_x & F_y & F_z \end{vmatrix}$$ Then you evaluate the determinant.

• Thank you - this gives me a starting point. I think I can see how to proceed. – Proioxis Nov 5 '14 at 1:43
• Not a problem. You can now equate the i, j, and k pieces and work from there. – dustin Nov 5 '14 at 1:44