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.

  • 2
    $\begingroup$ Approaching what question? $\endgroup$ – anon Nov 5 '14 at 1:20
  • $\begingroup$ Sorry - I was editing the equations. The post should make sense now. $\endgroup$ – Proioxis Nov 5 '14 at 1:26
  • 2
    $\begingroup$ Have you even tried using the hint? $\endgroup$ – anon Nov 5 '14 at 1:30
  • $\begingroup$ Is $G$ stated correctly? It's not divergence-free and so cannot be written as the curl of another vector field. $\endgroup$ – Semiclassical Nov 5 '14 at 1:34
  • $\begingroup$ Sorry - I realised I missed out a z term (I was trying to get to grips with the notation). I have corrected it now. $\endgroup$ – 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.

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

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