Multivariable calculus - scalar field

I don't know how to solve this problem.

Determine if $\mathbf{F}$ is or not the gradient of a scalar field. If it is find the corresponding potential function f.

$\mathbf{F}(x,y,z)= 3y^4 z^2\,\mathbf{i} + 4x^3 y^2\,\mathbf{j} - 3x^2 y^2\,\mathbf{k}$

Any help?

-
What have you tried? There's a standard method here. –  Ted Shifrin Jun 26 at 20:55
Do you have any ideas? –  Ice Boy Jun 26 at 21:04

2 Answers

Hint: $F$ is the gradient of some scalar field if and only if its curl is zero.

-
Before posting hints, maybe we should let the OP express his/her ideas :-) –  Ice Boy Jun 26 at 21:06
@skullpatrol my assumption, if absolutely no context is provided, is that OP wasn't able to make any progress on his/her own, so that the most appropriate course of action is the smallest possible nudge in the right direction –  Omnomnomnom Jun 26 at 21:09
@Omnomnomnom My assumption, if absolutely no context is provided, is that absolutely no context is provided. –  Pedro Tamaroff Jun 26 at 22:05

F has a potential if and only if the rotation (curl) of F is zero everywhere. The rotation is defined in 'classical' vector calculus as:

$rot(F(x,y,z)) = (\frac{dQ}{dz}-\frac{dR}{dy})i-(\frac{dP}{dz}-\frac{dR}{dx})j+(\frac{dP}{dy}-\frac{dQ}{dx})k$

where

$F(x,y,z) = (P, Q, R)$

All you have to do is to check if the rotation is zero. If it is not then there is no potential. Otherwise you have to solve an easy system of partial differential equations:

$\frac{\partial F}{\partial x} = P$

$\frac{\partial F}{\partial y} = Q$

$\frac{\partial F}{\partial z} = R$

-