# Multivariable calculus - scalar field [on hold]

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?

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## put on hold as off-topic by Michael Albanese, Jack D'Aurizio, Jeel Shah, Avitus, Pedro TamaroffDec 16 at 22:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Michael Albanese, Jack D'Aurizio, Jeel Shah, Avitus, Pedro Tamaroff
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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

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

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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:

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

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

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

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