# Finding a Potential for a field

• Nov 22nd 2009, 10:53 AM
TheUnfocusedOne
Finding a Potential for a field
So im trying to find the potential for the following gradient

$\displaystyle F = <2xy,+3, x^2-3z, -3y>$

I've gotten a bunch of different answers, but none of them have been right

Ive tried the method of integrating one component with its respective variable, (as in intergrating 2xy+3 with respect to x) and then taking the partial with respect to another variable and setting it equal to another term in the gradient, eventually solving for the new functions g(y,z) then h(z).

$\displaystyle \phi = x^2y-3yz$

the online software is telling me this is wrong, and im not sure if im just punching it in wrong or if im doing it wrong

thanks guys
• Nov 22nd 2009, 10:48 PM
simplependulum
Quote:

Originally Posted by TheUnfocusedOne
$\displaystyle F = <2xy,+3, x^2-3z, -3y>$

thanks guys

Is the field $\displaystyle F = <2xy +3, x^2-3z, -3y>$ or

$\displaystyle F = <2xy, x^2-3z, -3y>$ ??

If the $\displaystyle +3$ vanishes , i think it may go to your answer

From $\displaystyle \frac{\partial \phi}{\partial x} = 2xy$

$\displaystyle \phi = x^2 y + h(y,z)$

$\displaystyle \frac{\partial \phi}{\partial y } = x^2 + h_y(y,z) = x^2 - 3z$

$\displaystyle -3z = h_y(y,z)$

Also

$\displaystyle \frac{\partial \phi}{\partial z} = h_z(x,y)= -3y$

Finally , i obtain $\displaystyle h(y,z) = -3yz + C$

If you omit the constant $\displaystyle \phi = x^2y - 3yz$

But if $\displaystyle +3$ is present ,

$\displaystyle \phi = x^2y - 3yz + 3x$ not sure but you still can take a look !

ps : Is it a 4-D field !?
• Nov 23rd 2009, 03:12 AM
HallsofIvy
Quote:

Originally Posted by TheUnfocusedOne
So im trying to find the potential for the following gradient

$\displaystyle F = <2xy,+3, x^2-3z, -3y>$

I've gotten a bunch of different answers, but none of them have been right

Ive tried the method of integrating one component with its respective variable, (as in intergrating 2xy+3 with respect to x) and then taking the partial with respect to another variable and setting it equal to another term in the gradient, eventually solving for the new functions g(y,z) then h(z).

$\displaystyle \phi = x^2y-3yz$
That is correct except for the missing constant: $\displaystyle \phi(x,y,z)= x^2- 3yz+ C$ for any constant C. Perhaps your online software is expecting that.