# Thread: Finding a Potential for a field

1. ## Finding a Potential for a field

So im trying to find the potential for the following gradient

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

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).

my answer has been

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

which then gives you back your gradient

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

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

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

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

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

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

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

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

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

Also

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

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

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

But if $+3$ is present ,

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

ps : Is it a 4-D field !?

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

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

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).

my answer has been

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

which then gives you back your gradient

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