What is the Potential of a Given Function Independent of Path?

[SlideMan]

Homework Statement

I need to determine the potential of the following function:

F = [2x(y^3 - z^3), 3x^2y^2, -3x^2z^2]

The equation is given to be independent of path, and F \cdot dr = 0

The Attempt at a Solution

\frac{\partial f}{\partial x} = 2xy^3 - 2xz^3 \Rightarrow f(x,y,z) = x^2y^3 - x^2z^3 + g(y,z)

\frac{\partial f}{\partial y} = 3x^2y^2 = 3x^2y^2 + \frac{\partial g}{\partial y} \Rightarrow g(y,z) = h(z)

\frac{\partial f}{\partial z} = -3x^2z^2 = \frac{\partial h}{\partial z} \Rightarrow h(z) = -x^2z^3

So, f(x,y,z) = x^2y^3 - 2x^2z^3

This answer doesn't check out. Taking the partial of f with respect to x, y, and z does not yield the initial equation. What am I missing? Is there a better way to go about this?

The correct answer turns out to be x^2y^3 - x^2z^3, which is my initial equation for f without h(z).

Thanks!

 

[Antineutron]

1.

\frac{\partial f}{\partial z} = -3x^2z^2 = \frac{\partial h}{\partial z} \Rightarrow h(z) = -x^2z^3

I think you got this line wrong

 

[SlideMan]

^ OK...what am I missing?

Working backwards...

\frac{\partial}{\partial z}(-x^2z^3) = -3x^2z^2

 

[Antineutron]

\frac{\partial f}{\partial z} = -3x^2z^2 = -3x^2z^2 + \frac{\partial h}{\partial z} \Rightarrow h(z) = 0

Im having trouble with latex, but your partial f over partial z which is -3x^2z^2 = -3x^2z^2 + partial h over partial z. so, h(z) = 0=g(y,x)

 

[SlideMan]

^ Ahh...got it. I really shouldn't be doing this so late at night. :) Thanks!