# Homework Help: Determination of a Potential

1. Nov 5, 2007

### SlideMan

1. The problem statement, all variables and given/known data

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 $$F \cdot dr = 0$$

3. 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!

Last edited: Nov 5, 2007
2. Nov 5, 2007

### Antineutron

I think you got this line wrong

3. Nov 5, 2007

### SlideMan

^ OK...what am I missing?

Working backwards...

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

4. Nov 5, 2007

### 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)

Last edited: Nov 6, 2007
5. Nov 6, 2007

### SlideMan

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