# Conservative Vector Fields and Associated Potential

1. Jan 10, 2013

### sandy.bridge

1. The problem statement, all variables and given/known data
Having issues determining what I am doing wrong, so perhaps one of you can pin point it. I have the solution, and I am extremely close to the same result, however, I am nonetheless wrong.

Find the conservative vector fields potential.

$$\vec{F}(x, y, z)=[(2xy-z^2), 2yz+x^2), y^2-2zx)]$$

3. The attempt at a solution
$$\vartheta=\int(2xy-z^2)dx=x^2y-z^2x+C(y, z)$$
then we have
$$2yz+x^2=x^2+∂C(y, z)/∂y$$
therefore,
$$C(y, z)=zy^2+C(z)$$
It's at the following step that I mess something up.
$$y^2-2zx=y^2+∂C(z)/∂z$$
In the solutions however, they have,
$$y^2-2zx=y^2-2zx+∂C(z)/∂z$$
However, wouldn't the "-2zx" term go with ∂C(z)/∂z?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 10, 2013

### CAF123

Your scalar function $\phi = x^2y - z^2x + c(y,z)$ Then take the partial of this with respect to $z$ and equate with the $z$ component of $\underline{F}$. Did you mistakenly take the partial with respect to z of something else? You can write $$\frac{\partial c(z)}{\partial z} = c'(z).$$

3. Jan 10, 2013

### SammyS

Staff Emeritus
I'm betting that you're going to kick yourself for this !

So, at this point you have $\displaystyle \ \ \vartheta=x^2y-z^2x+C(y, z)=x^2y-z^2x+zy^2+C(z)\ .$

Now look at $\displaystyle \ \ ∂\vartheta/∂z\ .$

4. Jan 10, 2013

### sandy.bridge

Perfect! Not entirely sure how I didn't see that. Figured I should ask rather than stare at it for long. Thanks!