Conservative Vector Fields and Associated Potential

[sandy.bridge]

Homework Statement

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)]

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?

 

[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 $$\frac{\partial c(z)}{\partial z} = c'(z).$$

 

[SammyS]

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

Homework Statement

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)]

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.

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\ .

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?

 

[sandy.bridge]

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