- #26

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@TSny I had a mistake in the power of y. Now I get:

[tex]

g(y,z) = \int 0 dy + h(z) = 0 + h(z) \\

\implies \frac{\partial \phi}{\partial z} = \frac{\partial h(z)}{\partial z} = -cz^3 \\

\implies h(z) = -c \int z^3 dz = \frac{-c}{4} z^4 + C \\

\implies \phi = -\frac{a}{2}x^2y^3 - \frac{c}{4} z^4 + C[/tex]

@haruspex Now trying to do the integral properly:

[tex]

\phi = - \int \vec{F} d\vec{r} = - (\frac{a}{2} x^2y^3 + \frac{a}{2} x^2y^3 + \frac{c}{4} z^4 + C) \\

= - ax^2y^3 - \frac{c}{4} z^4 + C [/tex]

Now that's only a little different from the other one, but I'm still searching for the mistake.

Thanks a lot to both of you for helping me.

Julien.

[tex]

g(y,z) = \int 0 dy + h(z) = 0 + h(z) \\

\implies \frac{\partial \phi}{\partial z} = \frac{\partial h(z)}{\partial z} = -cz^3 \\

\implies h(z) = -c \int z^3 dz = \frac{-c}{4} z^4 + C \\

\implies \phi = -\frac{a}{2}x^2y^3 - \frac{c}{4} z^4 + C[/tex]

@haruspex Now trying to do the integral properly:

[tex]

\phi = - \int \vec{F} d\vec{r} = - (\frac{a}{2} x^2y^3 + \frac{a}{2} x^2y^3 + \frac{c}{4} z^4 + C) \\

= - ax^2y^3 - \frac{c}{4} z^4 + C [/tex]

Now that's only a little different from the other one, but I'm still searching for the mistake.

Thanks a lot to both of you for helping me.

Julien.

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