I am given the vectorfield $22\frac{x}{z}i+18\frac{y}{z}j+(1-\frac{11x^2+9y^2}{z^2})k$ and I want to find the potential function that doesn't contain any constants.
I'm assuming it's conservative as I haven't learned how to calculate curl yet. My problem arrises when I try to start out with $\frac{\partial f}{\partial x}$ has to satisfy $f = \int 22\frac{x}{z} = \frac{11x^2}{z}+h(y,z)$. Next step is to take the derivative with respect to $y$. This needs to satisfy $\frac{\partial f}{\partial y} = 18\frac{y}{z}$. But if I differentiate my new function with respect to $y$, I'm left off with only $h(y,z)$. I don't understand what to do from here. In all the other examples I've seen, it is simply deduced that $h(y,z)$ must be $0$ because because the first integral ends up satisfying the second condition.
Have I done a mistake considering my vectorfield?
You are not left with only $h(y,z)$ but with a partial derivative of it. Taking it from $$ f=11x^2/z+h(y,z) $$ and $$ \frac{\partial f}{\partial y}=18y/z=\frac{\partial h}{\partial y} $$ so $$ h=9y^2/z+g(z) $$ and $$ \therefore \, f=11x^2/z+9y^2/z+g(z) $$ finally $$ \frac{\partial f}{\partial z}=(1-\frac{11x^2+9y^2}{z^2})=-\frac{11x^2}{z^2}-\frac{9y^2}{z^2}+\frac{dg}{dz} $$ $$ \Rightarrow g=z+c $$ Putting it all together $$ f(x,y,z)=11x^2/z+9y^2/z+z+C $$ Note there is still a constant.