Find the general solution to Laplace's equation for spherical symmetry (everything can only depend on $r$, the radius), cylindrical symmetry (everything can only depend on $s$, the radius), and planar symmetry (everything can only depend on $z$).
My attempt:
For spherical symmetry, Laplace's equation is $\nabla^2V=\frac{1}{r}\frac{\partial}{\partial r}(r^2\frac{\partial V}{\partial r})+\frac{1}{r^2sin\theta}\frac{\partial}{\partial\theta}(sin\theta\frac{\partial V}{\partial\theta})+\frac{1}{r^2sin^2\theta}\frac{\partial^2V}{\partial\phi^2}=0.$ Since we have spherical symmetry, $V$ depends on $r$ only, so $\nabla^2V=\frac{1}{r}\frac{\partial}{\partial r}(r^2\frac{\partial V}{\partial r})=0\implies\frac{\partial}{\partial r}(r^2\frac{\partial V}{\partial r})=0\implies r^2\frac{\partial V}{\partial r}=C_1\implies V(r)=\frac{-C_1}{r}+C_2$, where $C_1$ and $C_2$ are arbitrary constants.
Repeating the same arguments for cylindrical symmetry, with cylindrical version of Laplace's equation, $\nabla^2V=\frac{1}{s}\frac{\partial}{\partial s}(s\frac{\partial V}{\partial s})+\frac{1}{s^2}\frac{\partial^2V}{\partial\phi^2}+\frac{\partial^2V}{\partial z^2}=0.$ I got $V(s)=ln(\frac{C_1}{s})+C_2$, where $s$ is the radius and $C_1,C_2$ are arbitrary constants.
And for planar symmetry, I got $\nabla^2V=\frac{\partial^2V}{\partial z^2}=0\implies\frac{\partial V}{\partial z}=C_1\implies V(z)=C_1z+C_2$, where $C_1,C_2$ are arbitrary constants.
Does this look correct?