I having problem how to find the differential surface element for a cylinder $x^2+y^2=r^2$ with height $l$.
The surface have three parts; top, cylinder and bottom.
I know how to parametrize and find the differential element for the cylinder, but not for the top and bottom.
(This is a problem in electrostatics. I know the top and botton cancel out, however I want to know the math behind).
The answers should be:
$$
d\mathbf{S}_{top}=\mathbf{\hat{z}}r'\,dr'\,d\phi, \quad r'\in[0,r], \quad \phi\in[0,2\pi] \quad \text{How? What is r'?}
$$
$$
d\mathbf{S}_{bottom}=-\mathbf{\hat{z}}r'\, dr' \, d\phi, \quad r'\in[0,r], \quad \phi\in[0,2\pi] \quad \text{Same here?}
$$
I know this:
$$
d\mathbf{S}_{cylinder}=\mathbf{\hat{r}}r\, d\phi \, dz, \quad z\in[-l/2,l/2], \quad \phi\in[0,2\pi]
$$
My solution for finding $d\mathbf{S}_{cylinder}$:
$$x=r\cos\phi, \quad y=r\cos\phi, \quad z=z
$$
$$
\mathbf{r}(\phi,z)=r\cos\phi{\mathbf{\hat{e}}_x} +r\sin\phi{\mathbf{\hat{e}}_y}+z{\mathbf{\hat{e}}_z}
$$
$$
\frac{\partial\mathbf{r}(\phi,z)}{\partial\phi}=-r\sin\phi{\mathbf{\hat{e}}_x} +r\cos\phi{\mathbf{\hat{e}}_y}
$$
$$
\frac{\partial\mathbf{r}(\phi,z)}{\partial z}={\mathbf{\hat{e}}_z}
$$
$$
\frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} \times \frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} =
r\cos\phi{\mathbf{\hat{e}}_x} +r\sin\phi{\mathbf{\hat{e}}_y}
$$
In cylindrical: $\frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} \times \frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} =
r\cos\phi{\mathbf{\hat{e}}_x} +r\sin\phi{\mathbf{\hat{e}}_y}=r\mathbf{\hat{r}}$
$$
d\mathbf{S}_{cylinder}=\frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} \times \frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} \, d\phi \, dz = r\mathbf{\hat{r}} \, d\phi \, dz
$$
Best Answer
Hint:
It seems that the $r'$ is the radial coordinate $\rho$ in cylindrical coordinates:
$$ x=\rho \cos\varphi \qquad y=\rho \sin \varphi \qquad z=z $$
In these system of coordinates the surface element in a surface of constant $z$ is $dS_z=\rho d\rho d\varphi$ ( see here)
Then, since the normal to the surface is directed outside, we have the results for $d\mathbf{S}_{top}$ (the outside normal is $\mathbf{\hat{z}}$) and $d\mathbf{S}_{bottom}$ (the outside normal is $-\mathbf{\hat{z}}$) in OP. And this shows because they cancel out in the calculus of the flux.