# Surface element for a cylinder, how?

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: $\frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} \times \frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} = r\cos\phi{\mathbf{\hat{e}}_x} +r\sin\phi{\mathbf{\hat{e}}_y}=r\mathbf{\hat{r}}$ $$d\mathbf{S}_{cylinder}=\frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} \times \frac{\partial\mathbf{r}(\phi,z)}{\partial\phi} \, d\phi \, dz = r\mathbf{\hat{r}} \, d\phi \, dz$$

Position your cylinder such that it is bounded below by $z = 0$ and above by $z = l>0$. To parametrize the top and bottom caps use;

$$G(r, \theta) = (r \cos \theta, r \sin \theta,i)$$

where $i = 0,l$. Let the cylinder have radius $R>0$. Your domain will be;

$$D = \{(r, \theta): 0 \leq r \leq R, 0 \leq \theta \leq 2 \pi\}$$

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.