1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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;

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.