# Verify the divergence theorem for a cylinder

1. May 5, 2014

### Feodalherren

1. The problem statement, all variables and given/known data

Verify the divergence theorem if $\textbf{F} = <1-x^{2}, -y^{2}, z >$ for a solid cylinder of radius 1 that lies between the planes z=0 and z=2.

2. Relevant equations
Divergence theorem

3. The attempt at a solution

I can do the triple integral part no problem. Where I run into issues is the surface integral part.

Parametrizing a cylinder is done by <rcost,rsint,z>, correct?

So looking at the top part I want it oriented in the positive k direction to get flow OUT of the cylinder - hence S: <cost,sint,2> because it lies in the plane z=2.
Similarly for the bottom S: <cost,-sint,0>

Now for the side, the side never has any k components so S: <cost,sint,0>

Now lets look at the top again. If I take the derivatives with repect to Z and try to cross them I end up with 0 net flow in every direction, it does not agree with the triple integral and isn't correct.
I have a feeling that I'm not parametrizing my cylinder correct, I remember it being a special case.

Another thing that I tried was parametrizing the top as <cost,sint,z>
Then the dS vector becomes
<cost,sint,0>
Now it's showing no component in the z direction - clearly it should have a component in the z direction?!

2. May 5, 2014

### Dick

Clearly dS along the top should be in the +z direction. It should be <0,0,1>dA. How are you getting that it isn't? The coordinates along the top are <rcos(t), rsin(t), 2> take the r derivative and the t derivative and cross them.

3. May 5, 2014

### Feodalherren

Ahh I see I thought it was
<1cost,1sint,z> I didn't realize that you allowed R to vary as it is fixed in the problem.

4. May 6, 2014

### Dick

The outer radius of the cylinder is fixed at 1. The coordinate r isn't fixed along the top surface. It ranges from 0 to 1. Only z is fixed. Along the sides r is fixed and z isn't.

5. May 6, 2014

### HallsofIvy

Staff Emeritus
No, S can have any value of z between 0 and 2: S: <cos(t), sin(t), z>. A two dimensional surface always requires two parameters! The NORMAL VECTOR has no k component:
writing $\vec{r}= cos(t)\vec{i}+ sin(t)\vec{j}+ z\vec{k}$, we have $\vec{r}_t= -sin(t)\vec{i}+ sin(t)\vec{j}$ and $\vec{r}_z= \vec{k}$ and their cross product is $cos(t)\vec{i}+ sin(t)\vec{j}$

The top is NOT "<cos(t), sin(t), z>". That has r fixed at 1 and both t and z varying so is the cylindrical side. The top has both r and t varying and z fixed at 2: <r cos(t), r sin(t), 2> (the bottom is <r cos(t), r(sin t), 0>).
The derivative with respect to r, for the top, is <cos(t), sin(t), 0> and the derivative with respect to t <-r sin(t), cos(t), 0>. Their cross product gives $\vec{n}dS= r \vec{k} drdt$. Similarly, $\vec{n}dS$ is $-r\vec{k}drdt$ for the bottom.

Last edited: May 6, 2014