Surface integral over a cylinder

• Jan 18th 2011, 08:44 AM
Faen
Surface integral over a cylinder
Here's a picture of the question:

http://img.photobucket.com/albums/v4.../question5.png

Here's the solution:

http://img.photobucket.com/albums/v4.../solution5.png

I can't really make complete sense of some things around this... Like how did the integral become 36pi based on what is written in the solution? Why is z equal to 0.. What is the double integral over surface S1 and why is it zero? How is the divergence 3 if z is zero? Also if i use stokes theorem and find the volum integral, it would equal to 3 times the volume of the sylinder since the divergence is 3. But it doesn't equal to 36 pi... So if somebody could enlighten me around these questions i'd really appreciate it :)
• Jan 18th 2011, 11:59 AM
alexmahone
Quote:

Originally Posted by Faen
Here's a picture of the question:

http://img.photobucket.com/albums/v4.../question5.png

Here's the solution:

http://img.photobucket.com/albums/v4.../solution5.png

I can't really make complete sense of some things around this... Like how did the integral become 36pi based on what is written in the solution? Why is z equal to 0.. What is the double integral over surface S1 and why is it zero? How is the divergence 3 if z is zero? Also if i use stokes theorem and find the volum integral, it would equal to 3 times the volume of the sylinder since the divergence is 3. But it doesn't equal to 36 pi... So if somebody could enlighten me around these questions i'd really appreciate it :)

$\displaystyle \vec{v}=\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\displaystyle div\ \vec{v}=\frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z}=3$

$\displaystyle V=\pi r^2h=\pi*2^2*(3-1)=8\pi$

Using the divergence theorem,

$\displaystyle \iint\limits_S \, \vec{v}(\vec{r})\,dS=\iiint\limits_V \, div\ \vec{v}\, dV=3*8\pi=24\pi$

In the divergence theorem, however, $\displaystyle S$ is a closed surface. So, we have to subtract $\displaystyle \iint\limits_{S_1} \, \vec{v}(\vec{r})\,dS$ where $\displaystyle S_1$ is the bottom surface. However this integral is zero because the two vectors in the dot product are perpendicular.

(Your solution erroneously assumes $\displaystyle z=0$ for the bottom surface, whereas actually $\displaystyle z=1$. So, the volume is erroneously calculated as $\displaystyle V=\pi*2^2*(3-0)=12\pi$ and so $\displaystyle \iint\limits_S \, \vec{v}(\vec{r})\,dS=3*12\pi=36\pi$.)
• Jan 18th 2011, 12:07 PM
Faen
Quote:

Originally Posted by alexmahone
$\displaystyle \vec{v}=\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\displaystyle div\ \vec{v}=\frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z}=3$

$\displaystyle V=\pi r^2h=\pi*2^2*(3-1)=8\pi$

Using the divergence theorem,

$\displaystyle \iint\limits_S \, \vec{v}(\vec{r})\,dS=\iiint\limits_V \, div\ \vec{v}\, dV=3*8\pi=24\pi$

(In the solution, the volume is erroneously calculated as $\displaystyle V=\pi*2^2*3=12\pi$ and so $\displaystyle \iint\limits_S \, \vec{v}(\vec{r})\,dS=3*12\pi=36\pi$

I see, thanks a lot :) By the way, how did you get that the divergence is 3? How can we know that the components are x, y and z.. Also in the solution z is set as 0, so that part confuses me a bit too.. Such that if z=0 then the derivative of z would be 0 and divergence would be 2?
• Jan 18th 2011, 12:20 PM
alexmahone
Quote:

Originally Posted by Faen
By the way, how did you get that the divergence is 3? How can we know that the components are x, y and z..

$\displaystyle \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$. This is the definition of $\displaystyle \vec{r}$.

Quote:

Also in the solution z is set as 0, so that part confuses me a bit too.. Such that if z=0 then the derivative of z would be 0 and divergence would be 2?
I have edited my post. Read the last paragraph.
• Jan 18th 2011, 12:54 PM
Faen
Ok now i understand it.. Thanks again for your help, it's greatly appreciated.