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Surface integral (Flux) with cylinder and plane intersections

 
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Nov27-12, 11:07 AM   #35
 

Surface integral (Flux) with cylinder and plane intersections


Quote by Dick View Post
Assuming we are still on polar coordinates ## \vec{r}=(r cos \theta, r sin \theta, 0)## , ##|\vec{r}_r\times \vec{r}_\theta|=r##, ##dA= dr d\theta##. So ##|\vec{r}_r\times \vec{r}_\theta| dr d\theta## IS ##r dr d\theta##.
I see that this implies that dS = r dr dθ. But in the example from the website they also have dA = r dr dθ. How does this follow if dA = dr dθ? How can it equal two different things?
 
Nov27-12, 11:16 AM   #36

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Quote by CAF123 View Post
I see that this implies that dS = r dr dθ. But in the example from the website they also have dA = r dr dθ. How does this follow if dA = dr dθ? How can it equal two different things?
They are two different uses of the symbol 'dA'. They are two different things.
 
Nov27-12, 11:20 AM   #37
 
Quote by Dick View Post
They are two different uses of the symbol 'dA'. They are two different things.
Ok, thanks. I think this also answers one of my other questions: The form dA=dr dθ is not really an area element because the dimensions don't check out. What is the difference between the two forms?
 
Nov27-12, 11:24 AM   #38

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Quote by CAF123 View Post
Ok, thanks. I think this also answers one of my other questions: The form dA=dr dθ is not really an area element because the dimensions don't check out. What is the difference between the two forms?
We've been over that several times already. dr dθ is the area element without the jacobian part 'r'. You use that in the cross product form of the surface integral because the cross product contains the jacobian part, 'r'. If you are just integrating a function in polar coordinates you want the full area element, i.e. with jacobian r dr dθ.
 
Nov27-12, 11:27 AM   #39
 
Quote by Dick View Post
We've been over that several times already. dr dθ is the area element without the jacobian part 'r'. You use that in the cross product form of the surface integral because the cross product contains the jacobian part, 'r'. If you are just integrating a function in polar coordinates you want the full area element, i.e. with jacobian r dr dθ.
Ok, everything makes sense. Thanks so much!
 
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