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Surface integral (Flux) with cylinder and plane intersections 
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#19
Nov412, 01:32 PM

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#20
Nov412, 01:35 PM

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#21
Nov412, 01:36 PM

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#22
Nov412, 02:08 PM

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#23
Nov2612, 04:54 PM

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Just a quick question about one of the things here. I said ##dA = dz d\theta##. Why is this the case? In polars, we have ##dA = r\,dr\,d\theta## which makes sense as an area since ##\theta## is dimensionless.



#24
Nov2612, 05:19 PM

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If you look at the above the dS has the dimensions of area. In the cross product form the extra r is coming from the cross product. 


#25
Nov2712, 02:38 AM

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#26
Nov2712, 06:23 AM

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#27
Nov2712, 08:10 AM

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I was looking at some other site (referenced below) and they take dA=r dr dθ. Why is this?
http://calculus7.com/sitebuildercont...ntesdblint.pdf 


#28
Nov2712, 10:01 AM

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#29
Nov2712, 10:06 AM

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What's the other form you mentioned? When I thought about it later, I see that they were just converting from Cartesian to polar and so the r was the Jacobian.



#30
Nov2712, 10:14 AM

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#31
Nov2712, 10:24 AM

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I see. So they are essentially using the form $$ dS = \frac{(r_u \times r_v)}{r_u \times r_v}r_u \times r_v dA = \hat{n} r\,dr\,d\theta$$
EDIT: if this is the case, then shouldn't ##r_u\times r_v dA = r\,dr\,d\theta, ##instead of just ##dA##? 


#32
Nov2712, 10:28 AM

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#33
Nov2712, 10:31 AM

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Did you catch my edit in the last post?



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Nov2712, 10:58 AM

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#35
Nov2712, 11:07 AM

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#36
Nov2712, 11:16 AM

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