Surface integral in cylindrical coordinates

  • #1
Hello everybody! Although this may sound like a homework problem, I can assure you that it isn't. To prove it, I will give you the answer: 40pi.

So.. I'm self-studying some electrodynamics. I'm using the third edition of Griffiths, and I have a quick question. For those who own the book and want to follow along, I am doing problem 1.42 on page 45.

Here is the problem as given to me:
http://img151.imageshack.us/img151/1237/85625524.png [Broken]

Here is my solution:
http://img547.imageshack.us/img547/404/77923328.png [Broken]

Although I get the right answer, I'm questioning my method a little bit. I know that the flux integrals across the rectangles in the xz and yz planes are 0, and my method gives that they are 0. However, I know that Griffiths often finds ds in cylindrical coordinates in a different way, and I'm wondering if my way is mathematically valid. Thank you all very much!

EDIT: I justified it to myself, and I now see how my surface element and Griffiths' are the same. The s-dependence in ds actually falls out (which, for some reason, I didn't notice before), and you are left with just phi-hat. I apologize for making a topic before thinking about it for long enough. However, if you have anything about the problem/life to contribute, feel free! :) If I knew how to close this, I would.
 
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Answers and Replies

  • #2
SammyS
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Hello everybody! Although this may sound like a homework problem, I can assure you that it isn't. To prove it, I will give you the answer: 40pi.

So.. I'm self-studying some electrodynamics. I'm using the third edition of Griffiths, and I have a quick question. For those who own the book and want to follow along, I am doing problem 1.42 on page 45.

Here is the problem as given to me:
http://img151.imageshack.us/img151/1237/85625524.png [Broken]

Here is my solution:
http://img547.imageshack.us/img547/404/77923328.png [Broken]

Although I get the right answer, I'm questioning my method a little bit. I know that the flux integrals across the rectangles in the xz and yz planes are 0, and my method gives that they are 0. However, I know that Griffiths often finds ds in cylindrical coordinates in a different way, and I'm wondering if my way is mathematically valid. Thank you all very much!

EDIT: I justified it to myself, and I now see how my surface element and Griffiths' are the same. The s-dependence in ds actually falls out (which, for some reason, I didn't notice before), and you are left with just phi-hat. I apologize for making a topic before thinking about it for long enough. However, if you have anything about the problem/life to contribute, feel free! :) If I knew how to close this, I would.

The normal to the surface in the xz plane is [tex]\displaystyle \hat{n}=-\hat{\varphi}\,.[/tex]
So the only contribution to [tex]\displaystyle \vec{v}\cdot \vec{ds}[/tex] for this surface is from the φ component of the field. Note, the φ component of the field is zero on both the xz plane and on the yz plane.

The normal to the surface in the yz plane is [tex]\displaystyle \hat{n}=\hat{\varphi}\,.[/tex] So by a similar argument to that above, the contribution to the surface integral is zero through this surface too.
 
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  • #3

The normal to the surface in the xz plane is [tex]\displaystyle \hat{n}=-\hat{\varphi}\,.[/tex]
So the only contribution to [tex]\displaystyle \vec{v}\cdot \vec{ds}[/tex] for this surface is from the φ component of the field. Note, the φ component of the field is zero on both the xz plane and on the yz plane.

The normal to the surface in the yz plane is [tex]\displaystyle \hat{n}=\hat{\varphi}\,.[/tex] So by a similar argument to that above, the contribution to the surface integral is zero through this surface too.
This is how the author did it. I'm not the best at visualizing spherical coordinates, so I employed x-hat and y-hat instead. According to the geometry of the problem, my solution will actually boil down to yours, and the integrals come out the same :)

Thanks for the help!
 

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