Surface integral in cylindrical coordinates

In summary, the conversation discusses a solution to an electrodynamics problem and the use of surface integrals in cylindrical coordinates. The author initially questions the validity of their method, but later realizes that it is mathematically sound and results in the same answer as the method used by the author in the book. The conversation ends with the author thanking for the help and mentioning their difficulty with visualizing spherical coordinates.
  • #1
scorpion990
86
0
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.
 
Last edited by a moderator:
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  • #2
scorpion990 said:
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.
 
Last edited by a moderator:
  • #3
SammyS said:

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!
 

1. What is a surface integral in cylindrical coordinates?

A surface integral in cylindrical coordinates is a mathematical tool used to calculate the flux through a surface in three-dimensional space. It takes into account the surface's shape and orientation in relation to the cylindrical coordinate system.

2. How is a surface integral in cylindrical coordinates different from a regular surface integral?

A surface integral in cylindrical coordinates is different from a regular surface integral because it uses a different coordinate system. In cylindrical coordinates, the surface is defined by a radius, angle, and height, while in a regular surface integral, the surface is defined by two variables.

3. What are the steps to solve a surface integral in cylindrical coordinates?

The steps to solve a surface integral in cylindrical coordinates are:

  1. Identify the surface and determine its orientation in relation to the cylindrical coordinate system.
  2. Set up the integral using the appropriate formula for a cylindrical surface.
  3. Substitute the appropriate values for the variables in the integral.
  4. Integrate the resulting function with respect to the variables.
  5. Evaluate the integral to find the flux through the surface.

4. When is it necessary to use a surface integral in cylindrical coordinates?

A surface integral in cylindrical coordinates is necessary when the surface being analyzed is not easily defined using a Cartesian coordinate system. Cylindrical coordinates are particularly useful for analyzing surfaces with circular or cylindrical shapes.

5. Can a surface integral in cylindrical coordinates be used for any type of surface?

No, a surface integral in cylindrical coordinates can only be used for surfaces that can be defined using cylindrical coordinates. It is not suitable for surfaces that cannot be described using a radius, angle, and height, such as a plane or a sphere. In those cases, other coordinate systems must be used.

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