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Simple integral in cylindrical coordinates

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data
    As a part of bigger HW problem, I need to calculate the integral:
    [itex]\oint[/itex][[itex]\hat{r}[/itex]+[itex]\hat{z}[/itex]]d[itex]\phi[/itex]

    2. Relevant equations



    3. The attempt at a solution

    In cylindrical coordinates:
    =[[itex]\hat{r}[/itex]+[itex]\hat{z}[/itex]] [itex]\oint[/itex]d[itex]\phi[/itex]
    =2∏[[itex]\hat{r}[/itex]+[itex]\hat{z}[/itex]]

    On the other hand if I convert it to Cartesian coordinates:
    [itex]\oint[/itex][cos[itex]\phi[/itex][itex]\hat{x}[/itex]+sin[itex]\phi[/itex][itex]\hat{y}[/itex]+ [itex]\hat{z}[/itex]]d[itex]\phi[/itex]
    =2∏[itex]\hat{z}[/itex]

    So, what is it that I am doing wrong in the case with cylindrical coordinates? I'm sure I'm missing something very basic.
    Thanks.
     
  2. jcsd
  3. Sep 16, 2012 #2

    SammyS

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    [itex]\hat{r}[/itex] is a function of [itex]\phi\ .[/itex]
     
  4. Sep 16, 2012 #3
    Sorry, but could you please explain this a bit?
    I can visualize that [itex]\hat{r}[/itex] is a function of [itex]\phi[/itex](unit circle on x-y plane centered at origin) for cartesian coordinates, but I'm drawing a blank when it comes to cylindrical coordinates.
     
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