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Simple integral in cylindrical coordinates

  • Thread starter Vagrant
  • Start date
  • #1
195
1

Homework Statement


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]

Homework Equations





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.
 

Answers and Replies

  • #2
SammyS
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Homework Statement


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]

Homework Equations





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.
[itex]\hat{r}[/itex] is a function of [itex]\phi\ .[/itex]
 
  • #3
195
1
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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