Simple integral in cylindrical coordinates

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SUMMARY

The discussion centers on calculating the integral \(\oint[\hat{r}+\hat{z}]d\phi\) in cylindrical coordinates. The user initially computes this integral as \(2\pi[\hat{r}+\hat{z}]\) but later converts it to Cartesian coordinates, yielding \(2\pi\hat{z}\). The discrepancy arises from the fact that \(\hat{r}\) is a function of \(\phi\), which the user fails to account for in the cylindrical coordinate system. Understanding the relationship between \(\hat{r}\) and \(\phi\) is crucial for accurate calculations in cylindrical coordinates.

PREREQUISITES
  • Understanding of cylindrical coordinates and their components (\(\hat{r}, \hat{\phi}, \hat{z}\))
  • Knowledge of vector calculus and line integrals
  • Familiarity with coordinate transformations between cylindrical and Cartesian systems
  • Basic understanding of integrals in multi-variable calculus
NEXT STEPS
  • Study the properties of cylindrical coordinates and their applications in vector calculus
  • Learn about the relationship between \(\hat{r}\) and \(\phi\) in cylindrical coordinates
  • Explore line integrals in different coordinate systems, focusing on cylindrical coordinates
  • Review examples of converting integrals from cylindrical to Cartesian coordinates
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Students and professionals in mathematics, physics, and engineering who are working with integrals in cylindrical coordinates, particularly those seeking to deepen their understanding of vector calculus and coordinate transformations.

Vagrant
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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.
 
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Vagrant said:

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]
 
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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