Can You Integrate Unit Vector ρ Without Switching to Cartesian Coordinates?

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Discussion Overview

The discussion revolves around the integration of the unit vector \(\hat{\rho}\) in cylindrical coordinates, specifically addressing whether it is necessary to convert to Cartesian coordinates for the integration process. Participants explore different methods of integration and their implications.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • Some participants suggest that integrating \(\int_{0}^{2\pi} \hat{\rho} d\phi\) directly yields \(2\pi \hat{\rho}\), while others argue that decomposing \(\hat{\rho}\) into \(\sin(\phi) \hat{i} + \cos(\phi) \hat{j}\) leads to an integral that evaluates to zero.
  • There is a question about whether it is possible to maintain the use of cylindrical coordinates without reverting to Cartesian coordinates while still achieving the same result.
  • One participant emphasizes that expressing \(\hat{\rho}\) as a function of the variable of integration is necessary, suggesting that the second method (decomposition) is the only viable option.
  • Another participant notes that using Cartesian coordinates would require expressing the integral in terms of \(x\) and \(y\).

Areas of Agreement / Disagreement

Participants express differing opinions on the necessity of converting to Cartesian coordinates, with some favoring direct integration in cylindrical coordinates and others advocating for decomposition. The discussion remains unresolved regarding the best approach.

Contextual Notes

Participants do not reach a consensus on the integration methods, and there are unresolved questions about the implications of using different coordinate systems.

ys98
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When doing integration such as \int_{0}^{2\pi} \hat{\rho} d\phi which would give us 2\pi \hat{\rho}, must we decompose \hat{ρ} into sin(\phi) \hat{i} + cos(\phi) \hat{j} , then \int_{0}^{2\pi} (sin(\phi) \hat{i} + cos(\phi)\hat{j}) d\phi , which would give us 0 instead?

Thanks
 
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ys98 said:
When doing integration such as \int_{0}^{2\pi} \hat{\rho} d\phi which would give us 2\pi \hat{\rho}, must we decompose \hat{ρ} into sin(\phi) \hat{i} + cos(\phi) \hat{j} , then \int_{0}^{2\pi} (sin(\phi) \hat{i} + cos(\phi)\hat{j}) d\phi , which would give us 0 instead?

Thanks

Which method looks correct to you?
 
PeroK said:
Which method looks correct to you?
The second one. My question is, is there anyway that I can keep using cylindrical coordinate without changing back to cartesian coordinate and get the same solution?
 
ys98 said:
The second one. My question is, is there anyway that I can keep using cylindrical coordinate without changing back to cartesian coordinate and get the same solution?

You must express ##\hat{\rho}## as a function of the variable with which you are integrating. Your second method looks the only viable option to me. That was using cylindrical coordinates. Using Cartesian coordinates would entail expressing the integral in terms of the Cartesian variables ##x## and ##y##.
 
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