Integrating unit vector ρ

  • #1
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When doing integration such as [itex] \int_{0}^{2\pi} \hat{\rho} d\phi[/itex] which would give us [itex] 2\pi \hat{\rho} [/itex], must we decompose [itex] \hat{ρ} [/itex] into [itex] sin(\phi) \hat{i} + cos(\phi) \hat{j}[/itex] , then [itex] \int_{0}^{2\pi} (sin(\phi) \hat{i} + cos(\phi)\hat{j}) d\phi [/itex] , which would give us 0 instead?

Thanks
 

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

Thanks

Which method looks correct to you?
 
  • #3
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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?
 
  • #4
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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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