Integrating unit vector ρ

  • #1
2
0
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
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
8,583
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
2
0
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
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
16,670
8,583
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##.
 

Related Threads on Integrating unit vector ρ

Replies
2
Views
6K
Replies
2
Views
7K
Replies
2
Views
764
Replies
1
Views
24K
Replies
2
Views
1K
Replies
3
Views
11K
Replies
18
Views
14K
Replies
10
Views
2K
Top