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Cartesian to polar integral help?

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations

    3. The attempt at a solution
    my only problem curently is in finding the angle θ. I do get the equation x^2 + y^2 =1 however am confused whether this would be a semi-circle on the positive axis or a full circle. because my teacher has notes that confuse me. for instance ∫[itex]\sqrt{1-y^2}[/itex] -[itex]\sqrt{1-y^2}[/itex] my formatting is a bit off but that would be higher/lower bound. that is the only case where she made a complete circle, instead of a semi circle.

    if it is a semi circle, theta would be 0≤ θ ≤ PI/2
    however if its a full circle, since from y=-1 to y=1 it would be -PI/2 ≤ θ≤ PI/2.
     
    Last edited: May 7, 2013
  2. jcsd
  3. May 7, 2013 #2

    LCKurtz

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    x goes from 0 to the positive square root of ##1-y^2##. What part of the circle do you get when you solve it for ##x## and take the positive root? That will tell you which ##\theta##'s to use.
     
  4. May 7, 2013 #3
    i think it would be the right hemisphere of the circle, since x=0 and increases, and the circle has radius of 1. so -PI/2 ≤ θ ≤ PI/2 what i got is right?
     
  5. May 7, 2013 #4

    LCKurtz

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    Yes.
     
  6. May 7, 2013 #5
    OK. then i got r3 by looking at f(x,y) so I ended up integrating ∫(r4 dr)dθ and got my final answer as PI/5.
     
  7. May 8, 2013 #6

    HallsofIvy

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    Integrating with r from 0 to 1, [itex]\theta[/itex] from [itex]-\pi/2[/itex] to [itex]\pi/2[/itex].

    Yes, that is correct.
     
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