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Domain for Polar Coordinate

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    f(x,y) = y([itex]x^{2}[/itex] + [itex]y^{2}[/itex])^-1
    y ≥ [itex]\frac{1}{2}[/itex], [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ 1


    2. Relevant equations



    3. The attempt at a solution
    Would you check my domain please?

    [itex]f^{pi}_{0}[/itex][itex]f^{sqrt(3)/2}_{-sqrt(3)/2}[/itex] sinθ drdθ
     
  2. jcsd
  3. Nov 13, 2011 #2

    Dick

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    Nope, all wrong. Did you draw a sketch of the region? The angle doesn't go from 0 to pi. And your r limits look more like x limits.
     
  4. Nov 13, 2011 #3
    Yes I did and it looks like the attachment. and the Coordinate was (-[itex]\sqrt{3}[/itex]/2, [itex]\frac{1}{2}[/itex]) to ([itex]\sqrt{3}[/itex]/2, [itex]\frac{1}{2}[/itex]) is it not?
     

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  5. Nov 13, 2011 #4

    Dick

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    Nice sketch! And like I said the sqrt(3)/2 stuff is the x limits. It's not the starting angle and ending angle. Those would be the angles your endpoints make with the origin, right?
     
  6. Nov 13, 2011 #5
    for dθ, do I have to use anti tangent(I think I forgot the proper name) of the coordinate to get the angle?

    for dr... I am lost on dr cuz... wait... is dr the distance from the origin to the each of the coordinate?

    edit - so dr domain is -1 to 1 right?
     
  7. Nov 13, 2011 #6

    Dick

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    Sure, you can use arctan to find the angles. Now to find the r limits think about a line through the origin at some angle theta. What's the radius value where it hits the circle x^2+y^2=1? That's easy. Now what's the radius value where it hits the line y=2? Time to use y=r*sin(theta).
     
  8. Nov 13, 2011 #7
    ok so for dθ, its pi/6 to 5pi/6

    for dr, the radius of x^2 + y^2 = 1 is 1 obviously. But where did y = 2 come from? Did you mean y = 1/2? Isn't dr domain based on the distance from the origin? like in the picture?
     

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  9. Nov 13, 2011 #8

    Dick

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    Sure. I meant y=(1/2). Sorry. And the r limit (that's not the same as 'dr') is the distance from the origin to boundary curve along the angle theta. So upper limit for r is 1, right.
     
  10. Nov 13, 2011 #9
    and the lower limit is -1 I hope?? :uhh:
     
  11. Nov 13, 2011 #10

    Dick

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    Noo. y=(1/2)=r*sin(theta). Solve for r. The lower limit depends on theta, doesn't it? Can't you see that from your sketch? As theta changes the distance from the origin to the point on the line changes.
     
  12. Nov 13, 2011 #11
    No this one I really don't get. So lower limit always depends on theta?
     
  13. Nov 13, 2011 #12

    Dick

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    Not always! As theta changes the distance to the circle x^2+y^2 stays 1. The distance to the line y=(1/2) does. If r is the distance along the angle theta from the origin to y=(1/2), what's r at theta=pi/2? What's r at theta=pi/6?
     
  14. Nov 13, 2011 #13
    well r at theta = pi/6 is 1
    r at theta = pi/2 is 0
    if not f*** me
     
  15. Nov 13, 2011 #14

    SammyS

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    r=0 is the origin. That's clearly not in the domain.

    (No intention to fool you.)
     
  16. Nov 13, 2011 #15

    Dick

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    You are really missing some essential part of the picture here. And I can't figure out what it is. If you draw a line from the origin at an angle of pi/2, where does it hit the line y=(1/2)??
     
  17. Nov 13, 2011 #16
    are you asking for this green part?
     

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  18. Nov 13, 2011 #17

    Dick

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    Exactly right. It's not 0, is it? And that's only the lower bound for r at theta=pi/2. Like I said before the lower bound is a function of theta. Can you write an expression for it?
     
  19. Nov 13, 2011 #18
    No it is not. I believe the distance from the origin to y = 1/2 is 1/2... that's the green part
     
  20. Nov 13, 2011 #19

    Dick

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    Ok, so the lower bound of r at pi/2 is 1/2 and the lower bound for r at pi/6 is 1. Now what is an expression for it at an arbitrary value of theta?? Once again, y=(1/2)=r*sin(theta).
     
  21. Nov 14, 2011 #20
    OHOHOHOHOHOHOHO i got it now I fully get it now Now I know
    THank you very much
    I'll buy you a beer when we meet :D
     
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