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Integration in Polar - polar coordinate

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    f(x,y) = xy
    x ≥ 0, y ≥ 0, [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ 4


    2. Relevant equations



    3. The attempt at a solution
    [itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] rcosθ * rsinθ * r drdθ

    I just wanted to check... is this right?? cuz I really don't think it is
     
  2. jcsd
  3. Nov 13, 2011 #2

    Dick

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    Why are you putting the upper limit for r to be 2*sec(theta)? The outer boundary is just a circle, isn't it?
     
  4. Nov 13, 2011 #3
    right! in terms of x, the bound should be from 0 to 2, but since x = 2 is rcos = 2, doesn't r becomes 2sec? or am I really really wrong and fault?
    OH WAIT
    is it just bounded by 0 to 2?

    OH OH WAIT I JUST HAD AN EPIC EPIPHANY!!! THIS IS U SUBSTITUTION!!!!
     
    Last edited: Nov 13, 2011
  5. Nov 13, 2011 #4

    Dick

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    Uh, the boundary isn't x=2. It's x^2+y^2=4. u-substitution?? I think you are overcomplicating this.
     
  6. Nov 13, 2011 #5
    [itex]f^{pi/2}_{0}[/itex][itex]f^{2}_{0}[/itex] rcosθ * rsinθ * r drdθ
    So it's not this?
     
  7. Nov 13, 2011 #6

    Dick

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    That's it.
     
  8. Nov 13, 2011 #7
    yes so I'll go head and calculate all this cuz there seems to be a problem and I don't know if it's calculation error or book error

    [itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] [itex]r^{3}[/itex]cosθ * rsinθ * drdθ

    =[itex]f^{pi/2}_{0}[/itex]([itex]r^{4}[/itex]cosθsinθ)/4 * drdθ

    =4[itex]f^{pi/2}_{0}[/itex]cosθsinθ dθ

    After u substitution...

    =4[itex]f^{1}_{0}[/itex]u du

    =4?
     
  9. Nov 13, 2011 #8

    Dick

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    The integral from 0 to 1 of u*du isn't 1.
     
  10. Nov 13, 2011 #9
    oh oh wow... How could I miss that...
    Thank you very much ;D
     
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