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Convert the double integral to polar coordinates

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Evaluate the double integral by converting to polar coordinates.
    ∫∫ arctan y/x dA; R is the sector in the first quadrant between the circles 1/4= x^2+y^2 and x^2+y^2=1 and the lines y=x/√3 and y=x.


    2. Relevant equations

    arctan y/x= θ

    3. The attempt at a solution

    D={(r,θ) l 1/4≤r≤1, ∏/6≤∏/4}

    ∫∫ θr drdθ

    Which gives me 25∏^2/3072 which is incorrect according to my book. The correct answer is 15∏^2/2304.
     
  2. jcsd
  3. Dec 9, 2011 #2

    Dick

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    Your lower bound for r is wrong. Think about it again.
     
  4. Dec 10, 2011 #3
    I'm not sure, I thought maybe 1/8 but that's not right.
     
  5. Dec 10, 2011 #4

    Dick

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    A circle around the origin has the equation x^2+y^2=r^2. If r^2=1/4, what's r?
     
  6. Dec 10, 2011 #5
    I tried 1/2 when I was originally having difficulty with the problem and still got the wrong answer. I guess I must be missing something else.
     
  7. Dec 10, 2011 #6

    Dick

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    Their given answer isn't in lowest terms for some reason. 15*pi/2304=5*pi/768. Is that part of the trouble?
     
  8. Dec 10, 2011 #7

    SammyS

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    Try again, This gives the right answer using the rest of what you have.
     
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