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Double integral circle limits

  1. Jul 31, 2011 #1
    1. The problem statement, all variables and given/known data
    Evaluate
    f(x,y)=y2[itex]\sqrt{1-x2}[/itex]
    over the region
    x2+y2< 1

    2. Relevant equations



    3. The attempt at a solution

    using x limits between -1 & 1 followed by the y limits of 0 & [itex]\sqrt{1-x2}[/itex]

    [itex]\int[/itex][itex]\int[/itex]y2[itex]\sqrt{1-x2}[/itex].dy.dx

    Evaluating this and multiplying be 2 to get the whole circle I get 0 as the 1 and -1 limits both give zero as the answer. Therefore I used the x limits of 0 & 1 and multiplied by 4 instead, this gave an answer of 4/9, however I have been given the answer of 32/45, and I have no idea how this could possibly be wrong.
     
  2. jcsd
  3. Jul 31, 2011 #2

    tiny-tim

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    Hi LASmith! :smile:
    Your intermediate integrand must be wrong. :redface:

    Show us what you had inside the ∫ when it was just ∫ … dx​
     
  4. Jul 31, 2011 #3
    [itex]\int[/itex] y3[itex]\sqrt{1-x2}[/itex]/3 .dx

    Then substituting the limits [itex]\sqrt{1-x2}[/itex] and 0 for y I obtained

    [itex]\int[/itex] (1-x2)2)/3 .dx

    Limits between 0 and 1 for this final integral
     
  5. Jul 31, 2011 #4

    HallsofIvy

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    The set [itex]x^2+ y^2< 1[/itex] is the interior of a circle with center at (0, 0) and radius 1. That means that a diameter, on the x-axis, goes from -1 to 1, and on the y-axis, from -1 to 1. By taking y from 0 (the x-axis) up to [itex]\sqrt{1- x^2}[/itex] (the circle) and x from -1 to 1, you are only integrating over 1/2 of the circle.

    Your y needs to go from the point where the vertical line crosses the circle below the x-axis to the point where it crosses the circle above the x-axis.
     
  6. Jul 31, 2011 #5

    tiny-tim

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    = ∫ (1 - 2x2 + x4)/3 dx …

    how did you not get a factor 1/5 after integrating this? :confused:
     
  7. Aug 1, 2011 #6
    Expanding the brackets and then integrating I obtained the correct answer, so thanks for that.

    However, looking back I had no idea how I did not get a factor of 1/5, I did not expand out the brackets, and just used the chain rule to obtain
    (1-x2)3/-18x
    Then putting in the limits 0 and 1 gets the answer 0-0, out of curiosity how does this method not work?
     
  8. Aug 1, 2011 #7

    tiny-tim

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    Hi LASmith! :smile:
    Just try differentiating (1-x2)3/-18x …

    you'd have to use both the chain rule and the product (or quotient) rule, and they don't cancel (why should they?).
     
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