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Iterated integrals

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to evaluate this
    [tex] \int_{-2}^{0} \! \int_{-\sqrt{9-x^2}}^{0} 2xy\,dy\,dx. [/tex]


    2. Relevant equations



    3. The attempt at a solution

    So I first tried integrating with respect to dy and I have:

    [tex] xy^2 [/tex] and evaluate that from [tex] -\sqrt{9-x^2}[/tex] to 0 , I will have [tex] -18x + 2x^3[/tex]. Then I will integrate this again over dx, then I have [tex] 9x^2 + \frac{x^4}{2} [/tex] and I evaluate this from -2 to 0. which results in -44. Where did I do wrong?
     
    Last edited: Feb 27, 2009
  2. jcsd
  3. Feb 27, 2009 #2

    Mark44

    Staff: Mentor

    You should have gotten -x(9 - x^2) here, or x^3 - 9x. You're off by a factor of 2.
    It looks like you made another mistake, this time a sign error. From your previous work, you should have gotten -9x^2 + 1/2 * x^4.

    The region over which integration takes place is a portion of the southwest quadrant of a circle of radius 3, centered at (0, 0). The region is bounded by the axes, the said circle, and the line x = -2.

    Just to check my work, I integrated over the same sized region in the northeast quadrant of the circle, this time bounded by the line x = 2. You can't do this in general, but the integrand in this problem, 2xy, is unaffected by changes in signs to both variables. Both integrations resulted in values of 14, but the second way was simpler, with fewer chances to make sign errors.
     
  4. Feb 27, 2009 #3
    So I should get 14 as a final result if I did everything correctly?
     
  5. Feb 28, 2009 #4

    Mark44

    Staff: Mentor

    Unless I made a mistake doing the problem in two different ways...
     
  6. Feb 28, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, 14 is correct.
     
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