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Iterated Integral

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following double integral over the closed region B:

    [tex]\int\int_B x^2 y^3 dx dy[/tex] where B is the region bounded by y = x^2 and y = x

    2. Relevant equations



    3. The attempt at a solution

    I think I set up the paramaters wrong:

    [tex]\int^{1}_{0}\int^{y}_{\sqrt{y}} x^2 y^3 dx dy[/tex]
     
    Last edited: Feb 9, 2009
  2. jcsd
  3. Feb 9, 2009 #2

    Dick

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    If y is in [0,1], as it looks like you've figured out for your integral, which is larger y or sqrt(y)? Is that what's confusing you?
     
  4. Feb 9, 2009 #3
    y>sqrt(y)

    The answer is 1/77, but I get -1/77, but taking it's absolute value should take care of that.
     
  5. Feb 10, 2009 #4
    What's sqrt(1/4)?
     
  6. Feb 10, 2009 #5

    HallsofIvy

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    Can you explain why "taking its absolute value" is justified?

    Once again, which is larger, y or [itex]\sqrt{y}[/itex] for y between 0 and 1?

    Which is larger 1/4 or [itex]\sqrt{1/4}[/itex]?
     
  7. Feb 10, 2009 #6
    I get it. Because y goes from 0 - 1, sqrt(y)>y in this case.

    but what if y went from 0 - 2, then would it be this:

    [tex]\int^{1}_{0}\int^{\sqrt{y}} }_{y} x^2 y^3 dx dy + \int^{2}_{1}\int^{y}_{\sqrt{y}} x^2 y^3 dx dy[/tex]
     
  8. Feb 10, 2009 #7

    Dick

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    Yes.
     
  9. Feb 10, 2009 #8
    Thank you all for your help.
     
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