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Double integrals interchanging order

  1. Jul 22, 2014 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_{1}^{4}\int_{1}^{\sqrt{x}}(x^2+y^2)dydx[/tex]





    3. The attempt at a solution

    I drew the region,

    I tried
    [tex]\int_{1}^{2}\int_{1}^{y^2}(x^2+y^2)dxdy[/tex]

    but it doesn't seem to work.

    when the order is changed

    [tex]1 \le y \le 2 [/tex]

    and [tex] \sqrt{x} = y \rightarrow x=y^2[/tex]

    so the change in x is [tex] 1 \le x \le y^2[/tex]

    but it doesn't work out to the correct answer.
     
  2. jcsd
  3. Jul 23, 2014 #2
    The region described by [itex]1\leq y \leq 2[/itex] and [itex]1\leq x\leq y^2[/itex] looks like this:
    attachment.php?attachmentid=71561&stc=1&d=1406094416.png
    However, the region specified in the first integral is this one:
    attachment.php?attachmentid=71562&stc=1&d=1406094416.png
    Can you tell what the proper bounds should be from this illustration ?
     

    Attached Files:

  4. Jul 23, 2014 #3
    I see it

    the new integral should be

    [tex] \int_{1}^{2}\int_{y^2}^{4} (x^2+y^2)dxdy = \int_{1}^{4}\int_{1}^{\sqrt{x}} (x^2+y^2)dydx [/tex]
     
  5. Jul 23, 2014 #4
    That's it. :)
     
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