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Another quick please tell me if my logic seems correct (change of variables)

  1. Mar 4, 2005 #1
    I'm trying to evaluate the double integral

    [tex]\int \int \sqrt{x^2 + y^2} \, dA[/tex]

    over the region R = [0,1] x [0,1]
    using change of variables.

    Well, after fooling around, I've got an answer. I set u = x^2, v =y^2, and then calculated the jacobian of T which was 1. The image transformation limits of integration for u and v turned out to be the same [0,1] x [0,1]

    So I did the following calculation (both integrals going from 0 to 1)

    [tex]\int \int \sqrt{u + v} * (1) dudv[/tex]


    which resulted in a value of roughly 3.238.

    Does my logic and answer seem sound here? Thanks in advance.
     
  2. jcsd
  3. Mar 4, 2005 #2
    Converting to rectangular coordinates would probably be easier
     
  4. Mar 4, 2005 #3
    That's cylindrical coordinates, sorry
     
  5. Mar 4, 2005 #4
    Answered in Calculus and Analysis.
     
  6. Mar 4, 2005 #5
    I'm totally at a loss here guys. I realized my Jacobian was computed wrong. Can someone please give me a clue as to what would be the most efficient integral setup? I'm completely dumbfounded. :( Thanks


    edit: more in-depth post in the calculus forum, thanks
     
    Last edited: Mar 4, 2005
  7. Mar 4, 2005 #6

    dextercioby

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    Please,DO NOT DOUBLE POST!!:mad:

    Daniel.
     
  8. Mar 6, 2005 #7
    Dexter...you need to drop the intensity down a notch. And to the ninja, just convert [tex] x^2 + y^2 [/tex] to [tex] r^2 [/tex] and integrate over the same area in cylindrical.
     
  9. Mar 6, 2005 #8

    dextercioby

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    He can't do that,the square [0,1]*[0,1] is not equivalent to the quarter of the circle you're implying...

    Daniel.
     
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