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Double integral problem

  1. Sep 28, 2011 #1
    8yv6i.png







    My idea was that the limits are
    BrQKm.png

    and that the anti-derivative of dy was

    xlog(1+y^2)

    but that seems wrong...

    maybe use these limits instead


    YKvJd.png

    and start with dx?

    gives us
    rDFwG.png

    then we take dy

    rgKWx.png

    guess, i figured it out eventually with the help of wolfram with the last integration
     
    Last edited: Sep 28, 2011
  2. jcsd
  3. Sep 28, 2011 #2

    SammyS

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    The reason your 1st attempt didn't work is that [tex]\int\frac{1}{1+y^2}\,dy=\tan^{-1}(y)+C\,.[/tex]
     
  4. Sep 28, 2011 #3

    HallsofIvy

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    Yes, it is wrong. It is standard "Calculus I" mistake to treat a function of the variable as if it were just the variable but you should be past that by the time you are doing multiple integrals. "[itex]1/(1+ y^2)[/itex]" is NOT the same as [itex]1/y[/itex] and its anti-derivative is not a logarithm. The anti-derivative of [itex]1/(1+ y^2)[/itex] is [itex]arctan(y)+ C[/itex]. That's a standard anti-derivative that you should have memorized.

    Yes, reversing the order of integration is the best way to handle this one. Integerating [itex]arctan(1)- arctan(x^2)= \pi/4- arctan(x^2)[/itex] is likely to be very difficult!
     
  5. Sep 28, 2011 #4

    SammyS

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    Yes. I agree !
     
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