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Trouble with an integral

  1. Jul 31, 2007 #1
    I have a book that has the following integral:

    [tex]6\int_\frac{-1}{2}^\frac{1}{2}\int_\frac{-1}{2}^\frac{1}{2}\frac{1}{r^2}a_r\cdot a_z dx dy[/tex]

    This integral gets converted to:

    [tex]3\int_\frac{-1}{2}^\frac{1}{2}\int_\frac{-1}{2}^\frac{1}{2}\frac{dx dy}{(x^2 + y^2 + 1/4)^\frac{3}{2}}[/tex]

    (z = 1/2 by the way...)

    I understand how it got to that point, but I'm having trouble understanding how it gets to this integral, I guess I don't understand the integration involved:

    [tex]3\int_\frac{-1}{2}^\frac{1}{2}(\frac{x}{(y^2 + \frac{1}{4})(x^2 + y^2 + 1/4)^\frac{1}{2}})\|_\frac{-1}{2}^\frac{1}{2}dy[/tex]

    can someone explain how it gets to there?
     
  2. jcsd
  3. Jul 31, 2007 #2

    HallsofIvy

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    If you don't tell us what ar and az are, I don't see how anyone can help you!
     
  4. Jul 31, 2007 #3
    oh, they are vectors, but they are not part of the problem. I understand how it goes from the first integral to the second integral, I'm having a problem figuring out how it goes from the second to the third integral here. The vectors have already been used up there.
     
  5. Jul 31, 2007 #4
    They're just doing the x-integration first. So they solve

    [tex]\int_{-1/2}^{1/2}\frac{dx}{(y^2+x^2+1/4)^{3/2}}[/tex]

    treating y as a parameter. This is a pretty standard integral, and its solution is in brackets in your last expression. Note that

    [tex]f(x)||_{-1/2}^{1/2}[/tex] means [tex]f(1/2)-f(-1/2)[/tex]

    so that your quantity in brackets is actually a sum of two terms, one with x=1/2 and the other with x=-1/2.
     
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