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Double Integrals

  1. Mar 1, 2005 #1

    Zurtex

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    Just a question of notation here, my lecturer will wright an integral like this:

    [tex]\int_3^6 dx \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}y \, dy[/tex]

    But mean this:

    [tex]\int_3^6 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}y \, dy \, dx[/tex]

    Is this standard notation? It seems rather odd to me.
     
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  3. Mar 1, 2005 #2

    Galileo

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    I encounter expressions where the [itex]dx[/itex] or [itex]dy[/itex] comes first often as well.
    [tex]\int_a^b dx f(x)[/tex]
    seems to be quite customary.

    To be consistent however, I'd use:
    [tex]\int_3^6 dx \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy \;y[/tex]
     
  4. Mar 1, 2005 #3

    dextercioby

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    Zurtex,it's highly reccomendable you use the first notation...There may be situations (like this one
    [tex] \int_{0}^{3}\int_{-4}^{\pi}\int_{-5}^{9}\sin(xyz+x\sqrt{y}\sqrt[\frac{\sqrt{3}}{2}]{z}) dx \ dz \ dy [/tex]

    ) in which you never know what integration to do first...

    Daniel.
     
  5. Mar 1, 2005 #4
    This integral is not possible, because the range for x is from 3 to 6, while you are integrating on a circle has the only x range (-2<x<2)

    and about the notation, first is better i advice to use it always.
     
    Last edited: Mar 1, 2005
  6. Mar 1, 2005 #5

    dextercioby

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    If i'm not mistaking,the square roots in "x" dissapear afer integrating wrt "y"...So the integral is possible...:wink:

    Daniel.

    P.S.Nothing is wrong.
     
  7. Mar 1, 2005 #6
    But it's not a logic integral on the area of that square, is it?
     
  8. Mar 1, 2005 #7

    dextercioby

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    Who gives a rat's a$$ what that integral represent,as long as it is correct?
    BTW,it's not an area at all...
    [tex]S=\iint_{D} dx \ dy [/tex]
    is the area of a plain domain from R^{2}.
    That integral is something else,as u may see...:wink:

    Daniel.
     
  9. Mar 1, 2005 #8
    Yes, Maybe I'm wrong, hehehe,
     
  10. Mar 1, 2005 #9

    Zurtex

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    But that makes more sense to me than the first, the first just seems confusing and looks like he is multiplying them.
     
  11. Mar 1, 2005 #10

    dextercioby

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    For "nice" cases,the theorem of iteration can be applied...But in this case,there's no multiplication/iteration,just an elegant way to saying what integration is performed first...

    Daniel.
     
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