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Triple integral, limits of integration

  1. May 18, 2013 #1
    Pretty general question.

    Integrate f(x,y,z) dxdydz over the area defined by:

    [itex]x^{2} + y^{2} + z^{2} \leq 4[/itex]
    [itex]x \leq 0[/itex]
    [itex]y \leq 0[/itex]
    [itex]z \leq 0[/itex]

    It is immidiately apparent that it is 1/8 of a sphere with r=2. So from that geometrical intuition we can do a variable substitution to spherical coordinates and use the following limits of integration.

    0 < r < 2
    0 < θ < pi/2
    0 < σ < pi/2

    Or something. What I'm wondering is: how would you go about finding these limits algebraically?? Let θ be the angle to the z axis and σ be the angle between the x and y axis and you would get

    0 < θ < pi/2

    But how would you figure out the angle between x and y?
     
  2. jcsd
  3. May 18, 2013 #2

    Zondrina

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    Spherical coordinates :

    ##x = rcosθsin \phi##
    ##y = rsinθsin \phi##
    ##z = rcos \phi##
    ##x^2 + y^2 + z^2 = r^2##
    ##|J| = r^2sin \phi##

    For starters, your limits for r are incorrect. Check them again.
     
  4. May 18, 2013 #3
    What? How are my limits for r incorrect?

    x^2 + y^2 + z^2 < 2^2
    r^2 < 2^2
    r < 2
     
  5. May 18, 2013 #4

    Zondrina

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    ##x^2 + y^2 + z^2 = r^2##

    So you get ##r^2 ≤ 4## which implies that ##|r| ≤ 2##.
     
  6. May 18, 2013 #5

    LCKurtz

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    No, you can't use those limits without knowing some kind of symmetry conditions on the integrand ##f(x,y,z)##. Changing to spherical coordinates is appropriate, but draw a 3d coordinate system and use it to figure out the appropriate ##(\rho,\phi,\theta)## limits for that back octant.
     
  7. May 18, 2013 #6

    tiny-tim

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    yes, as you say it's an octant, they're all the same, so you certainly can choose the "positive" octant and integrate to find the volume of that (if you don't already know the formula for the volume of a sphere! :wink:)
    isn't it obviously π/2 ? and haven't you already written 0 < σ < π/2 ? :confused:

    (perhaps i'm misunderstanding your question)​

    as an exercise, what do you think the limits are for the given ("negative") octant? :smile:
     
  8. May 18, 2013 #7

    LCKurtz

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    But they aren't all the same unless you have special properties in the integrand ##f(x,y,z)##.
     
  9. May 18, 2013 #8
    But it's a radius. It doesn't have a negative radius.
     
  10. May 18, 2013 #9
    Well yes it is obvious if you appeal to geometric intuition, I just wanted to know if there was a way to do it algebraically.
     
  11. May 18, 2013 #10

    LCKurtz

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    and if you ignore the fact that you can't do it that way.
     
  12. May 19, 2013 #11

    tiny-tim

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    Sorry, but I've no idea what you mean by "algebraically". :confused:

    How could the angle between x and y be anything other than π/2 ?
     
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