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Volume Integrals of a sphere

  1. Mar 30, 2008 #1
    Hey guys, could one of you explain why when doing a volume integral using spherical polar coordinates, you have the limits as 2 pi to 0 on phi but only pi to 0 on theta? Thanks.

    To clarify, I've been doing this all this time for questions, but it just occured to me that I Don't know why i do that :-p.
    Last edited: Mar 30, 2008
  2. jcsd
  3. Mar 30, 2008 #2


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    Hi Davio! :smile:

    (Of course, you can't go from o to 2π on both, because then you'd be covering every point twice!)

    Well, you can always do it the other way round … but the integrals that usually occur in practice just happen to be symmetric in phi, so integrating from 0 to 2π on phi is dead easy! :biggrin:

    In particular, if you're converting from (x,y,z), then you get ∫∫∫(r^2)sinthetadrdthetadphi … and that itself is symmetric in phi, so with luck the whole thing immediately becomes 2π∫∫r^2)sinthetadrdtheta. :smile:
  4. Mar 30, 2008 #3
    Hmmm, but wouldn't it be, for both phi and theta, just pi to zero - ie. half each? Or am imagining this wrong :-p. R is radius, so integrating along that place, in a circle, theta and phi are both angles, just in different directions? My maths is a bit poor (not good for a physics major)!
  5. Mar 30, 2008 #4


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    Hi, Davio!

    Take a line segment from the centre of the ball to its surface. That line segment makes the angle theta with the "z"-axis.

    In order to reach ALL points on the circle with the same angle to the "xy"-plane, you rotate the line segment around the z-axis, with phi then going from 0 to 2pi.

    Now, you will have covered ALL these circles (and hence all points) for theta-values going from 0 (i.e, the line segment runs along the positive z-half axis) to pi, (i.e along the negative half axis for z)
  6. Mar 30, 2008 #5


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    … don't ignore America …

    Yup, arildno is right!

    Start with a very long string, fixed at the North pole. Take the other end down the Greenwich meridian from theta = 0 to π. Now you're at the South pole.

    So far so good … :smile:

    Now sweep the string round from phi = 0 to π. You'll cover most of Europe and Africa and the whole of Asia and Australia.

    But you'll stop at the International Date Line!

    :rolleyes: What about America? :rolleyes:
  7. Mar 30, 2008 #6


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    One simple way to see it. Latitude goes from -90 to 90, while longitude goes from -180 to 180.
  8. Mar 31, 2008 #7


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    It's those blasted physicists again! They keep swapping [itex]\theta[/itex] and [itex]\phi[/itex] on us!
  9. Apr 1, 2008 #8
    I'm going to sit down and think about the replies in a minute, I'm on to a question about conical cones now, does anyone have any good resources for understand the images behind integration? I can integrate etc, but can't quite understand the limits of weird shapes, or even normal shapes!
  10. Apr 1, 2008 #9


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    … painting problem …

    Hi Davio! :smile:

    Do you mean comical cones?

    Now, they are weird! :biggrin:

    But keep this in perspective … this isn't an integration problem … it's only a painting problem.

    Imagine you have to program a robot to paint a sphere … do you tell it to paint from 0 to π, or 2π, for each coordinate?

    If you give it the wrong instructions, it'll either waste paint or not use enough! :frown:

    That's your only problem … making sure that everything is covered which should be! :smile:
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