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Integral of f over the surface of a sphere (in spherical polar coordinates)

  1. Nov 11, 2008 #1
    I have a function f(r, [tex]\phi[/tex], [tex]\vartheta[/tex]) = 3cos[tex]\vartheta[/tex].

    Evaluating the repeated integral of this function over the surface of a sphere, centered at the origin, with radius 5, I have come up with 0 as my result. I'm not sure if this is correct. I've double checked my calculations, and tried subdividing the surface S into smaller subsections and summing the integrals of each section, and I get the same result. Intuitively it makes some sense that the answer would be 0 since cos[tex]\var{theta}[/tex] takes on values either side of 0 for [tex]\vartheta[/tex] in the intervals [0, pi] and [pi, 2pi].

    Still, for some reason I'm uncomfortable with this result. Can anybody shed some light on this?
     
    Last edited by a moderator: Nov 12, 2008
  2. jcsd
  3. Nov 11, 2008 #2
    To find the repeated integral of f over the surface S with respect to dA, where dA is the limit of the small areas on S, I used the fact that [tex]dA = R^2 sin\theta \,d\phi \,d\theta[/tex].

    From there I calculated:

    [tex]
    \int \int_{S} f dA = \int^{\pi}_{0} \int^{2\pi}_{0} (3cos\theta 5^2 sin\theta) \, d\phi \, d\theta

    = \int^{\pi}_{0} \int^{2\pi}_{0} (75 cos\theta sin\theta) \, d\phi \, d\theta

    = \int^{\pi}_{0} [ 75 \phi cos\theta sin\theta ]^{2 \pi}_{0} \, d\theta

    = 150\pi \int^{\pi}_{0} cos\theta sin\theta \, d\theta

    = 150\pi [(1/2) sin^2\theta]^{\pi}_0
    [/tex]

    ... which of course equals 0 as [tex]sin\pi = sin0 = 0[/tex].
     
  4. Nov 11, 2008 #3

    Ben Niehoff

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    It looks correct to me.
     
  5. Nov 12, 2008 #4

    HallsofIvy

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    For any point on the surface of the sphere, [itex](\phi, \theta)[/itex] (you are using "engineering" notation here: [itex]\phi[/itex] and [itex]\theta[/itex] are reversed from what would be in a mathematics text) the diametrically opposite point is [itex](2\pi- \phi, \pi- \theta)[/itex] and [itex]cos(\pi- \theta)= -cos(\theta)[/tex]. Your function is "anti-symmetric" so its integral over the entire surface of the sphere is 0. The integral over any region around a given point is cancelled by the integral over its diametrically opposite point.
     
    Last edited: Nov 14, 2008
  6. Nov 14, 2008 #5
    Thanks for your help.
     
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