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Average of COSIN

  1. Aug 31, 2013 #1

    how can I calculate average of cos2x ?
    I want to take average over a sphere

    I tried to do like this: <cos2X>= 1/2π ∫cos2xdx
    and I get 1/2

    but in my books, wrote that average of cos2x , taken over a sphere, is 1/3
  2. jcsd
  3. Aug 31, 2013 #2


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    What sphere are you trying to average it over?

    1/2 is the average of cos2(x) on the interval [0,2pi], which is something that nobody would call a sphere.
  4. Sep 1, 2013 #3
    in fact, my question is - how can I take average over sphere?..
  5. Sep 1, 2013 #4
    What's x, is it something "specific"? Because if it happens to be, for example, the polar angle in spherical coordinates, then my guess is that you're supposed to calculate a surface integral over a sphere, [itex] A^{-1} \iint_A \cos^2(\theta) \mathrm{d}A[/itex], where A is the surface area of a sphere and dA is the area element. The radius will cancel out. This gives you the correct answer, but it could obviously be something else as well. But, like Office_Shredder said, just integrating over the interval [0,2π] won't do, you're certainly not taking the average over a sphere that way.
    Last edited: Sep 1, 2013
  6. Sep 1, 2013 #5


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    By integrating over the sphere and dividing by the surface of the sphere- it looks like what you did was integrate over a circle and divide by [itex]2\pi[/itex], the length of a circle.

    To further confuse things, your function, cos(x) depends only on a single variable, x. Is that what you intended or did you mean to have a function of all three variables, x, y, and z- or in polar coordinates, [itex]\theta[/itex] and [itex]\phi[/itex]?

    And do you mean the three dimensional ball or the surface of the sphere. The volume of a ball of radius R is [itex](4/3)\pi R^3[/itex] while the surface area is [itex]4\pi R^2[/itex].
  7. Sep 1, 2013 #6
    we have a function:


    X is angle between two vector, if the vectors direction is random, then averaging over all directions would be yield

    G(t) = 1/3 +2/3*cos(wt)

    I just do not understand, how to get it :)
  8. Sep 1, 2013 #7
    maybe you can try this:

    <cos2(x)>= 1/2 ∫cos2(x) sin(x) dx

    and with appropriate limits...

    and similar with sinus if necessary

    I thought it was in the HW section
    Last edited: Sep 1, 2013
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