Average of COSIN

  1. hi,

    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. Office_Shredder

    Office_Shredder 4,487
    Staff Emeritus
    Science Advisor
    Gold Member

    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. in fact, my question is - how can I take average over sphere?..
  5. 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. HallsofIvy

    HallsofIvy 41,260
    Staff Emeritus
    Science Advisor

    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. 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. 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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