Average of COSIN

  • Thread starter lepori
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  • #1
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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
 

Answers and Replies

  • #2
Office_Shredder
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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.
 
  • #3
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in fact, my question is - how can I take average over sphere?..
 
  • #4
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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:
  • #5
HallsofIvy
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in fact, my question is - how can I take average over sphere?..
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].
 
  • #6
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we have a function:

G(t)=cos(x)^2+sin(x)^2*cos(wt)

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 :)
 
  • #7
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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
maybe you can try this:

<cos2(x)>= 1/2 ∫cos2(x) sin(x) dx

and with appropriate limits...

===
and similar with sinus if necessary
====
edit;

I thought it was in the HW section
 
Last edited:

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