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Average of COSIN

by lepori
Tags: average, cosine
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lepori
#1
Aug31-13, 05:24 PM
P: 8
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
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Office_Shredder
#2
Aug31-13, 05:38 PM
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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.
lepori
#3
Sep1-13, 02:30 AM
P: 8
in fact, my question is - how can I take average over sphere?..

DeIdeal
#4
Sep1-13, 03:34 AM
P: 128
Average of COSIN

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.
HallsofIvy
#5
Sep1-13, 07:04 AM
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Thanks
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P: 39,682
Quote Quote by lepori View Post
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].
lepori
#6
Sep1-13, 07:52 AM
P: 8
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 :)
janhaa
#7
Sep1-13, 08:34 AM
P: 97
Quote Quote by lepori View Post
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


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