# Average of COSIN

1. Aug 31, 2013

### lepori

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. Aug 31, 2013

### Office_Shredder

Staff Emeritus
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. Sep 1, 2013

### lepori

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

4. Sep 1, 2013

### DeIdeal

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, $A^{-1} \iint_A \cos^2(\theta) \mathrm{d}A$, 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
5. Sep 1, 2013

### HallsofIvy

Staff Emeritus
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 $2\pi$, 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, $\theta$ and $\phi$?

And do you mean the three dimensional ball or the surface of the sphere. The volume of a ball of radius R is $(4/3)\pi R^3$ while the surface area is $4\pi R^2$.

6. Sep 1, 2013

### lepori

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)

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I just do not understand, how to get it :)

7. Sep 1, 2013

### janhaa

maybe you can try this:

<cos2(x)>= 1/2 ∫cos2(x) sin(x) dx

and with appropriate limits...

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and similar with sinus if necessary
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edit;

I thought it was in the HW section

Last edited: Sep 1, 2013