- #1

- 12

- 0

how can I calculate average of cos

^{2}x ?

I want to take average over a sphere

I tried to do like this: <cos

^{2}X>= 1/2π ∫cos

^{2}xdx

and I get 1/2

but in my books, wrote that average of cos

^{2}x , taken over a sphere, is 1/3

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- Thread starter lepori
- Start date

- #1

- 12

- 0

how can I calculate average of cos

I want to take average over a sphere

I tried to do like this: <cos

and I get 1/2

but in my books, wrote that average of cos

- #2

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 4,434

- 517

1/2 is the average of cos

- #3

- 12

- 0

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

- #4

- 140

- 16

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.

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- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

By integrating over thein fact, my question is - how can I take average over sphere?..

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

- 12

- 0

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

- 97

- 3

maybe you can try this:

how can I calculate average of cos^{2}x ?

I want to take average over a sphere

I tried to do like this: <cos^{2}X>= 1/2π ∫cos^{2}xdx

and I get 1/2

but in my books, wrote that average of cos^{2}x , taken over a sphere, is 1/3

<cos

and with appropriate limits...

===

and similar with sinus if necessary

====

I thought it was in the HW section

Last edited:

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