Calculating Average of COS2X over a Sphere

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SUMMARY

The average of cos²(x) over a sphere is definitively calculated as 1/3, contrary to the incorrect average of 1/2 obtained by integrating over the interval [0, 2π]. To compute the average over a sphere, one must perform a surface integral, specifically A⁻¹ ∫∫_A cos²(θ) dA, where A represents the surface area of the sphere. This method accounts for the spherical coordinates and ensures the correct averaging over all directions.

PREREQUISITES
  • Understanding of spherical coordinates and their application in integrals
  • Knowledge of surface integrals and their computation
  • Familiarity with trigonometric functions, specifically cos²(x)
  • Basic calculus concepts, including integration techniques
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  • Learn about surface integrals in spherical coordinates
  • Study the derivation of averages over different geometric shapes
  • Explore the properties of trigonometric functions in multi-variable calculus
  • Investigate the relationship between volume and surface area in three-dimensional geometry
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Mathematicians, physicists, and students studying calculus or geometry who need to understand averaging functions over spherical domains.

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