hi, 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
What sphere are you trying to average it over? 1/2 is the average of cos^{2}(x) on the interval [0,2pi], which is something that nobody would call a 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, [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.
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].
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 :)
maybe you can try this: <cos^{2}(x)>= 1/2 ∫cos^{2}(x) sin(x) dx and with appropriate limits... === and similar with sinus if necessary ==== edit; I thought it was in the HW section