Averege of Cosin

gagnidze

Dear mathmeticas,

how can I take average of cos(x)^2 over sphere?...

not to (0,2 pi) interval, but sphere

sigmundv

What do you mean by average in this case? How is the problem formulated?

Michael Collins

But [latex] \cos^2(x) [/latex] appears to be a function of one real variable, with the domain being the real line, so your question can't be answered. You'd need to describe a function which has a surface as a domain, preferably defined in terms of a polar and azimuth angle, i.e.

[latex] \displaystyle (\theta, \phi) \mapsto f(\theta, \phi) [/latex],

then, as sigmundv says a defintion of average would be helpful, although if I had to guess I'd imagine you mean the standard average which is the answer to the question: what constant function would integrate to the same value over the same domain.

ray giraffe

A poorly defined question, typical of a poor lecturer.

I believe x is intended to be the polar angle [latex]\theta[/latex]. (If x is the azimuth angle [latex]\varphi[/latex], then it's much easier!)

[latex] \frac{ \displaystyle \int_{S} f(\theta,\varphi) \, \mathrm{d}S_r }{\displaystyle \int_{S} (1) \, \mathrm{d}S_r } [/latex]
=
[latex] \displaystyle \frac{ \displaystyle \int_{\varphi=0}^{2 \pi} \int_{\theta=0}^{\pi} f(\theta,\varphi) \, r^2 \sin \theta \,\mathrm{d}\theta\ \mathrm{d}\varphi}{\displaystyle \int_{\varphi=0}^{2 \pi}\int_{\theta=0}^{\pi} (1) \, r^2 \sin \theta \, \mathrm{d}\theta\ \mathrm{d}\varphi} [/latex]
=
[latex]\displaystyle \frac{\displaystyle \int_{\varphi=0}^{2 \pi} \int_{\theta=0}^{\pi} (\cos^2 \theta) \, \sin \theta \, \mathrm{d}\theta\ \mathrm{d}\varphi}{\displaystyle \int_{\varphi=0}^{2 \pi} \int_{\theta=0}^{\pi} (1) \, \sin \theta \, \mathrm{d}\theta\ \mathrm{d}\varphi}[/latex]
=
[latex]\displaystyle \frac{\displaystyle \int_{\theta=0}^{\pi} (\cos^2 \theta) \, \sin \theta \, \mathrm{d}\theta\ }{\displaystyle \int_{\theta=0}^{\pi} (1) \, \sin \theta \, \mathrm{d}\theta\ }[/latex]

We are 'weighting' the function [latex]\cos^2 \theta[/latex] by the 'amount' of surface at polar angle [latex]\theta[/latex], which is proportional to [latex]\sin(\theta)[/latex].

MathsManiac

ray giraffe wrote: »

A poorly defined question, typical of a poor lecturer.

...or perhaps typical of a student who thinks they are telling us what they were asked to do but are actually giving us an "edited" or otherwise mangled version of what they were asked to do!

sigmundv

MathsManiac wrote: »

...or perhaps typical of a student who thinks they are telling us what they were asked to do but are actually giving us an "edited" or otherwise mangled version of what they were asked to do!

I believe this to be the case to be honest!

ray giraffe

MathsManiac wrote: »

...or perhaps typical of a student who thinks they are telling us what they were asked to do but are actually giving us an "edited" or otherwise mangled version of what they were asked to do!

You're right, it could well be the student's fault.

But I had a professor in college who gave us poorly-defined questions like that pretty often. It was a pain in the a**.

Maybe he was so used to conversing at a high level with other professors who were able to interpret him. Or maybe just lazy.