# Taking averages

Suppose I have a function $f(\theta)$ that is a function of the angle $\theta\in [0,2\pi)$.

Why is the average of $f$ over a large collection of randomly oriented objects:

$$\int f(\theta)\sin \theta\space d\theta ?$$

The $\sin{\theta}\space d\theta$ might be connected to the Jacobian for spherical coordinates? but I am not sure it makes sense to use that here?

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If $f$ is the constant function 17, then the average of $f$ is 17, but the integral is zero, so it would appear that you do not have a question. –  Gerry Myerson Jun 23 '12 at 12:10
@GerryMyerson: THanks, I think the average should just be ${1\over 2\pi}\int f(\theta)\,\,\,d\theta$ or if there is some sensible pdf, but clearly $\sin$ is not a valid pdf, so I really don;t understand what that is doing! O well... –  raymond Jun 23 '12 at 12:23
How is the angle $\theta$ distributed? –  ncmathsadist Jun 23 '12 at 12:41
@ncmathsadist: I would think uniformly, since the objects are "randomly distributed"? –  raymond Jun 23 '12 at 12:42
My guess is that this notion of averaging makes sense in the context of whatever source you encountered it in. Since you provided no context for the question, it is impossible to say whether this guess is true or not. –  user31373 Jun 23 '12 at 16:50