I am supposed to find the average value of the area of circles whose radii vary from $0$ to $R$.
I know that the formula for the arithmetic mean of a function $f$ on an interval $[a,b]$ is
$$\frac{1}{b-a} \int_a^b f(x)\,dx.$$
So for these circles it seems like it would now be this:
$$\begin{align}
\mu &= \frac{1}{R} \int_0^R \pi r^2\,dr \\
&=\frac{\pi}{R} \left(\frac {R^3}{3}\right) \\
&=\frac{R^3 \pi}{3}
\end{align}$$
So I try and verify this using three circles that are easy to calculate, of radius $1$, $2$ and $3$.
Using $$\frac{ \pi + 4 \pi + 9 \pi}{3}$$
gives me $\frac{14\pi}{3}$
Using my formula gives me $9 \pi$.
This makes me reevaluate what I am doing; it seems like my previously discovered formula accounts for a circle of radius $0$. I compensate for that Using $$\frac{ \pi + 4 \pi + 9 \pi}{4} = \frac{7\pi}{2}$$
Still wrong, what did I do incorrectly?
Best Answer
As expalined in the comments, you have an extra $R$, so that the formula should read: $$R^2\pi \over 3$$ I'd like to relate this to the discrete calculation you did. For $n$ circles of radius $1,2..n$, the average area is: $${\pi \over n} \sum_{i=1}^n i^2 = {\pi\over6} (n+1) (2 n+1)= {\pi n^2\over 3} + O(n)$$ Thus, you see that as the number of concentric circles becomes larger, approximating the integral by discrete concentric circles becomes closer to your integral formula.