Struggling to solve a seemingly simple average problem using integral calculus.
We are given a circle of radius $1$, given by the equation $x^2+y^2=1$. The average value of $x$ over the circle is 0. What is the average value of $x^2$ over this circle?
Because this is a curve we are supposed integrate with respect to the arc length element $dL$. Following parametrization is suggested:
$x=\cos t$, $y=\sin t$, where $1 \le t \le 2\pi$
I came up with the following $dL$ element: $dL = \sqrt {\sin^2 t + \cos^2 t} \space dt$. Because the sum of $\sin^2 t+\cos^2 t=1$ it becomes just $dL=dt$. The integral of that is just $t$.
At this point I'm lost. It appears that I should use root mean square method [1] but I don't understand how can I translate what I came up with for $dL$ to $x$.
Best Answer
Hint. You are on the right track. The average value of a function $f(x,y)$ over the circle $x^2+y^2=R^2$ is given by $$\frac{1}{2\pi R}\int_{0}^{2\pi}f(R\cos(t),R\sin(t))\cdot Rdt= \frac{1}{2\pi}\int_{0}^{2\pi}f(R\cos(t),R\sin(t))\cdot dt.$$ In your case $f(x,y)=x^2=R^2\cos^2(t)$.