Compute average value of $x^{2}$ of a circle (calculus)

Question

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 $x$.

[1] http://calculus.seas.upenn.edu/?n=Main.Averages

Answer 1

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 $f(x,y)=x^2=R^2\cos^2(t)$.

Answer 2

For the circle $C$ with the origin as center and radius $1$ the sought average of $x^2$ is: \begin{align} a &=\frac{\int\limits_C x^2 \, ds}{\int\limits_C ds} \\ &=\frac{1}{2\pi} \int\limits_C x^2 \sqrt{dx^2 + dy^2} \\ &=\frac{1}{2\pi} \int\limits_C x^2 \sqrt{1 + y'(x)^2} dx \end{align}

For $y'(x)$ we get: $$ x^2 + y^2 = 1 \iff \\ y = \pm\sqrt{1-x^2} \Rightarrow \\ y'(x) = \mp\frac{x}{\sqrt{1-x^2}} $$ which gives \begin{align} a &=\frac{1}{2\pi} \int\limits_C x^2 \sqrt{1 + \frac{x^2}{1-x^2}} dx \\ &=\frac{1}{2\pi} \int\limits_C \frac{x^2}{\sqrt{1-x^2}} dx \\ &=\frac{2}{\pi} \int\limits_0^1 \frac{x^2}{\sqrt{1-x^2}} dx \\ \end{align} where the last equation used the symmetry of the integrand over the circle.

We have \begin{align} \int \frac{x^2}{\sqrt{1-x^2}} dx &= \frac{1}{2} \left( \arcsin(x) -x \sqrt{1-x^2} \right) + C \end{align} for some integration constant $C$, as \begin{align} (\arcsin(x) - x \sqrt{1-x^2})' &= \frac{1}{\sqrt{1-x^2}} - \sqrt{1-x^2} + \frac{x}{2\sqrt{1-x^2}}(2x) \\ &= 2\frac{x^2}{\sqrt{1-x^2}} \end{align} So the end result is $$ a = \frac{2}{\pi} \frac{1}{2} \frac{\pi}{2} = \frac{1}{2} $$