0
$\begingroup$

I have function:

$\sin^4\left(\frac{r\pi}{R}\right)\frac{r}{R}$

where $R$ is a constant and $r$ has a maximum value $r_{max}=R$. I want to average the function over area $\pi R^2$. How can I do this ?

The function $f(r)$ is a radial function and $x^2+y^2.

  • 0
    Yes, function$f(r)$is a radial function.2012-03-23

1 Answers 1

1

I'm going to assume that your function $f(r):=\sin^4 (\pi r/R)\ (r/R)$ is the radial expression of the radial function of two real variables $f(x,y):= \sin^4 (\pi \sqrt{x^2+y^2}/R)\ (\sqrt{x^2+y^2}/R)$ defined in the disc $D:=\{ (x,y)\in \mathbb{R}^2:\ x^2+y^2\leq R^2\}$, because otherwise it will have no meaning to average $f(r)$ over an area.

The average of such an $f(x,y)$ over $D$ is then given by: $\overline{f} := \frac{1}{\operatorname{area}(D)}\ \iint_D f(x,y)\ \text{d}x\text{d}y\; ;$ the integral in the RH side can be computed using polar coordinates, hence: $\begin{split} \overline{f} &:= \frac{1}{\pi R^2}\ \int_0^R \int_0^{2\pi} f(r)\ r\ \text{d}r\text{d}\theta \\ &= \frac{2}{R^2}\ \int_0^R f(r)\ r\ \text{d}r\\ &= \frac{2}{R^2}\ \int_0^R \sin^4 \left(\pi \frac{r}{R}\right)\ \frac{r^2}{R}\ \text{d}r\\ &\stackrel{t=\pi r/R}{=} \frac{2}{\pi R^2}\ \int_0^\pi t^2\ \sin^4 t\ \text{d}t\; . \end{split}$ The latter integral can be computed using standard integration by parts techniques, which (after lenghty and tedious computations) yield: $\begin{split} \int_0^\pi t^2\ \sin^4 t\ \text{d}t &= \frac{1}{256} \left( 32 t^3+ 32(2t^2-1)\sin 2t +(8t^2-1)\sin 4t -64t\cos 2t +4t\cos 4t\right)\Bigg|_0^{\pi}\\ &= \frac{\pi}{64}\ (8\pi^2-15)\; , \end{split}$ therefore: $\overline{f} = \frac{8\pi^2-15}{32R^2}\; .$

  • 0
    This is a great answer. It helps a lot. I was looking for an average by taking square of the function and multiplying by $2*\pi r$. Which was not clear to me.2012-03-23