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Average Radial Distance of Points within a Circle

Date: 03/26/2003 at 03:35:51
From: Dashiel
Subject: The Average Radial Distance of Points Within a Circle

I'm trying to determine the average value of a circular/radial 
gradient that is at full value (white, call it 100% brightness) in the 
center, and drops in a linear fashion to zero (black, call it 0% 
brightness) at the radius.

Frankly, I'm just not sure where to start, or what kind of math I'm 
getting into here. My math education leaves off at college level 
statistics, so if this is getting into calculus Iím probably out of 
my league, but I have a suspicion that it can be reduced at least to 
something within my level of understanding, and hopefully thatís 
where you good folks come in.

Mathematically speaking, I'm trying to determine the average radial 
distance (distance from the center) of all points within a circle 
with a specified radius. Essentially what I need is a simple equation 
for this that can be evaluated for various radii.

Date: 03/26/2003 at 12:24:58
From: Doctor Douglas
Subject: Re: The Average Radial Distance of Points Within a Circle

Hi, Dashiel,

Thanks for submitting your question to the Math Forum.

Yes, the most straightforward way of determining the answer does in
fact use calculus. I will try to explain where this comes from, and 
identify it with concepts that you're probably familiar with from your 
statistics course.

The average value of something (say a radius r) is given by

         r1 w1 + r2 w2 + ...
   <r> = ---------------------- 
           w1 + w2 + ...

where the r's are the radii of various pieces, and the w's are weights 
that are used in the averaging. This should be very familiar to you 
from your college statistics - it looks like an expectation of r:  

In the case of your problem, the weights are the *areas* of pieces 
with a given radius. So you identify these areas with a "distribution 
function" (of r):

   f(r) = (2*pi*r)               0 < r < R

This distribution function weights each radius according to the
circumference of a (very thin) annulus with that radius. Now, the
integral of a distribution function is 1, because the total
probability over all possible outcomes is 1. For this problem, the 
integral is

   Integral(r=0,r=R) f(r) dr = Integral(r=0,r=R) 2*pi*r dr
                             = pi*r^2 |(r=0,r=R)
                             = pi*R^2.

This should be expected - the sum of all of the individual areas had 
better be the total area of the circle, i.e., pi*R^2.

Now, let's calculate the quantity you want - i.e., the average radius.  
To do this, we simply compute its expectation:

   Integral(r=0,r=R) f(r) r dr = Integral(r=0,r=R) 2*pi*r^2 dr
                               = pi*(2/3)r^3 |(r=0,r=R)
                               = 2*pi*R^3/3

This is not quite the expectation, because this is only the numerator 
of the sum - we still have to divide by the sum of the weights (which 
we've already calculated above as pi*R^2). So, in the end we obtain

         Integral(r=0,r=R) f(r) r dr   2*pi*R^3/3    2R
   <R> = --------------------------- = ---------- = ----
          Integral(r=0,r=R) f(r) dr      pi*R^2       3

There's your formula for the "average" radius among all the points in 
a circle of radius R. It's certainly reassuring that this number is 
between zero and R. In fact we also know that the outer radii of the 
circle possess more area than the innermost radii - so that the final 
answer should be somewhere between R/2 (which simply splits the 
difference between the minimum and maximum radii), and R.

I hope this helps!

- Doctor Douglas, The Math Forum
Associated Topics:
High School Calculus
High School Statistics

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