# Representing the area of a circle as the sum of circumferences

I had this idea of calculating the area of a circle as the sum of circumferences where the radius is being shortened by an infinitesimal amount until it reaches 0. I was told that this was impossible, because an area can only be represented as a sum of smaller areas, but not lines. I tried to make a program that calculates the area of a circle by using this method, here's the algorithm:

This works, I've tested it with numerous examples. The bigger the n, the greater the precision.

My 2 questions are:

1. Why is this working if an area cannot be represented as a sum of lines?
2. Why must I divide the result given by this method by the reciprocal value of the value I shorten the radius with? As I said, the greater the n, the greater the precision. Thus this will be the exact area when n approaches infinity, but wouldn't I be dividing by infinity later? How does this give me an exact result?
• an area can only be represented as a sum of smaller areas, but not lines Whoever told you this hasn't learned calculus. – 6005 Dec 24 '16 at 14:54

Your algorithm computes the following: $$P = \frac{1}{10^n} \sum_{\substack{i \ge 0 \\ r - 10^{-n} i > 0 }} 2 \pi (r - 10^{-n} i)$$ For simplicity, let's assume that $r$ is an integer multiple of $10^{-n}$; for example, $r$ is an integer. Then we can write this as $$P = \frac{1}{10^n} \sum_{j = 1}^{10^n r}2 \pi j$$ This is a Riemann sum (and thus a good approximation for) for the integral $$\int_{0}^r 2 \pi r' \; dr' = \pi r^2,$$ and the approximation gets better the more subintervals you have, i.e. the higher $n$ is.
Your clever algorithm approximates the circle by a union of annuli of radius $r/10^n$. The area of each annulus is its circumference times its width.