The Area of a Circle as a Limit of Finite Approximations

We already know the exact area of a circle:

$$A=\pi R^2.$$

So the purpose of this experiment is not to derive the familiar formula again. Instead, we can use the circle as a case where the exact answer is already known and ask a different question: can we recover that exact value from a sequence of finite approximations?

One way to do this is to divide the circle into concentric rings. If we mentally cut and straighten a ring, we can approximately represent it as a long strip resembling a trapezoid. We replace it with a rectangle whose height corresponds to the length of the ring's inner circumference. Part of the area is then left unaccounted for, so the value obtained for each ring is smaller than its actual area.

We add the areas of all these rectangles to approximate the area of the entire circle. This transforms the geometry of the circle into a sum of simple rectangular areas.

The same idea extends far beyond circles: when the boundary of a region is curved and its area cannot be obtained by a simple geometric formula, we can divide it into increasingly small pieces, sum finite approximations, and study what happens as their size approaches zero.

Here, because we already know the exact answer, we can watch this process happen and compare every approximation with $\pi R^2$.

How does increasing the number of rings affect the accuracy of the result? What happens as the rings become increasingly thin? Can a sequence of finite approximations recover the exact area of the circle even though every finite approximation $S_n$ remains smaller than the exact area?

Constructing the approximation

As an example, consider a circle of radius

$$R=1.$$

We divide it successively into $4$, $10$, $1\,000$, and $1\,000\,000$ concentric rings.

If the circle is divided into $n$ rings of equal thickness, the thickness of each ring is

$$\Delta r=\frac{1}{n}.$$

Let the inner radius of a ring be

$$r_k=\frac{k}{n}, \qquad k=0,1,\ldots,n-1.$$

Thus, the inner radii of the rings are

$$0,\quad \frac1n,\quad \frac2n,\quad \frac3n,\quad\ldots,\quad\frac{n-1}{n}.$$

The length of each ring's inner circumference is

$$C=2\pi r.$$

When a ring is replaced by a rectangle, the rectangle's height equals the length of the inner circumference:

$$h=2\pi r,$$

and its width equals the thickness of the ring:

$$\Delta r=\frac1n.$$

Therefore, the area of one rectangle is

$$A_{\text{rect}}=2\pi r\Delta r.$$

To approximate the area of the entire circle, we add the areas of all the rectangles (using a left Riemann sum, since in our case the height of each rectangle is determined by the value of the function at the left endpoint of each interval):

$$S_n=\sum_{k=0}^{n-1}2\pi r_k\Delta r.$$

Here $k$ indexes the rings from $0$ to $n-1$. For each ring, its inner radius is

$$r_k=\frac{k}{n},$$

and the width of each interval is

$$\Delta r=\frac{1}{n}.$$

The summation notation expresses the same operation performed for every ring: calculate the area of its rectangle and add all the resulting areas.

Substituting the inner radii gives

$$S_n= 2\pi\frac1n \left( 0+\frac1n+\frac2n+\cdots+\frac{n-1}{n} \right).$$

Factoring out $1/n$ gives

$$S_n= \frac{2\pi}{n^2} (0+1+2+\cdots+(n-1)).$$

The sum in parentheses can be simplified by writing it once in ascending order and once in descending order:

$$0+1+2+\cdots+(n-2)+(n-1)$$ $$(n-1)+(n-2)+(n-3)+\cdots+1+0$$

Adding the two rows term by term gives $n$ terms, each equal to $n-1$. Thus two copies of the original sum equal $n(n-1)$, so

$$0+1+2+\cdots+(n-1)=\frac{n(n-1)}2.$$

Then

$$\begin{aligned} S_n &=\frac{2\pi}{n^2}\cdot\frac{n(n-1)}2\\ &=\pi\frac{n-1}{n}\\ &=\pi\left(1-\frac1n\right). \end{aligned}$$

Approximation error

To understand how far the result is from the actual area of the circle, we calculate the absolute and relative errors.

The absolute error shows how far the calculated value is from the exact value:

$$E_n=A-S_n.$$

Since, for $R=1$,

$$A=\pi,$$

we obtain

$$E_n=\pi-S_n=\frac{\pi}{n}.$$

The relative error shows what fraction of the exact value is represented by the absolute error:

$$e_n=\frac{\pi-S_n}{\pi}.$$

In general,

$$\text{relative error} = \frac{\text{absolute error}}{\text{exact value}}.$$

In our case,

$$\begin{aligned} e_n &=\frac{\pi-S_n}{\pi}\\ &=\frac{\pi/n}{\pi}\\ &=\frac1n. \end{aligned}$$

Results

For $n=1\,000\,000$, our approximation accounts for 99.9999% of the area, leaving only 0.0001% unaccounted for.

If $n$ is increased by a factor of 10, both the absolute and relative errors decrease by a factor of 10.

What happens for finite $n$?

No matter how large a value of $n$ we choose, for any finite number of rings the result remains smaller than the actual area.

Since

$$A=\pi R^2=\pi\cdot1^2=\pi,$$

and

$$S_n=\pi\left(1-\frac1n\right),$$

for every finite $n$,

$$\frac1n>0.$$

Therefore,

$$S_n<\pi=A.$$

No finite approximation becomes the exact area of the circle.

For $n=1$, we obtain

$$S_1=0.$$

This is not an error: the only rectangle is constructed using the inner radius $r_0 = 0$, so its height and area are both zero.

The limit

However, when

$$n\to\infty,$$

we have

$$\frac1n\to0.$$

Therefore,

$$\begin{aligned} \lim_{n\to\infty}S_n &=\lim_{n\to\infty}\pi\left(1-\frac1n\right)\\ &=\pi. \end{aligned}$$

In the sequence

$$S_1,S_2,S_3,\ldots,$$

each $S_n$ corresponds to a finite number of rings. There is no separate element $S_\infty$ in this sequence.

Thus,

$$\lim_{n\to\infty}S_n=\pi$$

does not mean that there is some final, infinitely accurate approximation. It is a statement about the behavior of the entire sequence: by increasing the finite value of $n$, we can make $S_n$ arbitrarily close to $\pi$.

So we can see that no finite approximation in the sequence gives the exact area of the circle. Yet the sequence itself has an exact limit. The exact value is determined not by a final infinitely precise approximation, but by the limiting behavior of the sequence of finite approximations.