The use of rectangles and triangles to calculate the areas of curved planar shapes goes back thousands of years.

For example, the area of a triangle is pretty easy to calculate: it's just one half the base times the height. But imagine taking a right triangle (red) and deforming the hypotenuse into a parabola (blue).

The blue region fits nicely inside of the red one, which allows us to come up with an upper bound on the area of the blue region. What upper bound on the area does this observation give us?

In the last problem, we used a triangle to estimate a parabolic area, but it turns out that **limits** help us compute the area exactly!

In this unit, we'll explore in modern terms a technique similar to what the ancient mathematician Archimedes used to compute the parabolic area. We'll actually do this calculation much later in the course, but we won't leave you in suspense: the answer will turn out to be \( \frac13.\)

Instead of tackling this difficult problem right now, we'll learn the basic ideas underlying **integral calculus**, the second major pillar of our course, by solving an easier **area problem**. Specifically, we're going to compute the area of a circle from scratch.

This unit will rely quite a bit on basic geometry and trigonometry.

Here's the bird's-eye view of the unit. We start with a regular 3-gon (or triangle) inscribed inside a large blue circle of radius \(r_{\text{outer}}.\) This triangle has a smaller circle (red) inscribed within. The radius of this circle is \(r_{\text{inner}}.\) (See the gif below if you have a hard time visualizing this!)

As we add more vertices to make a regular 4-gon, 5-gon, etc., the circles get closer and closer together and the \( n \)-gon looks more and more like a circle itself.

Over the next few problems, we'll find the area formula of the \(n\)-gon and use it to find the area formula for a circle in the limit as \( n\) gets very, very large.

For the same figure, if the area of the solid black triangle is \( A_{\triangle}, \) find \[\frac{A_{\triangle}}{r_{\text{outer}}^2}. \]

**Hint:** Trigonometry tells us that the side opposite \( \theta \) has length \( 2 r_{\text{outer}} \sin \left( \frac{\theta}{2} \right), \)
and the height of the triangle formed by the solid black line opposite \(\theta\) and the two blue dashed lines is \( r_{\text{outer}} \cos \left( \frac{\theta}{2} \right). \)

Since the solid black triangle is “between” the red and blue circles, we know that \[ A_{\text{red circle}} \leq A_{\triangle} = \frac{3 \sqrt{3}}{4} r_{\text{outer}}^2 \leq A_{\text{blue circle}}. \]

For any \(n\)-gon, we'll have \[ A_{\text{red circle}} \leq A_{n} \leq A_{\text{blue circle}}. \]

Furthermore, \( A_{\text{red circle}} \to A_{\text{blue circle}} \) as \( n \to \infty.\) Here, “\( \to \)” means “approaches.” So if we take \(n\) very large, \( A_{n} \) will give us the area formula for a circle!Now that we understand the triangle case, let's look at the figure for \( n = 5.\)

If the area of the solid black pentagon is \( A_{5}, \) find \[\frac{A_{5}}{r_{\text{outer}}^2}. \]**Hint:** Use \( \sin(36^{\circ}) \approx 0.6\) and \(\cos(36^{\circ}) \approx 0.8. \)

From the gif, we know that \( A_{\text{red circle}} \leq A_{n} \leq A_{\text{blue circle}} \) and we expect that \( A_{\text{red circle}} \to A_{\text{blue circle}} \) as \( r_{\text{outer}} \to r_{\text{inner}} \) for \(n\) very, very large.

The Python codex below calculates \( \frac{A_{3}}{r_{\text{outer}}^2} , \frac{A_{3}}{r_{\text{outer}}^2}, \dots , \frac{A_{N_{max}}}{r_{\text{outer}}^2} \) given \( N_{max} \). Remember, \[ \frac{A_{n}}{r_{\text{outer}}^2} = n \sin \left( \frac{180^{\circ}}{n} \right) \cos \left( \frac{180^{\circ}}{n} \right).\] Choose various (large!) values for \( N_{max} \) and run the code.

What value do these numbers appear to approach?

It's amazing to think that Eudoxus and Archimedes used similar methods long before Newton and Leibniz introduced the world to calculus.

It's tempting to put \( n = \infty \) into the formula \( \frac{A_{n}}{r_{\text{outer}}^2} = n \sin \left( \frac{180^{\circ}}{n} \right) \cos \left( \frac{180^{\circ}}{n} \right) \), but as we'll see, \( \infty \) is not a number and we can't just plug it into a function as if it were. In other words, writing \[ \infty \sin \left( \frac{180^{\circ}}{\infty} \right) \cos \left( \frac{180^{\circ}}{\infty} \right) = \pi \] makes no sense at all.

However, we'll learn that a proper way of summarizing what we've uncovered here is to write \[ \lim\limits_{n \to \infty} n \sin\left(\frac{180^{\circ}}{n}\right) \cos \left( \frac{180^{\circ}}{n} \right) = \pi, \] which just means that the pattern in the numbers \( n \sin \left(\frac{180^{\circ}}{n}\right) \cos \left( \frac{180^{\circ}}{n} \right) \) as \( n\) tends to large values is \( \pi.\)

We've uncovered the area formula for a circle \( A = \pi r^2. \)

Let's see if we can find the *circumference* of the circle in terms of the radius.

Using the same sort of reasoning and assuming that \(\lim\limits_{n \to \infty} n \sin\left(\frac{180^o}{n}\right) = \pi, \) find the formula for the circumference of a circle in terms of its radius.

**Note:** You may wonder about what happened to the \( \cos\left(\frac{180^o}{n}\right) \) term in the limit \(\lim\limits_{n \to \infty} n \sin\left(\frac{180^o}{n}\right) = \pi. \) It turns out that this term approaches 1 as \( n \to \infty.\) We'll prove it later in the course, but experimenting with some large values of \(n\) should be pretty convincing at this stage.

Each of the \(n\)-gons can be thought of as a collection of \(n\) congruent triangles. We found the area of the circle by taking the sum of these triangular areas and then letting \( n\) tend to \( \infty.\)

It's natural to wonder if a similar idea can find other areas, like the area bounded by \(y=x^2, y = 0, x = 0, x = 1\) that we considered at the beginning of this unit.

With just a few rectangles we can find a pretty decent approximation to the area. With a few more, the approximation gets even better, just as adding more and more vertices to our \(n\)-gon gave us a shape closer and closer to a circle!

This unit laid out the approach we'll take in later chapters to solve the **area problem**: use \(n\) straight-edge figures to approximate the total area, then take the limit of this sum as \( n \rightarrow \infty.\)

The area problem opens up the vast vistas of **integral calculus**, which we'll explore in Chapters 6-9. It's surprising to realize how this simple notion allows us to compute a wide range of interesting quantities from physics, finance, chemistry, and engineering.

We've reached the end of the beginning. This short chapter gave us a glimpse of the fundamental ideas of calculus. Through calculus, we can compute rates of change, which give us the chance of finding extreme function values. Calculus also opens up the way to computing areas and understanding what an infinite sum truly means.

Behind all of these marvelous notions is the limit, to which we turn our attention in the next chapter.

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