Skip to main content

Section 1.4 The area problem

After thinking about the tangent problem, we turn to the other foundational problem in calculus - the area problem.

Finding areas of different objects is a problem with a long history. Much of the earliest known mathematics focused on measuring lengths, areas, and volumes. Figure 1.4.1 shows an image of the so-called Moscow Mathematical Papyrus, a document dating to the Twelfth Dynasty of Egypt (~1850 BCE) that contains \(25\) problems with solutions. While many of the problems are algebraic in nature (ten of them are about calculating the strength of bread or beer), seven of them concern geometry.

Figure 1.4.1. Image credit: Hieroglyphic transcript by I. I. Perepelkin - Struve, Vasilij Vasilevič, and Boris Turaev. 1930. Mathematical papyrus of the State Museum of Fine Arts in Moscow, Public Domain.

Problem 10, for example, is about caculating the surface area of a basket. One translation reads:

Example of calculating a basket. You are given a basket with a mouth of 4 1/2. What is its surface? Take 1/9 of 9 (since) the basket is half an egg-shell. You get 1. Calculate the remainder which is 8. Calculate 1/9 of 8. You get 2/3 + 1/6 + 1/18. Find the remainder of this 8 after subtracting 2/3 + 1/6 + 1/18. You get 7 + 1/9. Multiply 7 + 1/9 by 4 + 1/2. You get 32. Behold this is its area. You have found it correctly!

If we work through this solution in more familiar notation, we find that the author had both the correct formula for the area and a pretty good approximation of \(\pi\text{,}\) \(\frac{256}{81}\approx 3.16\text{.}\) Approximating \(\pi\) is of particular importance as it is crucial to finding circumferences and areas of circles and surface areas an volumes of spheres.

Like the Egyptians, many cultures had approximations of \(\pi\) and some of them, like that of the Kerala School in India and the method of Zu Chongzhi (祖沖之) during the Southern Qi dynasty in China, have survived in the historical record. Archimedes, perhaps the greatest Greek mathematicians in antiquity, created what he called the method of exhaustion that is a wonderful illustration of one of the themes of the course: if you don't know how to solve a problem, approximate the solution using a problem you know and refine your result.

Archimedes framed the calculation of \(\pi\) as finding the circumfernce of a circle of radius one. While he didn't know a formula to calculate this, he did know how to find the perimeters of regular polygons. So, first he drew an inscribed square in the circle and calculated its perimeter. This is Archimedes first estimate of \(\pi\) and because the square is inside the circle, it must be below the actual value of \(\pi\) Then, he doubled each of the sides to form an octogon and calculated its perimeter. As the octogon is between the square and circle, it's perimeter is bigger than the square's but still less than the circle. Archimedes' trick was to keep doing this to get better and better lower bounds for \(\pi\) (see the animation in Figure 1.4.2). To get an upper bound he did similar iterations with circumscribed polygons whose perimeters will always be longer than the circle's. This cleverly squeezed the value of \(\pi\) between two numbers that were getting closer and closer together.

Figure 1.4.2. An animation of Archimedes' method of exhaustion.

We might call the method of exhaustion a solution to the length problem for the circle. We could also pose and solve a version of the area problem for the circle by calculating areas of the polygons rather than their perimeters. But how do we solve for areas of more complicated shapes?

A first observation is that no matter how complex a shape we have, we can always break it up into pieces that look like graphs of functions. This breaking up of our problem into lots of little problems allows us to solve each of them and put all the answers together to get the full area. But, we are still left with finding the area under a curve, which we don't know how to do!

Like Archimedes, we will approach the area problem using the solution to problems we already know to create an approximation of the area. The simplest geometric object for which we can compute the area is the rectangle - once we know the width and height, we can multiply them together to get the area. So, we will tile the area under a curve as best we can with rectangles.

Looking at Figure 1.4.3, we see the graph of the function \(f(x)=x^3-3x^2+8\) want to find the area under the curve between zero and four. We will construct rectangles that all have the same width, say one, and have heights so that the upper left-hand corner of the rectangle touches the curve. If we use the functions to calculate the heights at \(\{0,1,2,3\}\) we get \(\{8,6,4,8\}\text{.}\) Our approximate area is then

\begin{equation*} (8+6+4+8)\cdot 1=26. \end{equation*}

Figure 1.4.3. An animation of approximating areas by rectangles.
But, it is not quite right there are some places under the curve that aren't covered and some places above the curve that are. How can we fix this? Well, just like Archimedes, we can refine our estimate - instead of increasing the number of sides of a polygon, we will increase the number of rectangles.

A way to do this is to repeat the process with a smaller width. If we move to width one-half, the area estimate is 28.5 and we can see visually that the under- and over-counting is smaller. If we continue these iterations our approximations get better and better. If we could do this infinitely many times - which amount to the limit of the approximate areas as the width goes to zero or, equivalently, that the number of rectangles goes to infinity, we get the exact area we are looking for.