# The Area of a Square

26 Jul 2019

Here's a funny idea. Suppose someone forgot the formula for the area of a square, known since antiquity to be

It's ok, we forget things all the time. So without knowing this, let's try to figure this out with calculus. First, let's loosely define a square to be a two-dimensional, four-sided shape with all sides of equal length, whose sides meet at 90-degree (right) angles.

Figure 1. Plot of a square of sides \(s\), with solid black representative rectangle of height \(s\) and width \(dx\) inscribed in the square, located in the first quadrant of the Cartesian plane, starting from the origin.

**INTEGRATION** will be used to find the *area under the curve*. In this case, a flat (and rather boring) curve serves as the top boundary of the square, denoted by the solid black line, \(s\). Next, the area must be explicitly defined to include only the region contained inside the square, so the following boundary conditions are imposed:

We've restricted the area to be the closed (inclusive of endpoints) interval of \(0\) to \(s\) on both the \(x\) and \(y\)-axes, written compactly as \([0,s]\). The dotted line marks the right-most boundary of the square, closing off the area we want. Now, what's the black strip contained within the square? Let's imagine it to be a plank of Australian buloke wood flooring inside the square. It has a height \(s\), since it fits perfectly inside the square from top to bottom, and a width \(dx\). All we can say for now is that the plank has an infinitely small width of \(dx\) that can't be given an exact numerical value such as \(0.1\), \(0.01\), or even \(0.0000000001\). Also, the location of the plank highlighted inside Figure 1 is unimportant—it can be placed anywhere inside the bounded area of the square. So while \(dx\) can't be used immediately, it will be a critical component during the problem's construction.

Figure 2. Similar to Figure 1, but with representative rectangle removed, and replaced with many narrow bands inscribing and filling the square. The bands represent an infinite number of very small areas that are added together to find the whole area of the square. Don't bother counting or measuring the bands. That's not the point!

The big idea with integration is that by dividing the area of the curve into infinitely many planks, calculating the area of each plank, and adding them together, you've then computed the total area under the curve. Think of each plank as a "piece of square." Combine and add all the pieces of the square together like a jigsaw puzzle, and you've got the whole square. Likewise, just as a pizza is divided into eight slices with each of your friends getting one slice, they can all be said to have a piece of the pizza. Now if everyone were to combine all the slices together (though highly unlikely), it would again form the whole pizza.

Alright, time to construct the math. Line 1 sets up the integral that says starting from zero and ending with \(s\) along the \(x\)-axis, calculate each infinitely small plank of the area that has a height \(s\) and width \(dx\).

Line 2 is interesting. While \(s\) is a variable, it represents a number, any number you want that's \(\ge 0\). To be constant also means that it remains the same throughout the problem, and *does not depend on the \(x\)-axis*. Since \(s\) is independent of \(x\), it can be pulled outside of the integral and be used after the integral has been evaluated.

For line 3, recall what was said about the piece of square or piece of pizza. If you combined all the pieces of the square or pizza together, you get the whole square and whole pizza, respectively. In the same way, adding up all the pieces of \(dx\) becomes \(\Large{x}\).

In line 4, observe that in line 3, \(x\) is assigned a value of \(s\) on the top and \(0\) at the bottom. These are the limits of integration, or the upper and lower bounds of the interval, which is also placed at the ends of the fancy s-shaped integral sign (\(\int\)). Simply plug-in the top value into \(x\), and subtract from it the bottom value after it's been plugged-in for \(x\) as well, leaving a simple subtraction problem with \(s-0=s\).

While a trivial exercise, it still provides a fundamental understanding of what integration aims to achieve. Break up what you're trying to solve into very small pieces, get an answer for every small piece, then add all the pieces up to get an answer. The remaining lines reveal the area of the square to simply be the product of its sides, \(s^{2}\). Thankfully, no surprises there!