How to prove that pie are squared

Hello, yeah, I understand it’s been like ages since I last posted. That’s because for the past few months I’ve virtually been taking 20 credit hours worth of classes, so I had to sacrifice most of the time I could have spent keeping my friends and family in touch with my incredibly interesting life.

What I want to talk about now is something related to geometry, trigonometry, and calculus, namely the area of a circle. Most people that have taken basic geometry in grade school can tell you the area of the circle is πr^2, but ask the average person why that is so and they wouldn’t have a flippin’ clue. I’m about to prove how exactly you solve for the area of a circle using analytical geometry and calculus.

In Figure 1, we see the geometry of the circle. The variable r is the radius of the circle, a is any arbitrary distance from the center of the circle, L is the length of the chord the distance a away from the center of the circle, and da represents a minute thickness of the segment dA with length L that contributes to the overall area of the circle. The summation of all of the dA’s (horizontal chords of minute thickness) is the exact area of the circle.

The area of the segment dA is L da. We know by the Pythagorean theorem that the length of L is twice the length sqrt(r^2-a^2). Thus:

To find the summation the total area of the circle, we will sum up all of the dA’s from a=−r to a=r:

This being a complex integral, the use of trigonometric substitution is necessary for solving it. In figure 2, the triangle is shown in more detail. The value for a could be replaced by a trigonometric equality, a=r sin θ.The integral therefore becomes:

One of the essential trigonometric identities we come across is 1-(sin θ)^2=(cos θ)^2. Since there are mixed variables in the integral, we must somehow make a substitution. A correlation between da and can be determined by the equation a=r sin θ.

When changing the variable, it is necessary to change the limits of the integral to correspond to the new variable. To find the value the limits change to, we solve for θ:

Making the final substitutions:

To complete the integration, we must recognize another trigonometric equality, namely
(cos θ)^2=(1 + cos 2θ)/2. We then derive:

And there you have it, the area of a circle. You might have thought it would be a little more simple than that, but you were wrong.

So, I realize I just made myself look like a nerd, but I don’t really care, because I just had a great time doing it. The end.