Note: Over the course of this summer, I have taken both Geometry and Precalculus, and I am very excited to be taking Calculus 1 next year (Sophomore for me). In this question, I will use things that I know from Calculus, but I emphasize that I have not taken the course, so please bear with me. This will be long.
Among other Geometric formulas that I have learned recently, I am currently trying to prove that the surface area of a sphere is $4\pi r^2$. Intuitively, this seems fairly straightforward for me, since I have already proven the volume of a sphere using integration to be $\frac{4}{3} \pi r^3$, and that the integral of the circumference of a circle is its area. Using these two facts together, it makes sense that the integral of the surface area of a sphere should be its volume, leading me to believe the formula for surface area stated above.
However, this is a very weak argument, since I make the connection between circumference to area and area to volume without much proof. As an alternative, I made a proof involving infinite sums of the lateral areas of cylinders squeezed into a sphere. One can use this method of infinite cylinders to prove the volume of a sphere, but when I tried almost the exact same thing for surface area, I found that the value of pi is 4.
I have only heard of one other strange proof that leads to this result, and it can be found here: Is value of $\pi = 4$. It was promptly disproved in numerous ways, some of them here:How to convince a layman that the π=4 proof is wrong?
As far as I know, there's nothing wrong with my math -- I most likely simply set up the problem incorrectly. Here's what I did:
Consider a sphere of radius $r$, aligned with the coordinate plane in a way that its center is the origin. To approximate the surface area (or volume) of this sphere, we can imagine fitting $n$ stacked cylinders into it, each with a center on the line $x = 0$. Each cylinder has a height $h$, or $\Delta y$.
Since the height of the sphere is $2r$,
$h = \frac{2r}{n}$
Looking at a cross-section of this sphere, the radius of each cylinder must satisfy the equation:
$r_i^2 + y^2 = r^2$
where $r_i$ is the radius of the cylinder with index $i$, and the radius of the sphere is $r$. Now, we sum the lateral areas of the cylinders to get an approximation for the surface area of the sphere. As $n \rightarrow \infty$, our approximation gets better, so
$A = \lim_{n\rightarrow\infty}{\sum_{i=1}^n{2\pi r_i (\frac{2r}{n})}} = \lim_{n\rightarrow\infty}{\sum_{i=1}^n{2\pi r_i \Delta y}}$
In other words:
$\int_{-r}^r{2\pi r_i dy}$
And as we already know,
$r_i = \sqrt{r^2-y_i^2}$
So we substitute and take out constants:
$2\pi \int_{-r}^r{\sqrt{r^2-y^2} dy}$
From here, we have two options: 1) Take the integral and find something very messy; 2) Recognize that the expression inside the integral is just a semicircle, and that a semicircle has half the area of a circle.
Choosing option 2), I end up with:
$2\pi \frac{\pi r^2}{2} = \pi^2 r^2$
Which is clearly not the surface area of a sphere, but I can't figure out why. Strangely, another "proof" I did also led to this result. Take the semicircle $y = \sqrt{r^2 - x^2}$ with arclength $\pi r$ and rotate it about the x axis $2\pi$ radians. We now have a sphere, with surface area $2\pi^2 r^2$. Something is clearly wrong here, but it gets stranger. If we simultaneously accept Archimedes proof of the surface area of a sphere, we find:
$\pi^2 r^2 = 4\pi r^2$
And by "solving for $\pi$," we find that: $\pi = 4$.
I'm not looking for a better proof or someone to convince me that Archimedes is right, as I fully accept the textbook formula and have used other proofs to show it. I have a feeling that I may have approached the sphere in a "non-smooth" way, since the zig-zag shape that the cylinders make is eerily like the method used in the classic $\pi = 4$ proof, and when I used polygonal approximations I got a valid answer. Thanks for reading all the way through this, and does anyone know how I messed up?