# False proof: $\pi = 4$, but why?

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?

• What's wrong with the answers given to the questions you linked to? – nbubis Aug 9 '15 at 16:10
• Wait, why is $2\pi\frac{\pi r^2}{2} = \pi^2r^2$ blatantly false? I would consider it obviously true, as you get the right hand side by cancelling the $2$ on the left hand side. I certainly cannot derive any value of $\pi$ from that equation. – celtschk Aug 9 '15 at 16:14
• They were responding to a different question. I don't have enough reputations to post a picture, it seems, but the cylinders in my question are partially inside of the sphere. Many of the answers to the questions I linked (at least the ones I could understand) explained that the "zig-zag" line is always outside of the circle, so it is always slightly longer than the circle's circumference. In my question, the cylinders are not always on the exterior of the sphere. I'm not sure that the same answers apply. – biffletsbq Aug 9 '15 at 16:19
• @celtschk, sorry for the confusion. I meant that it is blatantly false that a sphere has that surface area. – biffletsbq Aug 9 '15 at 16:20
• Some fantastic answers below. Here is something you may find fashionable at this point: en.wikipedia.org/wiki/Coastline_paradox – Colm Bhandal Aug 9 '15 at 19:09

## 5 Answers

One thing that's true about volume but not surface area (or, down a dimension, true about area but not perimeter) is that if shape $A$ contains shape $B$, then $A$ has a larger volume than $B$. In symbols: $$A\supseteq B\implies\operatorname{volume}(A)\ge\operatorname{volume}(B)$$ This means that it's fairly easy to approximate volumes — just find the volume of something that contains $A$ to get an upper bound, and find the volume of something that's contained in $A$ to get a lower bound. Now, to get the volume exactly: if we can get better and better approximations, then we can just take the limit as the two approximations get increasingly better. Hopefully, they approach the same limit (call it $a$) — and, except for weird cases, they usually do approach the same limit — so we have that the volume of $A$ is bounded above and below by $a$. That is, $a\le\operatorname{volume}(A)\le a$. Thus, we have that the volume is exactly $a$.

This all goes out the window with surface area. Let's say $A$ is a smooth shape, and $B$ is some really spiky thing contained in $A$. Now, it's entirely possible that $B$ has a larger surface area than $A$. So we get no upper or lower bounds. This means that we have no reason to expect our approximations to be accurate whatsoever. And, as you've found out, they aren't always accurate.

There is a way to save this. If $A$ and $B$ are both convex, then we do have that $A$ has a larger surface area than $B$, and our approximations do work. And we can use limits to get the volume exactly. However, I'm pretty sure your union of cylinders isn't convex.

• Thanks for the answer. This also explains why polygonal approximations give a more reasonable result, since regular polygons are inherently convex. – biffletsbq Aug 9 '15 at 16:28
• @biffletsbq By the way, the way you approximate the length of arbitrary curves is to break it up into lots of small parts, replace each part with a line connecting the endpoints, and sum the length of the lines. It's basically the polygon method but more general (because the curve doesn't need to be a loop). This works even if the curve isn't convex. – Akiva Weinberger Aug 9 '15 at 17:00
• Note that analogies don't always immediately go for higher dimensions. One can approximate any curve with a polygonal chain, and the chain total length tends to a curve length when the longest segment in a chain tends to zero length. However, when we approximate a surface with a polyhedral mesh, making the mesh faces tiny doesn't suffice to get a surface area approximation – IIRC we need also keep faces 'round' enough; for triangular mesh it means keeping all triangles' angles greater than some constant minimum value. – CiaPan Aug 10 '15 at 10:14
• When you talk about the smooth shape $A$ containing the spiky thing $B$, do you mean that $A$ and $B$ are three-dimensional, with $B$'s volume contained in $A$'s? I can see how that could lead to $B$ having an arbitrarily large surface area but I don't see it if they're two-dimensional shapes. – David Richerby Aug 10 '15 at 12:40
• My most voted-for answer is one that I wrote while partially sleep deprived… How did this happen? – Akiva Weinberger Aug 10 '15 at 14:57

The layman will be convinced by considering the 2D version of your approach.

Decompose a circle as a stacking of $n$ rectangles (this is what you see when taking the section of your stacked cylinders).

The 2D version of your claim would be that the length of the circumference is the sum of the "lateral lengths", i.e. the sum of the rectangle widths. This is obviously false as this sum equals the length of the diameter, whatever the value of $n$. There is no convergence to the length of the curve. • I think another diagram showing that the circumference is correctly approximated (in the limit) by connecting the corners of successive rectangles would make this a most convincing "layman" explanation. Clearly each of the hypotenuses is bigger than the corresponding green segment in your diagram. – Floris Aug 10 '15 at 12:17
• @floris: the question is about the wrong approach, not the right one. – Yves Daoust Aug 10 '15 at 13:20
• Yes, but sometimes it is easier to show A is wrong by showing that B is right, and A ≠ B. – Floris Aug 10 '15 at 13:41
• @Floris: this argument doesn't work here, as you would need to prove that taking the hypothenuses would yield the right arc length evaluation; this is far from being obvious. Like you say: "clearly each of the hypotenuses is bigger ". How much bigger ? What's the relation between the length of the polyline and that of the arc ?? On the opposite, showing that the sum of the widths is the diameter is obvious and bulletproof. – Yves Daoust Aug 10 '15 at 13:47
• @floris: Correct me if I'm wrong, but it seems that the length of the hypotenuses connecting corners of rectangles varies based on which two rectangles are connected. I think that this is really interesting, but hard to convince someone of, since the arcs between corners aren't always congruent. With a polygonal approximation, both the central angles and lengths of segments are all the same. This might make it easier to prove. – biffletsbq Aug 10 '15 at 14:22

Actually, you can not use the following equation $$dx = dy + dz,$$where $dx$is the hypotenuse, $dy$ is the opposite and $dz$ is the adjoint.

The limit you got is totally wrong. What you get is actually a square not a cycle by doing the way you stated.

I think you initial premise is wrong. The volume of a cube of diameter d is d^3, but its surface area is 6*(d^2). Therefore, you cannot in general, get the volume by integrating the surface area. What you can do is calculate the volume using a triple integral and the area with a double integral, but in general the term being integrated will be different and depends on the topology.

• Actually, it is pretty standard to relate the volume of a sphere to the integral of $A \mathrm{d}r$ over a family of spheres starting from the original and shrinking to a point. – Hurkyl Aug 10 '15 at 13:38
• Consider the "radius" $r$ (half of the side length). The cube has volume $8r^3$ and surface area $24r^2$. You were just differentiating by the wrong thing. Can you think of why differentiating the volume by the radius would give you the surface area? – Akiva Weinberger Aug 10 '15 at 13:42
• (This won't work for, say, an ellipsoid. I think that it needs to be able to be inscribed in a sphere?) – Akiva Weinberger Aug 10 '15 at 13:48
• I seem to have shot myself in the foot here. But I don't think it is obvious that by summing a large quantity of thin discs to find the volume, that the net result must be that the integral of the single object surface area is equal to its volume. So the statement needs to be qualified by the topologies that it applies to. – jrrk Aug 10 '15 at 13:55

The area you should be integrating increases as $$y$$ increases. There is a slope in the circle that you are not considering which makes the perimeter increase with respect to each cylinder as you move away from the origin.

You can correct this by dividing by $$\cos(\theta)$$ (the angle formed where you are integrating). This is: $$\cos(\theta) = \frac{\sqrt{r^2 - y^2}}{r}$$.

The integration is then: $$\int_{y=-r}^{y=r}{2\pi r_i\frac{dy}{\cos \theta}} = 4\pi r^2$$

So $$4\pi r^2 = 4\pi r^2$$