Are there two $\pi$s?

Question

The mathematical constant $\pi$ occurs in the formula for the area of a circle, $A=\pi r^2$, and in the formula for the circumference of a circle, $C= 2\pi r$. How does one prove that these constants are the same?

Answer 1

The formula $C = 2\pi r$ is the definition of $\pi$. That means when people ask what $\pi$ is, the answer is $\frac{C}{2r}$.

So the real question here is why is the area of a circle $\frac{1}{2}Cr$? For an intuitive answer imagine cutting a circle into pizza slices and stacking then as in this picture:

$\hspace{5.5cm}$

If your pizza slices are thin enough then that shape is almost a rectangle and we can get it's area by length times width. The width is the radius and the length is half the circumference. Thus $A = \frac{1}{2}Cr$.

Answer 2

I believe Archimedes argued that as far as area is concerned, a circle is equivalent to a triangle with the circumference as a base, and the radius as altitude on that base.

Answer 3

One way to see it is if you consider a circle with radius $r$ and another circle with radius $r+\Delta r$ (where $\Delta r\ll r$) around the same point, and consider the area between the two circles.

As with any shape, the area is proportional to the square of a typical length; the radius is such a typical length. That is, a circle of radius $r$ has the area $Cr^2$ with some constant $C$. Now the area in between the two circles has the area $\Delta A = C(r+\Delta r)^2-Cr^2\approx 2Cr\,\Delta r$. That relation gets exact as $\Delta r\to 0$.

On the other hand, the distance between the two circles is constant, and therefore for sufficiently small $\Delta r$ you can "unroll" this shape into a rectangle (again, the error you make when doing this vanishes in the limit $\Delta r\to 0$). That rectangle has as one side the circumference, $2\pi r$, and as the other side $\Delta r$. Since the area of a rectangle is the product of its side lengths, we get as area $\Delta A = 2\pi r\,\Delta r$.

Comparing the two equations, we get $2Cr\,\Delta r=2\pi r\,\Delta r$, that is, $C=\pi$.

Answer 4

Here's something I like to call the "boundary rule": $$ \begin{align} & \phantom{={}} [\text{rate of motion of boundary}]\times[\text{size of boundary}] \\[10pt] & = [\text{rate of change of size of bounded region}] \end{align} $$

Apply this to a growing circle. The rate of motion of the boundary is $\dfrac{dr}{dt}$. The size of the boundary is $C$. The rate of change of size of the bounded region is $\dfrac{dA}{dt}$.

Since $A$ is the area and $r$ is a distance, we must have $$A=(\text{some constant}\cdot r^2).\tag{1}$$ Hence $$ \begin{align} C\frac{dr}{dt} & = \frac{dA}{dt} \text{ by the boundary rule} \\[6pt] & = \frac{dA}{dr}\cdot\frac{dr}{dt}. \end{align} $$ Canceling, we get $$ C = \frac{dA}{dr}. $$ Applying $(1)$ and differentiating gives you equality of the two constants.

Now the hard question: In what contexts would this qualify as a "proof"?