Why is the area of a circle not $2 \pi R^2$?

Question

Let $$ C=\{(x,y,z)\mid x^2+y^2=1, 0 \le z \le 1\} $$

Counter-calculus, my intuition says that the area of a circle with radius $r$ should be $2 \pi \cdot R^2$, because I think that if all radii segments are to be raised perpendicularly to the perimeter of the unit circle (at plane $xy$), then the area is preserved.

I know I am wrong but I don't (qualitatively) understand why (note I never learnt measure theory).

Answer 1

It is true that if you draw lines from each point on the circumference to the center, you will cover the entire interior of the circle without overlap (except at the center).

However, that is not the limit of a process of ever finer divisions that cover the interior of the circle more and more accurately. On the contrary, if you divide the circumference into $n$ intervals, and construct strips toward the center, you will find that they always overlap for any finite $n$, and what's more, the degree of overlap remains more or less consistent, approaching (in the limit as $n \to \infty$) a factor of $2$.

That is why this process yields a limiting total coverage of $2 \pi r^2$, even though the actual area of the circle is only half of that. The process that covers the interior of the circle more and more accurately is to draw wedges (not strips) from the circumference toward the center point. In the limit, these wedges produce zero overlap, and since they have a collective area equal to half that of the strips, they yield the correct total area: $\pi r^2$.

Answer 2

Your intuition is almost correct. Draw in a whole bunch of little radii from the center of the circle to the edge. This splits the circle into a bunch of thin regions. And all of these regions have height $R$. But the regions are approximately triangles, not rectangles. So we should get \begin{align*} \text{Total Area} &= \frac12 (\text{Total of all bases}) (\text{Height}) \\ &= \frac{1}{2} (2\pi R) (R) \\ &= \pi R^2. \end{align*} You got $2 \pi R^2$ instead with your method, because you forgot the factor of $\frac12$, treating the small triangular regions as rectangular instead.

The following picture may also help:

In the above picture on the left, you can see drawn in all the approximately-triangular regions. On the right, it is shown that the total are of these triangles is approximately $\pi R^2$, by lining them up to form an approximate rectangle with base $\pi R$ and height $R$.

Answer 3

If we cut a circle in $N$ equal sectors ($N=12$ in the above picture) and rearrange such pieces we get something that is close to a rectangle (closer and closer as $N$ increases) with height $R$ and base given by half the perimeter of the circle. It follows that the area of a circle is half the product between its length and its radius, i.e. $\pi R^2$.

An alternative approach is the following: let $L(R)$ be the length of a circle with radius $R$ and $A(R)$ be its area. We have $L(R)=2\pi R$ and $A(R)=C R^2$ for some (at the moment) unknown constant, since by duplicating the radius we have $\frac{A(2R)}{A(R)}=4$. The area of the annulus between two concentric circles with radii $R,R+\varepsilon$ can be approximated with the area of a rectangle having height $R$ and base $L(R)$, hence: $$ A(R+\varepsilon)-A(R) \approx 2\pi R \varepsilon. $$ Consequently, $$ \frac{d}{dR} A(R) = 2\pi R = \frac{d}{dR}CR^2 = 2C R $$ hence $C=\color{red}{\pi}$.

Answer 4

I will leave this as an answer because even though it is in the comments above, all other answers (in my humble opinion; no offense to anyone) involve more structures than are needed to answer the question; essentially the concept of volume and area.

In one dimension, we have the concept of length. In two, the concept of area. In three, the concept of volume.

When you raise all radii segments tangent to the perimeter then you are involving the third dimension, and thus the concept of volume encoded by them. Let their length be $R$ that is of the radii segments. Then the volume enclosed is $ 2 \pi R^2$. Now divide out by the third dimension, namely, $D=2R$ (since when you raise the radii, you are raising them twice because $D=2R$, instead of just once as per the usual concept of volume) and then you are left with the true area, that is $\pi R^2$.

Answer 5

I have created an animation that shows what I think the OP means by "all radii segments are to be raised perpendicularly to the perimeter of the unit circle":

If this is indeed the intended interpretation, then let me make a few observations:

1. The radii segments, when "raised up perpendicularly to the unit circle", do indeed seem to form a cylinder (with no top or bottom); the lateral surface area of that cylinder is, as the OP says, $2\pi R^2$.

2. The problem with this visualization can be best understood by imagining each of the radii segments, when in the "upright" position, as being thickened into a rectangular strip. In the upright position, those strips form a cylindrical surface. But as they fold down, the strips will overlap one another. See animation below:

3. How to prevent the rectangular strips from overlapping one another as they fold down? Cut each one into a triangular shape with its base on the perimeter of the circle, and then they will fit together nicely as they fold down. (See animation below.)

4. Of course, if you cut the rectangular strips into triangles then they will no longer form the cylindrical surface --- each one has had half of its area removed, so the area that remains is just $\pi R^2$.

So at an intuitive level, that's where the factor of $1/2$ comes from: the need to prevent the infinitesimal rectangular strips from overlapping each other as they fold downwards.