# Why is $\pi r^2$ the surface of a circle

Why is $\pi r^2$ the surface of a circle?

I have learned this formula ages ago and I'm just using it like most people do, but I don't think I truly understand how circles work until I understand why this formula works.

So I want to understand why it works and not just how.

Please don't use complicated symbols.

• I believe you can't possibly understand this until you learn a huge amount of mathematics. – Git Gud Aug 13 '13 at 17:34
• Do you know calculus? just integrate $1$ over a circle. – JLA Aug 13 '13 at 17:39
• @GitGud I think you're right.. – user1534664 Aug 13 '13 at 17:39
• @Shodan...ehm, user1534664: you can use, for example, an intuitive "limit to infinity" approach, by considering the area of all polygons inside the circle, when the number of their sides increases until "infinity". This implies to understand trigonometry limits and, most of all, the equality $360^\circ=2\pi$, though. – Avitus Aug 13 '13 at 17:39
• ...or use @Prahlad Vaidyanathan answer: you can express the area of the regular polygon as a function of the number $n$ of its sides. Heuristically, let $n$ grow and try to imagine what does it happen to the area itself. – Avitus Aug 13 '13 at 17:51

The simplest explanation is that the area of any shape has to be in units of area, that is in units of length squared. In a circle, the only "number" describing it the the radius $r$ (with units of length), so that the area must be proportional to $r^2$. So for some constant $b$, $$A=b r^2$$

Now, to find the constant $b$, I think the easiest way is to look at this Wikipedia diagram:

This shows how when you subdivide the circle into many equal small triangles, the area becomes a rectangle with height $r$ and length equal to half the circumference of the circle, which is $\pi r$, by the definition of $\pi$.

• The length is half the circumference, which is $\pi r$. – Tony Huynh Aug 13 '13 at 23:42
• @TonyHuynh - thanks, fixed. – nbubis Aug 14 '13 at 2:53
• how in the world does the diagram end up at $\pi r$ ? I see no relation to the picture. In fact the last triangle's tip is over about 3.5r – zerosofthezeta Aug 14 '13 at 3:30
• @euclid that's because it's only a hexagon, so half the outer perimeter is only $3$. It gets better if you use more pieces. Going into the complexities of the number $\pi$ itself is probably beyond the question being asked. – Robert Mastragostino Aug 14 '13 at 3:57

Much depends on how you define $\pi$. It is intuitively easy to see that for circles the ratio between the area and $r^2$ is constant, and you could define $\pi$ to be this constant.

If however you defined $\pi$ in the more usual way as the (also constant) ratio of the circumference and the diameter $2r$ of circles, there is an intuitive way to see that this same constant should also give the ratio between the area and$~r^2$. If you approximate the circle by any polygon (not necessarily regular), all of whose sides are tangent to the circle, then the area of the polygon will be exactly $\frac r2$ times the circumference of the polygon. This is because the polygon can by cut up into triangles, each having as base a side of the polygon, and all sharing the center of the circle as vertex. Thus all these triangles have height$~r$, and if $b$ is the basis of such a triangle (a side of the polygon) its area is $\frac r2\times b$.

The total area then is is $\frac r2$ times the sum of the bases of the triangles, which is the circumference of the polygon. Now as one increases the number of sides of the polygon indefinitely, its area and circumference both tend to those of the circle. So the area of the circle should also be $\frac r2$ times its circumference, which circumference is $2\pi r$ by definition, and this gives an area of$~\pi r^2$.

I would like to note that this is an intuitive, not a formal argument. The delicate point in that the area and circumference of the polygons both tend to that of the circle. For the area this is fairly easy to justify, as the neighbourhood of any point outside the circle eventually does not contribute to the area of the polygons either. But for the circumference it is not obvious, as it is possible to obtain a curve as a limit of a set of polygonal lines while the lengths of the latter do not tend to the length of the curve. The fact that this is not the case here is related to the fact that the polygon sides all tend to be parallel to the circle arc near it.

• See illustrations at en.wikipedia.org/wiki/Area_of_a_disk#Archimedes.27s_proof. – lhf Aug 14 '13 at 3:03
• @jhf: Thanks, I've inserted the most relevant image. However the proof I suggest (which does not pretend to be formal) is not that of Archimedes and involves no inequalities at all. Instead it is similar to the argument in the lead of the Wikipedia article, except that using regular circumscribed polygons is unnecessary. – Marc van Leeuwen Aug 14 '13 at 5:30

One way to prove it is the following : Embed a regular polygon (ie. all sides are equal, all angles are equal) in a circle (ie. its vertices are on the circle. Divide this polygon into $n$ triangles by drawing radii out to the vertices. Calculate the area of this triangle and multiply by $n$, and you will get the area of the polygon.

Now, if you can assume the following limit : $$\lim_{x\to 0} \frac{\sin(x)}{x} = 1$$ Then when you take a limit as $n$ goes to infinity, you will end up with $\pi r^2$.

So all that remains is proving the above formula :)

• +1, but I think that the tricky part in this construction is to move from degrees to radians in measuring angles to justify the presence of $\pi$. – Avitus Aug 13 '13 at 17:41
• I guess you could use the "definition" of $\pi$ as the circumference divided by diameter. Then the central angle of any such triangle will be $2\pi/n$. That would work right? – Prahlad Vaidyanathan Aug 13 '13 at 17:46
• Yes, I agree: defining $\pi$ as the coefficient stating the proportionality between the diameter and length of the unit circle surely works; $\pi$ has to be defined somehow, after all :-) – Avitus Aug 13 '13 at 17:49
• let $\pi = 22/7$ ... – Prahlad Vaidyanathan Aug 13 '13 at 17:58

A number of different ways of showing this exist. First notice that the area has to be $(\text{constant}\cdot r^2)$ because the area of a region of any shape in a plane must be proportional to the square of the distances. E.g. if you multiply all distances by $3$, then the area is multiplied by $9$. And "constant" in this case means it's the same number regardless of what $r$ is. So now the question is: Why must the "constant" be the same as the ratio of circumference to diameter?

As $r$ increases, we have \begin{align} \text{rate of growth of area} & = \text{size of boundary} \times\text{rate of motion of boundary} \\ & = \text{circumference} \times \text{rate at which $r$ is changing} \\ & = 2\pi r \times \text{rate at which $r$ is changing}. \end{align} From calculus, recall that $$\text{rate of change of r^2} = 2 r \times\text{rate of change of r}.$$ So we have $$\text{rate of growth of area} = \pi\times \text{rate of change of r^2}.$$ So the area grows at the same rate at which $\pi r^2$ grows. That makes them always equal if they're equal when $r=0$. And it's easy to see that they're equal when $r=0$.

That's only one way to do it; there are others.

PS: Supposing we don't know calculus; how would we know that $$\text{rate of change of r^2} = 2 r \times\text{rate of change of r}\text{ ?}$$ We could do that as follows. A square whose side has length $r$ is growing because its north side is moving northward and its east side is moving eastward. Then \begin{align} & \text{rate of growth of $r^2$} \\ = {} & \text{rate of growth of square's area} \\ = {} & \text{size of moving boundary}\times\text{rate of motion of boundary} \\ = {} & 2r \times \text{rate of change of $r$}. \end{align}

My knowledge is base on a removed(?) video from Dr. James Grime on the YouTube channel "Numberphile"

Introduction:

Properties of the Area(s) of a triangle:

I. Properties #1-#3: II. Property #4: Constant area:

Let have a circle with the radius $$r$$. We split the circuit into equal $$d$$ parts (in best case: infinitesimal short parts); now we imaginate a cut once into the center- so we get kinda a handheld fan with a lots of quasi-triangles. last move is to use property #4:

I also recommending to check the rings version on YouTube from:

1. MinutePhycics

2. BYJU'S- Area of a Circle in a Triangle

3. Eddie Woo-Quick Visual Proof: Area of a Circle