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The area of a circle radius $R$ is $\pi R^2$ which is quite easy to prove with integral calculus.

Consider a ring of radius $\mathrm{d}r$ at a distance $r$ from the centre. This ring has area $2\pi r \mathrm{d}r$.

Integrating,

$$\int_0^R2\pi r \mathrm\,{d}r=\pi R^2$$

But calculus is a relatively new tool, while the area of a circle has been known for what I presume since Archimedes at least. So what is the historic derivation for this?

I think it is something related to the method of exhaustion which is in essence a primitive form of integration, the Wikipedia article states

Archimedes used the method of exhaustion as a way to compute the area inside a circle by filling the circle with a polygon of a greater area and greater number of sides. The quotient formed by the area of this polygon divided by the square of the circle radius can be made arbitrarily close to $π$ as the number of polygon sides becomes large, proving that the area inside the circle of radius $r$ is $πr^2$, $π$ being defined as the ratio of the circumference to the diameter $\frac{C}{d}$ or of the area of the circle to the square of its radius $A/r^2$.

I am unclear as to how Archimedes would have proved that these two definitions of $\pi$ are in fact equivalent and refer to the same constant. Maybe in his proof of $\pi r^2$ the equality emerges but I have been unable to find any reference to or a description of how Archimedes proved this.

I don't know what to tag this, apart from reference request, so feel free to re-tag.

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From the Wikipedia article Area of a disk:

CircleArea.svg

Once you know the circumference of a circle with radius $r$ is $2\pi r$, you can use what is described in the image to approximate the area of the circle by a parallelogram with height $r$ and base $\pi r$.

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  • $\begingroup$ Wow. This proof might not be historic but it looks great. $\endgroup$ – Guy Mar 21 '14 at 9:14
  • $\begingroup$ @Sabyasachi Yes, this is not the proof Archimedes gave, but in my opinion it is the most intuitive argument why $A=\pi r^2$ is the right formula. The historic proof can be found in the article as well. $\endgroup$ – Christoph Mar 21 '14 at 9:19
  • $\begingroup$ Okay I will read the article. This proof make a lot of intuitive sense though. Beautiful :) For the proof represented by$2\pi r dr$ you can imagine the individual rings 'uncurled' into a triangle of base $2\pi r$ and height $r/2$ $\endgroup$ – Guy Mar 21 '14 at 9:22
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The proof is straightforward.

1) You need to know that similar figures have constant ratios between corresponding lengths, so call the ratio between a circle's circumference and its diameter $\pi$. In other words, the circumference $= 2\pi r$ where $r$ is the radius (half the diameter).

2) If you construct a polygon inside the circle it can be viewed as a number of triangles arranged round the centre of the circle. The area of each triangle is $1/2 (\operatorname{base length})r'$ (where $r'$ is the perpendicular height) and if you add up all the areas you have $1/2 (\operatorname{perimiter length})r'$.

3) As the polygon gets more sides then it gets closer to the circle so that in the limit the area of the polygon is the area of the circle, $r' = r$, and the perimiter is the circumference of the circle.

4) Putting everything together, the perimiter = circumference = $2\pi r$ and the area is then $1/2 \cdot 2\pi r\cdot r$, i.e. $\pi r^2$

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Your question quotes a text on Archimedes: 'Archimedes used the method of exhaustion as a way to compute the area inside a circle by filling the circle with a polygon of a greater area and greater number of sides.' Regardless of how he originally did it, here is how it can be done now for the unit circle using more modern techniques. Start with the formula for the area of a regular polygon: $$A=n\sin\left(\frac{1}{2}\frac{2\pi}{n}\right)\cos\left(\frac{1}{2}\frac{2\pi}{n}\right)=\frac{1}{2}n\sin(\frac{2\pi}{n})$$ where $n$ is the number of sides. Since the circumference $C = nL$ where $L$ is the side length we have: $$A=\frac{C/L}{2}\sin(\frac{2\pi}{C/L})$$ and since for the unit circle $C=2\pi$ we have: $$A=\frac{2\pi}{2L}\sin(\frac{2\pi L}{2\pi})$$ $$A=\pi\frac{\sin(L)}{L}$$ If $L\rightarrow0$ (because $n\rightarrow\infty$) we have $A=\pi$. Non-unit circles can of course be analyzed with a simple modification of this technique. I haven't seen this proof in exactly this form before but some of the credit must go to Friedrich Philip (see here) who pointed out that the $sin(x)/x=1$, $x\rightarrow0$ result may be relevant. This proof can be seen as an improvement to the argument used here in section 4.2 of the Tau Manifesto. For a proof of the $sin(dx)/dx$ result see here.

EDIT: $L$ is a proportion of $r$! The general (non unit) starting formula is: $$A=\frac{1}{2}nr^2\sin(\frac{2\pi}{n})$$ which converts to: $$A=\pi r^2\frac{\sin(L/r)}{L/r}$$ if $C=nL$ and $C=2\pi r$ - so this only applies to the circle, which in turn implies that $L\rightarrow0$.

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  • $\begingroup$ discussing how to prove $\frac{\sin(l)}{l} \to 1$ isn't the point, but Archimedes probably said that $\frac{\text{chord}}{\text{arc}} \to 1$ which is "obvious" geometrically. and you should write $A(l)$ and $A = \lim_{l \to 0} A(l)$ $\endgroup$ – reuns Jun 15 '16 at 1:45

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