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A circle's area could be defined as $\pi$ times $r$ times $r.$ However, how do we get this valuation? This video shows one way we can find the area of a circle. I highly encourage you to watch it as it will be vital to understand it before continuing.

In the video, they talk about how we can create a circle within the circle, and even smaller circles within that circle. Basically like an onion, with multiple rings. You would then take all the rings, and place them in such a way that it would form a triangle. The height would be the circumference, or $\pi$ times two times $r$, and the length would be $r.$

So using the two values, we can find the area of the triangle, and therefor find the area of the circle.

However, I have a question on this. When you draw out the rectangles, no matter how many there are, they should look something like this:

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Notice the height difference? Of course, the more rectangles we create, the smaller the gap, but the gap will always be there. Does that mean, for this proof at least, that the area of the triangle is a little smaller than $\pi$ times $r$ times $r$?

And if it is, could we express it as a limit? Perhaps as the limit of the number of rectangles reaches infinity, the area becomes closer and closer to a circle's?

Simon Fraser
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    The idea presented in that video does, in fact, involve limits. In particular, it could be expressed as $$\int_{0}^{r} 2\pi\rho ,\mathrm{d}\rho, $$ where $2\pi\rho$ is the circumference of a circle of radius $\rho$. Note that the integral is a limit of a sequence of Riemann sums. – Xander Henderson Jan 02 '20 at 01:59
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    Just a remark: if you arrange that the last rectangle has height $2\pi r$, then the rectangle approximation is actually an overestimate of the circle's area, not an underestimate. This is because each ring has an outer radius and an inner radius, and a rectangle whose width is the thickness of the ring and whose height is the outer radius of the ring will have area slightly more than that of the ring. – Will Orrick Jan 02 '20 at 02:17
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    Some of you guys suggested another answer, but I have no clue about integral calculus... –  Jan 02 '20 at 02:32
  • Since you are looking for a non-calculus answer, and since the duplicate target and its answers are entirely calculus based, I think an additional comment may be helpful. The tricky aspect of this argument is the part where the rings get "unrolled" into rectangles. This can't be exactly right since the top and bottom sides of a rectangle have the same length, but the inner and outer circumferences of a ring do not. It can be shown that if the width of the rectangle is taken to be the outer circumference, the approximation is going to be too big and if it is taken to be the inner radius, ... – Will Orrick Jan 03 '20 at 16:39
  • ... the approximation is going to be too small. It turns out that if the width of the rectangle is taken to be the average of the outer circumference and the inner circumference, the calculation is going to be exactly right. You can easily verify this by working backwards from the known formula for the area, but justifying it without circular reasoning is more difficult and will probably involve limits or calculus. In conclusion, there is a "staircase triangle" whose area is exactly that of the circle, but its height is not $2\pi r$. Rather, if the circle is divided into $n$ rings of ... – Will Orrick Jan 03 '20 at 16:40
  • ... width $\Delta r=r/n$, then the $n$ stair heights need to be $\pi\Delta r$, $3\pi\Delta r$, $5\pi\Delta r$, ..., $(2n-1)\pi\Delta r$. – Will Orrick Jan 03 '20 at 16:40
  • Further comment: the average of the two rectangle areas in my comments above is the area of a trapezoid, suggesting that a ring can be unrolled into a trapezoid whose area is the same as that of the ring, whose widths are the inner and outer circumferences of the ring, and whose height is the thickness of the ring. Unlike rectangles, the resulting trapezoids combine to form a genuine triangle of exactly the right base and height, rather than a staircase approximation of a triangle. – Will Orrick Jan 09 '20 at 17:05
  • If you try to visualize unrolling a ring cut out of paper, or even out of a rubber sheet, it is not immediately clear--at least it wasn't to me--how it is to be done, or that the necessary distortions don't somehow affect the area. The reason area is preserved is that the unrolling process requires no stretching or compression in what was initially the radial direction, and only shear transformations in what was initially the tangential direction. If the question gets reopened, I will try to post some pictures providing some intuition for this. – Will Orrick Jan 09 '20 at 17:13

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Let's say we have the following picture after constructing the triangle... The large triangle has height $r$ and base $2\pi r$. enter image description here

If we have $n$ rectangles, then as in the figure, the error in our approximation will be $n\times \triangle CGE$. All the small triangles will have the same area as their heights are equal and their bases are proportional to their heights.

Now, the height of this small triangle is $CG=\dfrac{r}{n}$. The base, $GE$ can be found using proportionality, which is $GE=\pi \cdot CG$, which gives $GE=\pi\dfrac{r}{n}$.

So the total error is $n\times\frac{1}{2}\times\dfrac{\pi r}{n}\times\dfrac{r}{n}=\dfrac{\pi r^2}{2n}$.

Now as you've already realized, if we increase $n$, we decrease the error. In calculus, this $n$ is taken to be approaching infinity. As a result, the error approaches a perfect zero. In limit notation,

$\displaystyle \lim_{n \to \infty} \dfrac{\pi r^2}{2n}$

which is just $0$. Thus, the area of the triangle precisely equals that of the circle only when we take the limit $n \to \infty$.

Riz222
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