B Area of a circle without calculus (1 Viewer)

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mathman

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π is defined by the ratio of the circumference (R) of a circle to its diameter. The area of the circle is πR². Can this be derived without calculus (or Archimedes method)?
 

phyzguy

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It depends what you mean by "derive" and how rigorous you want to be. If you just imagine the radius of a circle sweeping around the full circle, the outer end travels a distance of 2πr, and the center travels a distance of zero, so if you average these two you find the total area swept out is just:
[itex] A = r * \frac{2 \pi r + 0}{2} = \pi r^2 [/itex]

Of course, this is really just an intuitive application of calculus.
 
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I suppose you could make a circular field by putting a pole in the ground then dragging a string around it.
 

Mark44

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I suppose you could make a circular field by putting a pole in the ground then dragging a string around it.
This would determine the extent of a circle, but it wouldn't be helpful to determine its area, which is what the OP is asking about.
 
This would determine the extent of a circle, but it wouldn't be helpful to determine its area, which is what the OP is asking about.
We just need a steady supply of patience and 1 mm grid paper. I have a vague memory we did something like this in middle school using different shapes and 1 cm grids (counting partials as 0.5 cm2).
 

Svein

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Or you could use the "Monte Carlo" method - Draw two random numbers scaled to [-1, 1]. Use the first as an x-coordinate and the second as a y-coordinate. Increase a "total tries" counter. If x2+y2≤1, then increase the "inside circle" counter. Repeat...

The ratio between the "inside circle" counter and the "total tries" counter will approximate the ratio between the areas of a circle with radius 1 and a square with side 2.
 
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All you need to do is consider there are thin circular strips inside the circle which completely fill the circle, now cut these circular rings through any one radius up to the centre and stretch them out to form a triangle.
Now,
triangle has base length =2πR
Height of the triangle =R
area (circle) = area (triangle)
= 1/2x base x height
=1/2 x 2πR x R
= πR^2
 

mathman

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All you need to do is consider there are thin circular strips inside the circle which completely fill the circle, now cut these circular rings through any one radius up to the centre and stretch them out to form a triangle.
Now,
triangle has base length =2πR
Height of the triangle =R
area (circle) = area (triangle)
= 1/2x base x height
=1/2 x 2πR x R
= πR^2
Construction is unclear - how do you get triangle?
 

mathman

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It looks like a neat construction. But how you prove that the shape of the final figure is a triangle, i.e. the sides are straight lines? It looks like a geometry attempt to mimic the usual elementary calculus proof.
 
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It looks like a neat construction. But how you prove that the shape of the final figure is a triangle, i.e. the sides are straight lines? It looks like a geometry attempt to mimic the usual elementary calculus proof.
I don't know whether it can be proved mathematically that the sides are straight lines
But, when the strips are very thin and as the radius undergoes a gradual decrease in when coming inside, so as the circumstances of inner circles also decreases the sides are going to form straight lines. My knowledge in mathematics is very primitive but, I will try to find a possible solution for this.
 

Mark44

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I don't know whether it can be proved mathematically that the sides are straight lines
But, when the strips are very thin and as the radius undergoes a gradual decrease in when coming inside, so as the circumstances of inner circles also decreases the sides are going to form straight lines. My knowledge in mathematics is very primitive but, I will try to find a possible solution for this.
In your first drawing in post #9, you are essentially using integration to find the area of a circle, using circular strips, or annuli.
Drawing1.jpg

The radius of the outer circle is 1, and is centered at the origin. If we divide the circle is divided into thin concentric annuli, each of thickness ##\Delta r##, and of radius r, the area of the annulus is about ##2\pi r \cdot \Delta r##. If ##\Delta r## is reasonably "small" there's not much difference between the inner and outer radii of the annulus. In the drawing I show only one annulus, in blue.
To get an approximate value for the circle's area I can add the areas of all the annuli like so:
##A \approx \sum_{i = 1}^n 2 \pi r \Delta r##, where r is the radius to some point in the annulus, and ##\Delta r## is the thickness of the annulus, which we can take as ##\frac 1 n##.

If we make the annuli thinner (and consequently more of them), the approximation will be better. Where calculus comes in is replacing the sum above (a Riemann sum) by a definite integral: ##\int_0^1 2 \pi r dr##, which turns out, unsurprisingly, to equal ##\pi##. Unfortunately, since the OP wanted to derive the area of a circle without calculus, this solution doesn't meet that requirement.
 

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The area of a circle can be approximated without calculus but any improvement in the estimate will surely have to be made by smaller and smaller increments which eventually approach the infinitesimal. This is basically a definition of calculus.
 
π is defined by the ratio of the circumference (R) of a circle to its diameter. The area of the circle is πR². Can this be derived without calculus (or Archimedes method)?
In this source it is claimed that Gilles de Roberval, a French mathematician, was the first to accomplish to calculate the area under a sine curve without Evaluation theorem. I don't know if the same method could be applied to a circle. We might search that through the history of mathematics. The ancient Greeks could calculate it with using limits but were they the first to calculate the area of a circle?

https://books.google.com.tr/books?id=eztUxtCfNXoC&pg=PA357&lpg=PA357&dq="french+mathematician"+"roberval"+"sine"&source=bl&ots=OVI_cjWozP&sig=fcBF1C1Ot3HlDjUqxRi1TOv09sI&hl=en&sa=X&ved=0ahUKEwjN38Dw7KHbAhWGESwKHdpjB1wQ6AEIMjAC#v=onepage&q="french mathematician" "roberval" "sine"&f=false

Thanks
 

stevendaryl

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I have a feeling that you can't actually give a proof without calculus (or some other notion of limits), but the conclusion is made very plausible by the following construction: Divide the circle into a large number of wedges of equal size (32 are shown). Then take the wedges and rearrange them into a parallelogram. *Now, it's not actually a parallelogram, because the wedges are not exactly triangles---their short edges are slightly curved, rather than straight, but...that's why it's a plausibility argument rather than a proof). The parallelogram has a base of length ##\pi R## (since the top and bottom edges together make the full circle). The parallelogram has a height that is approximately ##r##. So the parallelogram has area ##B \cdot H = \pi R^2##

circles.jpg
 

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Orodruin

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I have a feeling that you can't actually give a proof without calculus (or some other notion of limits)
I agree. Essentially any "intuitive" construction by splitting the circle in smaller segments is an approximation of a Riemann sum - including of course the construction you mention later.
 

lavinia

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π is defined by the ratio of the circumference (R) of a circle to its diameter. The area of the circle is πR². Can this be derived without calculus (or Archimedes method)?
Not sure what you mean by the method of Archimedes.

He did prove that the area of a circle equals the area of a right triangle whose base is the circumference and whose height is the radius. This proof does not use limits - just Euclidean geometry.

He also approximated pi which I would guess in those days was defined as half the circumference of a circle of radius one.
 
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He did prove that the area of a circle equals the area of a right triangle whose base is the circumference and whose height is the radius. This proof does not use limits - just Euclidean geometry.
Wikipedia has a description of Archimedes' proof here:

https://en.wikipedia.org/wiki/Area_of_a_circle#Archimedes's_proof

It appears to be a simple proof by contradiction (twice, to show that the circle's area is not greater than and not less than the triangle area).
 

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