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mathman

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mathman

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- #2

phyzguy

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[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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- #4

Mark44

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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.

- #5

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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 cm

- #6

Svein

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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.

- #7

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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

- #8

mathman

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Construction is unclear - how do you get 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

- #9

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The construction is shown the document attached within this.Construction is unclear - how do you get triangle?

Hope that helps.... :)

- #10

mathman

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- #11

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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.

- #12

Mark44

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In your first drawing in post #9, you are essentially using integration to find the area of a circle, using circular strips, orI 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.

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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- #14

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the circumference (R) of a circle

If the area is ##\pi R^2##, then ##R## is the circle's radius, not its circumference.

- #15

mathman

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My error - sorry.If the area is ##\pi R^2##, then ##R## is the circle's radius, not its circumference.

- #16

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In this source it is claimed that

https://books.google.com.tr/books?i...ench mathematician" "roberval" "sine"&f=false

Thanks

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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##

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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.I have a feeling that you can't actually give a proof without calculus (or some other notion of limits)

- #19

lavinia

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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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- #20

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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).

- #21

Stephen Tashi

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Before offering a mathematical proof about the "area" of a circle, one must have a definition of "area" that applies.

- #22

lavinia

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define"area" without using concepts from calculus.

Before offering a mathematical proof about the "area" of a circle, one must have a definition of "area" that applies.

It seems like Archimedes must have realized intuitively that whatever the general definition of area may be that there must al least be inequalities. If a planar region can be covered by a figure whose area you already know then its area must be less - or if the match is exact the same. This seems to imply a limit definition.

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I still don't see where the divided by 2 comes from. I would have thought the radius times the circumference would give the area of the circle. Clearly this isn't true, since the area of a circle is pi * r squared and it doesn't have a factor of 2 in it. Another example would be the area of a square, which is just the length of one side times the length of another side. No taking of an average involved and it makes intuitive sense.

[itex] A = r * \frac{2 \pi r + 0}{2} = \pi r^2 [/itex]

Of course, this is really just an intuitive application of calculus.

- #24

phyzguy

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[tex]A_{circle} = \int_0^r 2 \pi r dr = \pi r^2[/tex]

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