Area of a circle without calculus

In summary: Delta r##where ##r_i## is the radius of the ith annulus and ##\Delta r## is the same (small) interval between all of the radii.So, the total area under the curve is the sum of the areas of each of the rectangles (the sum from i = 1 to n) (where n is the number of rectangles).In summary, the value of π can be derived by using circular strips or annuli and applying basic geometric principles. This can be done without using calculus or Archimedes' method, but the accuracy of the result depends on the size of the strips and the number of annuli used.
  • #1
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)?
 
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  • #2
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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  • #3
I suppose you could make a circular field by putting a pole in the ground then dragging a string around it.
 
  • #4
rootone said:
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.
 
  • #5
Mark44 said:
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).
 
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  • #6
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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  • #7
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 anyone 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
 
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  • #8
jishnu said:
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 anyone 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?
 
  • #9
mathman said:
Construction is unclear - how do you get triangle?
The construction is shown the document attached within this.
Hope that helps... :)
1524021048356.jpg
 

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  • #10
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.
 
  • #11
mathman said:
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.
 
  • #12
jishnu said:
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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  • #13
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.
 
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  • #14
mathman said:
the circumference (R) of a circle

If the area is ##\pi R^2##, then ##R## is the circle's radius, not its circumference.
 
  • #15
PeterDonis said:
If the area is ##\pi R^2##, then ##R## is the circle's radius, not its circumference.
My error - sorry.
 
  • #16
mathman said:
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?i...ench mathematician" "roberval" "sine"&f=false

Thanks
 
  • #17
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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  • #18
stevendaryl said:
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.
 
  • #19
mathman said:
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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  • #20
lavinia said:
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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  • #21
A fundamental problem is whether we can 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.
 
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  • #22
Stephen Tashi said:
A fundamental problem is whether we can 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.
 
  • #23
phyzguy said:
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.
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.
 
  • #24
With a rectangle of sides L and W, imagine taking the side of length L and 'sweeping' it across a distance W. Each end of the side moves a distance W, so it sweeps out an area of L x W. Now imagine doing the same thing with the radius of a circle, 'sweeping' it around in a 360 degree circle, with the center fixed. The center end of the radius doesn't move at all, and the outer end of the radius moves a distance 2πr. So to calculate how much area the radius sweeps out, you need to average the distance moved by the center (0) with the distance moved by the outer end (2πr), then multiply by the length of the radius (r). In calculus terms, we would write these as:[tex]A_{square} = \int_0^W L dx = L \times W[/tex]
[tex]A_{circle} = \int_0^r 2 \pi r dr = \pi r^2[/tex]
 
  • #25
Yeah, I forgot to take into consideration that the "circumference" of the circle is different depending on where you cut off the radius. I get it now.
 
  • #26
1) draw a circle of radious 1
2) inscribe 6 equilateral triangles.
3) count the sides close to the circumference and you get 6 and divide this be 2 for the first approximation. it equals 3
4) divide a 60 degree segment in half and draw the 2 new chords. compute the length of the chords.
5) add the 12 chords and divide by 2 to get 3.105 for the 2nd approximation
6) Divide the 1/12 chords in half and compute the new chord length.
7 compute the total chord length of the 24 chords and divide in half for the 3rd approximation.
etc
 
  • #27
From the above approach I obtained the following

6 chords pi = 3
12 chords pi = 3.105828541
24 chords pi = 3.132628611
48 chords pi = 3.139350197
 
  • #28
arydberg said:
divide a ... segment in half and draw the 2 new chords - compute the length of the chords.
And how do you do that without calculus?
 
  • #29
Also:
  1. That method computes the area, but the approach is suggestively similar to a Riemann sum. It is just an area integral of a piecewise constant function.
  2. You obtain the area, but it does not show that the area is ##\pi##. If you have a numerical value for ##\pi## obtained from elsewhere, you can just see it approaching. You need to show that the limit of the series is equal to your given value (which probably is also a different series).
 
  • #30
Svein said:
And how do you do that without calculus?
It is a right triangle.
 
  • #31
arydberg said:
It is a right triangle.
As you described things, "it" isn't a right trangle. When you divide one of the original 60° sectors in two, the two new chords are presumably points on the circle. Each of the new triangles is isosceles, with a 30° angle at the top (radiating from the circle's center) and two base angles of 75°. These aren't right triangles. The two triangles whose hypotenuses are the new chords, and whose bases are half the length of the old chord (before splitting the equilateral triangle into two). Those two small triangles are right triangles, each with an acute angle of 15°.
 
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  • #32
Orodruin said:
Also:
  1. That method computes the area, but the approach is suggestively similar to a Riemann sum. It is just an area integral of a piecewise constant function.
  2. You obtain the area, but it does not show that the area is ##\pi##. If you have a numerical value for ##\pi## obtained from elsewhere, you can just see it approaching. You need to show that the limit of the series is equal to your given value (which probably is also a different series).
pi is simply the length of one half of a unit circle.

At the process goes on 1/2 the sum of the lengths of the chords approaches the length of half the circumference.
Mark44 said:
As you described things, "it" isn't a right triangle. When you divide one of the original 60° sectors in two, the two new chords are presumably points on the circle. Each of the new triangles is isosceles, with a 30° angle at the top (radiating from the circle's center) and two base angles of 75°. These aren't right triangles. The two triangles whose hypotenuses are the new chords, and whose bases are half the length of the old chord (before splitting the equilateral triangle into two). Those two small triangles are right triangles, each with an acute angle of 15°.
Yes everything you say is correct. My point is that you can use these right triangles to compute the length of the two new chords in terms of the old chord. This results in a approximation of pi and it can be done over and over to any degree of accuracy desired.
 
  • #33
arydberg said:
pi is simply the length of one half of a unit circle.

At the process goes on 1/2 the sum of the lengths of the chords approaches the length of half the circumference.

Yes. And how do you show that your sum actually approaches this number? As I said, you need to show that your series converge to the same number.
 
  • #34
Is finding the area of a circle not called “squaring the circle” ?
Was calculus not devised to make it possible to “square the circle” ?
What has changed that now makes calculus no longer necessary for the task it was created to solve ?
 
  • #35
Baluncore said:
Is finding the area of a circle not called “squaring the circle” ?

No. "Squaring the circle" refers to one of the classic problems in straightedge-and-compass Euclidean geometry, the problem of constructing, using straightedge and compass alone, a square with area equal to that of a given circle. This problem is not solvable by straightedge and compass alone (briefly, because ##\pi## is transcendental, and transcendental numbers cannot be constructed with straightedge and compass alone).

Baluncore said:
Was calculus not devised to make it possible to “square the circle” ?

No.
 
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<h2>1. How do you find the area of a circle without using calculus?</h2><p>To find the area of a circle without using calculus, you can use the formula A = πr^2, where A is the area and r is the radius of the circle. This formula is derived from the fact that the area of a circle is equal to the number of squares that can fit inside it, and each square has a side length equal to the radius of the circle.</p><h2>2. Can you explain the concept of pi and its role in finding the area of a circle without calculus?</h2><p>Pi, denoted by the Greek letter π, is a mathematical constant that represents the ratio of a circle's circumference to its diameter. In other words, for any circle, the circumference is always approximately 3.14 times the diameter. This constant is used in the formula A = πr^2 to find the area of a circle without calculus.</p><h2>3. Is it possible to find the area of a circle without knowing its radius?</h2><p>No, the radius is a necessary component in the formula for finding the area of a circle without calculus. However, if you know the diameter of the circle, you can divide it by 2 to find the radius and then use the formula A = πr^2 to calculate the area.</p><h2>4. Are there any alternative methods for finding the area of a circle without calculus?</h2><p>Yes, there are other methods for finding the area of a circle without using calculus. One method is to use the formula A = ½bh, where b is the base of the circle and h is the height. Another method is to divide the circle into sectors and use the formula A = (θ/360)πr^2, where θ is the central angle of the sector.</p><h2>5. How accurate is the formula for finding the area of a circle without calculus?</h2><p>The formula A = πr^2 is an approximation of the true area of a circle and becomes more accurate as the number of squares used to approximate the circle increases. However, it is not exact because it is impossible to fit an infinite number of squares inside a circle. As the radius of the circle becomes smaller, the accuracy of the formula also increases.</p>

1. How do you find the area of a circle without using calculus?

To find the area of a circle without using calculus, you can use the formula A = πr^2, where A is the area and r is the radius of the circle. This formula is derived from the fact that the area of a circle is equal to the number of squares that can fit inside it, and each square has a side length equal to the radius of the circle.

2. Can you explain the concept of pi and its role in finding the area of a circle without calculus?

Pi, denoted by the Greek letter π, is a mathematical constant that represents the ratio of a circle's circumference to its diameter. In other words, for any circle, the circumference is always approximately 3.14 times the diameter. This constant is used in the formula A = πr^2 to find the area of a circle without calculus.

3. Is it possible to find the area of a circle without knowing its radius?

No, the radius is a necessary component in the formula for finding the area of a circle without calculus. However, if you know the diameter of the circle, you can divide it by 2 to find the radius and then use the formula A = πr^2 to calculate the area.

4. Are there any alternative methods for finding the area of a circle without calculus?

Yes, there are other methods for finding the area of a circle without using calculus. One method is to use the formula A = ½bh, where b is the base of the circle and h is the height. Another method is to divide the circle into sectors and use the formula A = (θ/360)πr^2, where θ is the central angle of the sector.

5. How accurate is the formula for finding the area of a circle without calculus?

The formula A = πr^2 is an approximation of the true area of a circle and becomes more accurate as the number of squares used to approximate the circle increases. However, it is not exact because it is impossible to fit an infinite number of squares inside a circle. As the radius of the circle becomes smaller, the accuracy of the formula also increases.

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