area of circle?

[DivideByZero]

How did mathematicians come up with Area of circle = ( pi * r^2 ) ?

 

Why did they choose r^2 as one of the products?

Was pi found by brute forcing numbers until it worked?

[PhDP]

r². The 2 comes from the fact that we're looking for something in two dimension; an area. The formula for the perimeter is 2pi*r (one dimension = r^1), the volume of the sphere is (4/3)pi*r³ (3 dimensions), et cetera...

 

And yes, pi can only be found by brute force. It's an irrational number, it can't be described by any fraction A/B. If I remember correctly, the first 'good' values of pi were found using polygons. For example, if you inscribe a triangle in a circle, you will get a pretty bad estimate of the area of the circle, if you inscribe a square, it's a little better, so by inscribing regular polygons with an increasingly large number of sides, you'll get a better estimate of pi.

[YT2095]

I rem someone telling me that 22/7 was used for the longest time as Pi.

I don`t how historically accurate that is though?

[the tree]

How did mathematicians come up with Area of circle = ( pi * r^2 ) ?

This page explains it fairly well. It'll become gradually clearer the more you work with limits.

 

Why did they choose r^2 as one of the products?

"they" didn't, it just is.

 

Was pi found by brute forcing numbers until it worked?

You can get a fairly good approximation of π by drawing a circle and measuring out it's circumference with a bit of string, the practical method would have been the first approach.

 

Archimedes drew regular polygons on the inside and outside of circles to get a range that π would fall into, the more sides the polygons had the smaller the range, he apparently calculated by hand for 92-gons to determine that 3¹⁰⁄₇₁< π < 3¹⁄₇[Numerical Approximations of π - Wikipedia].

 

Since π has been studied so extensively, there are a lot of ways of calculating it, one of the most elegant (by no means the most efficient) has to be this sequence by Leibniz:

 

[math]1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \cdots = \frac{\pi}{4}[/math]

[YT2095]

aha, turns out what I heard is indeed correct: http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80

[Bignose]

There are many ways, but always liked the conversion from Cartesian to cylindrical coordinates:

 

[math]dxdy = rdrd\theta[/math]

 

To find the area of a space, you integrate over the limits of the space. In Cartesian coordinates, the limit is described by [math]x^2 + y^2 = R[/math], but in cylindrical coordinates, the equation of the circle is much simpler: [math] r = R[/math]

 

So, in cylindrical coordinates:

 

[math] A = \int^{2\pi}_0\int^R_0 rdrd\theta = \pi r^2[/math]

 

You get a completely analogous result in spherical coordinates in 3-D:

 

[math]V = \int^{\pi}_0\int^{2\pi}_0\int^R_0 r^2 \sin\theta drd\theta d\phi = \frac{4}{3}\pi r^3[/math]

 

The conversions from coordinate systems completely take care of themselves.

[Country Boy]

Why did they choose r^2 as one of the products?

 

"they" didn't' date=' it just is.[/quote']

That's not true. The area could as easily be written A= pi*d^2/4. Using the radius instead of the diameter is a CHOICE.

[the tree]

That's not true. The area could as easily be written A= pi*d^2/4. Using the radius instead of the diameter is a CHOICE.

Barglefarglebleh *stomps feet* you knew what I meant. A=pi*r^2 is still the area of a circle whether you use it or not.

[DivideByZero]

Thanks a lot for the replies!

Is is possible to find the area of the circle without pi?

like creating infinite triangles from the center to a part of the circumference and using limits to solve it?

[Royston]

Thanks a lot for the replies!

Is is possible to find the area of the circle without pi?

like creating infinite triangles from the center to a part of the circumference and using limits to solve it?

 

Well yes, that's essentially what Archimedes did, but without the integration. I guess you could use [math]\frac{C}{d}[/math] but that equals [math]\pi[/math] anyway.

 

This may help...

http://documents.wolfram.com/mathematica/Demos/Notebooks/CalculatingPi.html