There should be many isn't it?
The simplest proof - without any of the above is below.
The area of any regular polygon is (a*p)/2 where a is the apothem (the line drawn from the centre of the polygon, perpendicular to any of the sides) and p is the perimeter.
In the case of the circle a becomes the radius r snd p becomes 2(pi)r. Therefore the area is (pi)(r squared).
Also there's a simple, just a couple of steps proof using the line integral for the area inclosed of a simple, closed and smooth curve, use the polar form for the curve as r cos(theta), r sin(theta), for theta to 0 to 2(pi)
The more famous or popular one is the Onion Proof.
There should be many isn't it?
The simplest proof - without any of the above is below.
The area of any regular polygon is (a*p)/2 where a is the apothem (the line drawn from the centre of the polygon, perpendicular to any of the sides) and p is the perimeter.
In the case of the circle a becomes the radius r snd p becomes 2(pi)r. Therefore the area is (pi)(r squared).
Also there's a simple, just a couple of steps proof using the line integral for the area inclosed of a simple, closed and smooth curve, use the polar form for the curve as r cos(theta), r sin(theta), for theta to 0 to 2(pi)
The more famous or popular one is the Onion Proof.
That would be difficult...with integrals yes. with added differential (??). advanced methods yes, possibly using Gaussian curvatures with integrals - there is But to have them together in one (????)
Thanks.. I love mathematical problems too. - at least you get your brain into gear - rather than !@#$ politics.
That would be difficult...with integrals yes. with added differential (??). advanced methods yes, possibly using Gaussian curvatures with integrals - there is But to have them together in one (????)
Thanks.. I love mathematical problems too. - at least you get your brain into gear - rather than !@#$ politics.