18 Oct '04 21:01

How do you find the area of a circle without using calculus? It's the way the Greeks did it. If anyone knows the answer, don't post it...just post if you can figure it out without having heard the answer before.

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tinyurl.com/y9ls7wbl- Joined
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my head- Joined
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Chess Castle19 Oct '04 23:35

calculus? Area? I can do it in algebra. now if i could just find the pi sign on my keyboard...*Originally posted by AThousandYoung***How do you find the area of a circle without using calculus? It's the way the Greeks did it. If anyone knows the answer, don't post it...just post if you can figure it out without having heard the answer before.**- Joined
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tinyurl.com/y9ls7wbl20 Oct '04 05:18**If you're thinking of the one we worked out at lunch last year, that one really involved calculus, because we took a limit as the number of sides of the n-gon increased without bound.**

Hmm. Interesting point. The method I am thinking of is the ancient Greek method that sounds suspiciously like what you're talking about.

I don't know if it can be done without limits.- Joined
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my head20 Oct '04 19:03

i'm not, and as thousand-young seems to be thinking of that one, i'll give the one i learned in eight grade:*Originally posted by royalchicken***If you're thinking of the one we worked out at lunch last year, that one really involved calculus, because we took a limit as the number of sides of the n-gon increased without bound.**

we can mesure the radius (r) and cercumfrence (C) of the circle.

we know that any polygon can be devided into a number of isosolease triangles.

these triangles can be arranged into a paralelagram, the area of which we can find by multiplying the height times the whith.

thus a circle can be described as an ifinet number (I) of triangles with angles:90,90, and 360/I and sides: r,r, and C/I

if we arange these into a rectangel, then it will have hight r, and the cercumfrence will make up the two long sides, thus it will have length C/2

the area is r*C/2

to put this in conventenal terms:

pi=ðŸ™„=C/d

d=2r

so ðŸ™„*r=r*C/2r=C/2

so we add another r to get A=ðŸ™„r^2- Joined
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The Lord's Army20 Oct '04 19:50

Yeah that's the one I was thinking of, but it does involve the calculus.*Originally posted by royalchicken***If you're thinking of the one we worked out at lunch last year, that one really involved calculus, because we took a limit as the number of sides of the n-gon increased without bound.**- Joined
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Elsewhere20 Oct '04 20:54

If you were at at Kennebunk High School while I was, and you ever wanted to chat about maths for some reason, fearlessleader was the only one to speak to. He's also generally awesome ðŸ™‚.*Originally posted by telerion***Yeah that's the one I was thinking of, but it does involve the calculus.**

Unfortunately, his proof here is equivalent to our earlier one, and involves calculus.- Joined
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tinyurl.com/y9ls7wbl21 Oct '04 02:41fearlessleader, you've got a bunch of googly-eyed happy guys in your equations and I don't know what they are supposed to represent.

However you're discussion of infinities involves limits, because you can't use infinities in math...only the limit as something approaches infinity. Right?

This is the same problem with the ancient Greek method. I guess I meant to say, can anyone find the area of a circle without integrating - without excluding taking limits. Your method probably works fine if this is the rule we have to follow - if I knew what squiggly eye guy meant!- Joined
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my head- Joined
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my head22 Oct '04 20:361 edit

i'm sorry, those relations SHOULD read:*Originally posted by fearlessleader***area of:**

x^2+y^2+z^2=r^2

is the same as the area of:

x^2+y^2=r^2 AND r>z>-r NOT x^2+y^2=z^2

can anyone find a 2 equivilant? that might give an answer.

x^2+y^2+z^2<r^2

and

x^2+y^2<r^2 NOT x^2+y^2<z^2

respectivly.

and it should say volume, not area- Joined
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tinyurl.com/y9ls7wbl22 Oct '04 23:43**of triangles with angles:90,90, and 360/I**

I'm sorry, I don't understand. The angles of the triangles are 90 degrees over infinity x2 and 360 degrees over infinity? That makes no sense to me.

**x^2+y^2+z^2<r^2**

I thought the first version was correct - they are equal to each other.

Also, your post in general was very confusing. Could you explain clearer what you are saying and asking?- Joined
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my head23 Oct '04 14:151 edit*of triangles with angles:90,90, and 360/I*

three angles, two are 90 degrees, the third is an infinite fraction of 360

*x^2+y^2+z^2<r^2*

it has to be a volume, and thus it must be an inequality

my point with these inequalitines is just than if we can find some two dimensional equivilant, then it may help up find a non calculus way to get the area of a circle.

this is however, probably an impossible task, so lets instead try to find a proof that no such means can exist.- Joined
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The Lord's Army23 Oct '04 23:13

Think of centering the circle on a Cartesian plane. Basically, in the limit as I -> infinite, you are adding up the "area" of an infinite number of rays extending from the origin to a point on the circle, that is from (0,0) to some (x,y) such that x^2 + y^2 = r^2.*Originally posted by AThousandYoung***[b]of triangles with angles:90,90, and 360/I**

I'm sorry, I don't understand. The angles of the triangles are 90 degrees over infinity x2 and 360 degrees over infinity? That makes no sense to me.