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A helpful math formula involving circles.
Forum Index - Sunken Ghost Ship - Forum Graveyard - The New World of Insanity - A helpful math formula involving circles.
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SMW C:\\user_19007_shree\posts >> target.display('layout_mathos','Sea of Science')
Well, this is not much of a formula, much more a derivation.
Here goes:

Proof
(A=area, C=Circumference, r=Radius)

∏r2 = A

Multiplying both sides by 2,

2∏r2 = 2A

As 2∏r = C,

Cr = 2A

Cr/2 = A

Here's the trick:

We know that

2∏/2∏ = 1.

So,

Cr/2 x 1 = A.

Cr/2 x 2∏/2∏ = A.

So, we finally have,

Cr x 2∏/2 x 2∏ = A

Again, 2∏r = C,

So,
C x C/4∏ = A

C2/4∏ = A.

For those of you too lazy to read:





Hope this helps!

Go ahead and solve a few problems. They turned out perfect for me!

Oh, and sorry if this thread is too long, but you know math!

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Thanks! This piece of information which I didn't know before will be very useful to me! :D
Nice... now prove that the perpendicular bisectors of a rhombus are congruent.

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Me: Look, matey... it's pronounced ARK-EE-YUS. Got that?
Dude: Ok, you make Arse-ee-yus sound dumb when you say it that way.


Ruby, you make me feel so junior. I can barely understand what you mean

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This is so useful, that I can't understant anything you did post here. :D

Type everything very simply, because I need that one for an exam. :o
So for those who find pi * r^2 too hard, you've devised a formula that requires you to calculate the circumference? In my experience, most problems just give you the radius and makes you use that for most things - any backwards calculation is simple enough to do on your own.

Tell me, for what purpose is this useful?

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Originally posted by theraze2fan
Ruby, you make me feel so junior. I can barely understand what you mean

Then clearly you have not taken Geometry or Algebra II.#fim{^_^;}

About the proof you did there, that equation is not used for a reason. Like Incognito said, 1/2tau r^2 is used because it is much simpler and it also does not require as much calculations to derive as yours.

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Me: Look, matey... it's pronounced ARK-EE-YUS. Got that?
Dude: Ok, you make Arse-ee-yus sound dumb when you say it that way.




A formula to circle's area is more simple to prove when you integrate using polar coordinates. Like here:




Like others said, it's better to use pi*r^2 to evaluate a circle's area.

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2 integral signs? All the problems I've faced only used just one! Tho they were for Cartesian coordinates.
A later chapter in my math book has problems with three integral signs.

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Every single level I will ever make in SMM2 will be easier than Ultra Necrozma.
And here I am still doing circle theorems and inequalities in my 11th year of education. Nice...


Originally posted by Sokobansolver
2 integral signs? All the problems I've faced only used just one! Tho they were for Cartesian coordinates.
A later chapter in my math book has problems with three integral signs.


Simple, Sokoban: You have been studied simple one-dimensional integrals. I presented you some two-dimensional ones. The concept of Riemann's sums applies here too, but now is the sum of volumes, not areas. And the polar coordinates are a different manner of seeing our normal coordinates. This is to make integration over circles easier. And three integrals... they are triple integrals. The form of seeing them as Riemann's sums can be more abstract (some 4-D stuff). And we have other systems of coordinates for them (cylindrical, spherical).

And, finally, the line integrals (and surface integrals). They are more complicated to explain, but they are EXTREMELY important for you to understand some theorems at physics (like the Work, for example - Work can be defined as a line integral).

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Why not put 12.48 instead of 4Pi? If ind 12.48 easy... unless those expect you to end in that state.

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Because 'pi' is an exact value.
Are surface integrals surface area?

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How am I so creative? I think taking walks might have something to do with it.
Every single level I will ever make in SMM2 will be easier than Ultra Necrozma.


Originally posted by Sokobansolver
Because 'pi' is an exact value.
Are surface integrals surface area?


At some point, yes. A surface integral is an area, indeed. But the double integral is a volume, at first, but you can use it to calculate areas, too. You can use even line integrals to calculate some areas. Example: The Si(x) function is an area function, but you use a complex line integral to calculate it, because Si(x) is the integral of sin(x)/x, which doesn't have a primitive.

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