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Depending on if you want the derivation of the formula or just a mathematical proof of it, you can either get a really long answer or a really short and simple(if you understand calculus) answer. Are you looking for how people actually determined these formulas or just a way that proves that they are correct?
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Well I was wondering how people actually determined them.
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I think fish is nice, but then I think that rain is wet, so who am I to judge? It's gotten to the point where careful investigation is needed just to tell parody from reality. I think that means reality is broken.- Noclevername. |
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I like to look at it from the point of view of derivatives. Take the volume of a sphere, if you were to increment the radius, how much would the volume increase? In other words, what is the derivative of the volume?
d/dx (4/3 pi r3) But that is 4/3 pi 3r2, right? 4 pi r2, which is the surface area. And, what is the derivative of the area of a circle? ![]() |
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I'll try answering the first question: Why is A = pi*r2
First you have to remember that the circumference, C, of a circle is pi times its diameter. This is just a fact about circles. And since the diameter is twice the radius, then C = 2pi*r. Next imagine cutting the circle up into a whole bunch of identical wedges, like pieces of a pie (no pun intended, in greek pi is pronounced "pee"). As we make these wedges progressively thinner, the difference between the base, which is a section of the circumference, and a straight line becomes ever less. These narrow wedges approximate triangles. The area of a triangle is half its height times its base, or A = H*b/2. Now, as the base of these wedges approaches zero, the height aproaches the radius of the circle, so the area of one of them becomes A = r*b/2. Next, we could add up the area of all these triangles. They all have the same height, r, so we just need to change the dimension for the base to what it is for all of them. Well, that's the circumference, 2pi*r, right? So let's replace the value b for one triangle with the value 2pi*r for all of them in that equation above: A = r*2pi*r/2 The 2s cancel, and you're left with A = r*pi*r, or A = pi*r2 okay, I used that phrase "approaches zero". Sound familiar? |
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As far as the area of the circle (or disk) is concerned, you will find some suggestive pictures in this page (scroll down to where it says "History of Area of a Circle").
Of course, these are merely graphic motivations, not rigorous proofs. Here's a description of Archimedes' (more or less) rigorous proof by the method of exhaustion.
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"All your bias are belong to us." Ara Pacis "A witty saying proves nothing." Voltaire |
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I get it. I finally get it. They taught me that in primary school and I finally get it. If you chop up a circle into wedges and arrange them into a wobbly parallelogram, then even out the wobbles by bisecting the curves, you get a rectangle with the radius making up the width and the circumference split along the top and bottom sides. Since the circumference is 2pir, and the radius is r, then the area of the rectangle would be 2pir/2*r, which is pi*r2. Thank you!!
I'm a southpaw, so I'm not good with abstractions. I need to see things visually
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I think fish is nice, but then I think that rain is wet, so who am I to judge? It's gotten to the point where careful investigation is needed just to tell parody from reality. I think that means reality is broken.- Noclevername. |
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Quote:
If we consider a line segment to be a one-dimensional spheroid, the "surface contents" of the 1-, 2-, and 3-sphere are 2, 2*pi*r, and 4*pi*r2. Arguments similar to the above give for the "volume contents" of the above the values 2*r/1 = 2*r, (2*pi*r)*r/2 = pi*r2 and (4*pi*r2)*r/3 = (4*pi/3)*r3. Now what is the "surface content" of a Euclidean 4-sphere? Looking at the above sequence would you guess 2*pi2*r3? Maybe if you were Gauss you might, but most of us would never guess this. But knowing this you can confidently state that the "volume content" of the 4-sphere would be 2*pi2*r3*r/4 = (pi2/2)*r4. I know these things to be true, but I cannot think of an easy way to explain the "surface contents".
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Microsoft is over if you want it. The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high. |
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Simple math problem [courtesy of the town baker this morning, over an almond croissant and Cafe Americano]:
A bird hunter takes aim at a group of 8 birds perched together on a tree branch, and fires. Two fall to the ground. How many birds are left on the branch? ![]() [ETA "and fires", per parallaxicality's post below]
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* Last edited by sarongsong; 16-February-2008 at 04:42 PM. |
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None. They'd all scatter once they heard the shots. Assuming, of course, that the shots actually brought the birds down. The riddle doesn't actually say that the hunter shot them.
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I think fish is nice, but then I think that rain is wet, so who am I to judge? It's gotten to the point where careful investigation is needed just to tell parody from reality. I think that means reality is broken.- Noclevername. |
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