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Old 14-February-2008, 05:02 PM
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Default Some really tedious math questions

These are things I've always wondered but never really thought to ask because I imagined the answers would be incomprehensible to my cretinous mind.

But I''m going to do it anyway.

Why is the area of a circle pi r squared?

Why is the surface area of a sphere 4 pi r squared?

Why is the volume of a sphere 4 pi r cubed over 3?

Why is the volume of a cone one third the volume of a cylinder of the same radius and height? (and why is the volume of a pyramid one third the volume of a rectalinear polyhedron(?) of the same base area and height?)
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Old 14-February-2008, 05:11 PM
Planar WaveSherpa Planar WaveSherpa is offline
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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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Old 14-February-2008, 05:13 PM
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Well I was wondering how people actually determined them.
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Old 14-February-2008, 05:24 PM
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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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Old 14-February-2008, 07:49 PM
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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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Old 14-February-2008, 07:55 PM
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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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Old 14-February-2008, 07:57 PM
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My explanation above is Archimedes' Method, as illustrated in Disinfo Agent's link.
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Old 15-February-2008, 08:52 AM
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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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Old 15-February-2008, 12:20 PM
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Same proof we were taught in fifth grade, Archimedes' method then, wow, ontogeny recapitulates phylogeny
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Old 15-February-2008, 07:20 PM
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Old 16-February-2008, 04:52 AM
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Quote:
Originally Posted by parallaxicality View Post
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 2*pi*r, and the radius is r, then the area of the rectangle would be 2pi*r/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
Likewise, if you chop the sphere up into a bunch of pyramids and remember that V=(1/3)bh, the height h will tend toward r and b towards 4*pi*r2 hence the volume will be (4*pi/3)*r3. But why is it that the surface area of a sphere just happens to be exactly four times the area of a diametral circle? I can't think of an easy answer to that. The following argument will show you why.

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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Old 16-February-2008, 07:15 AM
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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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Old 16-February-2008, 09:24 AM
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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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Old 16-February-2008, 04:44 PM
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Quote:
Originally Posted by parallaxicality View Post
None...
Yup---the baker got me good on that one!
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