What is the volume of a sphere?

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The discussion centers on proving the volume of a sphere as (4/3)(pi)r^3, with one participant attempting to derive it using integrals of circular cross-sections. There is debate about the validity of this approach, with some suggesting the use of spherical polar coordinates for a more natural integration method. A key point of contention arises from a claim that the volume of a sphere is zero, which is clarified as a misunderstanding since a sphere is a two-dimensional surface, while the volume enclosed by it is indeed (4/3)(pi)r^3. Participants express frustration over miscommunication and the complexity of calculus, with references to Archimedes' historical contributions to the problem. Ultimately, the conversation highlights the importance of understanding both the mathematical concepts and the terminology involved in discussing geometric volumes.
  • #31
Thanks for the interesting link to Archimedes' method. I remember my first year calculus text referenceing the "area" problem as a motivation for integral calculus, but one that had been solved by the Greeks, using the notion of a limit. However, they didn't explicitly define that notion or coin the term, IIRC. It's very interesting stuff. My apologies for my poor explanation regarding the "r" in the integral. It is a variable after all...the method works for any solid sphere of radius r. What I meant was that since we were integrating with respect to x (or y), of which r was independent, it could be treated as a constant in terms of the actual integration. The variable that you integrate with respect to does not remain after integration. For any variables that do remain, the volume is expressed as a function of those variables. So you get a general formula of the volume of a spherical object as a function of radius: V(r). Plugging in a specific value of r gives you a specific volume for a sphere of that radius, as you already know. I hope that clear's things up Jameson. Apologies to Daniel for previous remarks. All I was trying to say was that you were pointing out to Jameson that his use of the term "sphere" was semantically incorrect (if that is the right word...I mean to say that it was the wrong terminology *strictly speaking*), but you didn't state that explicitly, only implied it. Given that it was pretty clear what he meant, ie. that he was trying to find the volume of a spherical solid, or "ball", or the volume enclosed by a spherical surface, or whatever, I thought that not explaining clearly what the error was would only lead to confusion. I'll be honest: I used to think of a cylinder as a can...with a circular base. The mathematical definition is far more broad, as we learned in 3D coordiniate geometry. Same thing with sphere---until we saw the formal definition, I equated the term "sphere" in my mind with the everyday use of the word.
 
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  • #32
well to equate limits with calculus sems a bit broad to me, as that poses the question whether it is calculus to asset e.g. that .333... = 1/3, or even if .9999... = 1.

most inclusively, calculus involves three parts, the limit calculation of area, and of tangency, and of the connection between the two.

basically what archimedes did was discover a formula for the sum of the first n squares. then if you write out the Riemann sum - integral formulas for the area under the parabola y = x^2, ot for the volume of a sphere (or ball, but I think the historical justification for this use is lacking), you will see this formula occurs in both. so all three problems are the same.
 
  • #33
assume your sphere has radius R, and you subdivide it into n equal parts each of length R/n, and circumscribe a stack of cylinders each of height R/n, inside the sphere.

Then by pythagoras, the radius r of the ith cylinder satisfies r^2 + [(i-1)R/n]^2 = R^2.

i.e. r^2 = R^2 - [(i-1)R/n]^2.

So the volume of the ith cylinder is pi r^2 (R/n) = pi (R/n) [R^2 - [(i-1)R/n]^2].

Hence the sum of all their volumes is the sum of these volumes as i goes from 1 to n.

which is pi R^3 - pi R^3 [1^2 + 2^2 +...+(n-1)^2]/n^3

now if we have the formula for the sum of these squares, namely (n^3)/3 +
terms of lower degree,

we see that as n goes to infinity, this quantity approaches pi R^3 - (1/3)(pi)R^3

= (2/3)pi R^3 = volume of a hemisphere (or semi - volume enclosed by a sphere).

thus indeed knowing the sum of the squares implies one can compute volume of a spherical solid.
 
  • #34
as to area under a parabola like y = x^2, one subdivides the x-axis between 0 and b, into n equal parts, each of length b/n, and circum scribes rectangles over the parabola in each subinterval, the ith one having area (ib/n)^2(b/n) = i^2 b^3/n^3.

adding up from i =1 to i=n, gives us b^3 [n^3/3 +...]/n^3, which as n goes to infinity, approaches b^3/3.

again the only ingredient of the problem is the sum of the first n squares.

so archimedes made a very intelligent use out of that one formula, plus his understanding of limits.

which pretty much makes him father of integral calculus, if you wish.

i do not understand sow ell how he compouted the surface area of a sphere, unless he did it the way we do in calculus, approximating it by cones. i guess that's it. one of his books was on cones i think. so he was sort of the master of quadratic mathematics.
 
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  • #35
this has nothing to do with limits, but i thought up this way of finding the sum of the squares, which i thought was pretty cool.
1^2+2^2+3^2+4^2+5^2...x^2

all you have to do is look at increasing three digit numbers, and the amount of combinations in an increasing three digit number is 1+2+3+4+5+6+7+(1+2+3+4+5+6)+(1+2+3+4+5)...

but there is any easier way to express this, it could be 9(8)(7)/6 because an increasing number is only one out of six.

now look 100(1)+/99(2)/+98(3)+/97(4)/+96(5)+/95(6)/...+/1(100)/= 102(101)(100)/6

and 99(1)+97(2)+95(3)+93(5)...+1(50)

so the terms in / / are going to be repeated again, and you divide that by two so you get the series below, and you get (102)(101)(100)/24 as the sum which is (2n+2)(2n+1)(2n)/24 which is equal to the sum of the nth squares.
kind of a cool way. to find the sum of cubes, use four digit numbers.
 
  • #36
huh?...
 
  • #37
an increasing number is a number where all its digits are increasing from left to right (123, 234..) no digit being 0.

If i denote the first digit as 1, the second as 2, then i have (3,4,5,6,7,8,9) for the choices of the third digit. If i have the first digit as 1, the second as 3, then i have (4,5,6,7,8,9) to choose from. And so on. what i end up with is 7+6+5+4+3+2+1 combos for a three digit number with the first digit denoted as one. And (7+6+5+4+3+2+1)+(6+5+4+3+2+1)...+1 is the amount of combos for a three digit increasing number.
If all the digits are distinct, then there are six numbers to express with those three different digits. (123, 213, 321, ...) only one of those is increasing from left to right. so for all the digits to be distinct, we have nine choices for the first digit (1,2,...9), 8 choices for the second, 7 choices for the third. 9(8)(7) is the amount of combos for the three digit numbers which has all their digits different. and 9(8)(7)/6 for all the combinations of which the digits are increasing in a three digit number.
So...

9(8)(7)/6= (7+6+5+4+3+2+1)+(6+5+4+3+2+1)+(5+4+3+2+1)...+1 or

9*8*7/6= 7*1+6*2+5*3+4*4+3*5+2*6+1*7
 
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  • #38
Is there a relationship, yes!

because the squares go by a similar pattern...

(1)+(1+3)+(1+3+5)+(1+3+5+7)...
 
  • #39
It's weird that a thread with a nonsensical title got so many replies,many of them even correct...

Daniel.
 
  • #40
Dang, isn't it easier to just anti-differentiate the formula for the area of a circle?
 
  • #41
You mean the area of the sphere...?

Daniel.

P.S.\int_{0}^{R} 4\pi r^{2} dr=\frac{4\pi R^{3}}{3}
 
  • #42
Speaking of sums of squares, I guess it was interesting when I was a HS freshman but if you integrate \frac{n(n+1)+1}{2} and multiply it by 2 you obtain the formula for the sum of squares. Similarly you can integrate the formula for the sum of the squares and multiply by 3 to get the formula for the sum of cubes.

It's probably just a case of simple integration rules but it fascinates me as long as I'm too ignorant to see the connection!
 
  • #43
By the way, "pleonastic" is a good one. I haven't heard that used since 1970, when Nat Hentoff, the jazz critic said that about Ringo Starr's drumming. Hentoff didn't like Jimmy Hendrix guitar playing either.

I'll use that in class today, if I really want to obfuscate something.
 
  • #44
mathwonk said:
assume your sphere has radius R, and you subdivide it into n equal parts each of length R/n, and circumscribe a stack of cylinders each of height R/n, inside the sphere.

Then by pythagoras, the radius r of the ith cylinder satisfies r^2 + [(i-1)R/n]^2 = R^2.

i.e. r^2 = R^2 - [(i-1)R/n]^2.

So the volume of the ith cylinder is pi r^2 (R/n) = pi (R/n) [R^2 - [(i-1)R/n]^2].

Hence the sum of all their volumes is the sum of these volumes as i goes from 1 to n.

which is pi R^3 - pi R^3 [1^2 + 2^2 +...+(n-1)^2]/n^3

now if we have the formula for the sum of these squares, namely (n^3)/3 +
terms of lower degree,

we see that as n goes to infinity, this quantity approaches pi R^3 - (1/3)(pi)R^3

= (2/3)pi R^3 = volume of a hemisphere (or semi - volume enclosed by a sphere).

thus indeed knowing the sum of the squares implies one can compute volume of a spherical solid.

Thanks Mathwonk. I've been trying to find out how he did that.
 
  • #45
Awareness

I attached a plot of a circle with 9 rectangles circumscribed in the upper half semi-circle representing 9 cylinders in the ball. The Archimedes analysis thus proceedes as follows:

Since:
x^2+y^2=R^2

then:
r_i^2=R^2-(\frac{iR}{n})^2
So that that volume of the i'th cylinder is:

V_i=\frac{\pi R}{n}[R^2-(\frac{iR}{n})^2]=\frac{\pi R^3}{n}-\frac{\pi R^3 i^2}{n^3}

Thus the total volume for the upper hemisphere with n partitions is:

V_{hem}=\sum_{i=1}^{n}[\frac{\pi R^3}{n}-\frac{\pi R^3 i^2}{n^3}]
V_{hem}=\pi R^3-\frac{\pi R^3}{n^3} \sum_{i=1}^{n}{i^2}

V_{hem}=\pi R^3-\frac{\pi R^3}{n^3}[\frac{n(n+1)(2n+1)}{6}]

V_{hem}=\pi R^3-\pi R^3[\frac{1}{3}+\frac{1}{2n^2}+\frac{1}{6n^3}]

Taking the limit of this process as n approaches infinity, obviously the last two terms in brackets go to zero leaving:

V_{hem}=\frac{2}{3}\pi R^3

Finally giving for the whole ball: \frac{4}{3}\pi R^3

You know, I spoke of "awareness" elsehere in the group. How nice to be "aware" of retracing the footsteps of a master. Now I understand why he's called "the Father of Integral Calculus".

(my thanks to MathWonk for explaining the process)
 

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  • #46
Thank you,i didn't know of this construction being assessed to Archimede's name.I liked the one with the area under of parabola.This is one is nice,too.

Daniel.
 
  • #47
well i have never actually read archimedes. i have read that he discovered the sum of the first n squares, and that he could compute the area under a parabola, and the volume of a spherical ball.

having taught these subjects many times, and having always sought ways to explain and simplify them, I finally noticed these can all be derived by basically the same argument.

thus i surmised that this is the reason archimedes was able to do all of them.

so my attribution is that of a mathematical detective, deduced from evidence, not that of a historian working from actual sources.

still it seems very persuasive, does it not?

it seems a worthy goal is always to understand mathematical phenomena as clearly as possible, and as simply as possible. and to become freed from the canned presentations found in books, and only be guided by the underlying ideas of the masters.

once one knows the facts and tools available to the masters, one can be pretty confident that they would have successfully deduced any available consequences which follow from them.


so when i teach calc, I call this archimedes method, as opposoed to Newton's method, which uses the technique of antiderivatives to evaluate the same limit.

notice the different focus of mathematics in archimedes day, one of simply calculating an answer, rather than struggling with a precise definition of that answer, and questions such as whether or not area and volume actually
"exist".
 
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  • #48
do your calculus students see the beauty in mathematics? I mean, do they like it so much as to study it for fun?
 
  • #49
well, some do.

many are afraid they will get a bad grade. and these have a fear of trying to understand the material. they think it is safer to memorize the answer to all possible problems, which of course is impossible.

our job includes at least showing them an example of someone who does love the subject.
 
  • #50
mathwonk said:
well i have never actually read archimedes. i have read that he discovered the sum of the first n squares, and that he could compute the area under a parabola, and the volume of a spherical ball.

having taught these subjects many times, and having always sought ways to explain and simplify them, I finally noticed these can all be derived by basically the same argument.

thus i surmised that this is the reason archimedes was able to do all of them.

so my attribution is that of a mathematical detective, deduced from evidence, not that of a historian working from actual sources.

still it seems very persuasive, does it not?

it seems a worthy goal is always to understand mathematical phenomena as clearly as possible, and as simply as possible. and to become freed from the canned presentations found in books, and only be guided by the underlying ideas of the masters.

once one knows the facts and tools available to the masters, one can be pretty confident that they would have successfully deduced any available consequences which follow from them.


so when i teach calc, I call this archimedes method, as opposoed to Newton's method, which uses the technique of antiderivatives to evaluate the same limit.

notice the different focus of mathematics in archimedes day, one of simply calculating an answer, rather than struggling with a precise definition of that answer, and questions such as whether or not area and volume actually
"exist".

Another words, maybe he didnt' do it that way. Sure took me a long time to figure it out even with your description. I can't imagine a better way to do it from "first principles" though. The leap is the limit!
 
  • #51
yes, maybe he didn't, but I think I have made a conjectural case that he very likely did. It is hard to know how he did it since I understand some monk erased his works a few centuries ago, and the historians are only now trying to reconstruct it one symbol at a time, if you saw the tv show on it.

it seems well established for example that he did compute the area of a circle by approximating it by polygons with more and more sides, thus "exhausting" the circle in the limit. So it seems a small jump to exhaust a sphere by cylinders.

But yes, there is no guarantee he did it this way. What do you think?

The reason I think this, may sound strange if you are not a mathematician, but once the ideas are there, and all it takes is putting them together, then one often finds that all mathematicians do this independently in the same way.

I.e. if using these ideas that Archimedes had in his possession, I was able to construct these proofs, then it is not too much to expect that archimedes could certainly also do the same thing.

More bold perhaps, if Archimedes did it in some other way, then with my advantages of hindsight, I would also eventually succeed in doing it too. Since no one has suggested another way to deduce these results, probably Archimedes did not have one either.

Laypersons may believe for example that Fermat actually had a marvellous proof of his "last theorem" but I doubt any mathematician believes this. If such an elemetary argument has not been found in 350 years, then I think none exists.

All mathematicians share a grasp of logic, and an ability to reason by analogy. moreover the solution of a problem is most often not really created, but discovered, so if they are looking at it with the same tools, and in the same place, they will find it in the same way. That is why researchers hurry when they have made progress on a problem, because they know that anyone who hears what they have done, may be able to push it further in the same way as they are able.

But to be honest, since I am a mathematician and not a historian, I have no interest in doing research on Archimedes by reading parchments. I prefer to do it by thinking along what seem to be the same lines, and rediscovering the ideas myself.

It may seem odd, but i believe this is actualy more likely to lead to an understanding of what he did, than any other method available.
 
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  • #52
mathwonk said:
But yes, there is no guarantee he did it this way. What do you think?

My only problem is the inscription on his tomb which was a SINGLE rectangle inside a circle. Seems if this was his most cherished discovery and he did so as you suggest (which I believe also), then would they not have inscribed the tomb with the same figure I drew in the attachement above (the pancakes in the circle)?
 
  • #53
Here's, BTW, a link to the organization owning the palimpsest, where "The Method" is preserved:
http://www.thewalters.org/archimedes
 
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  • #54
saltydog, wasn't the inscription on his tomb rather a sphere inside a cylinder?

and when you say, arildno, that the palimpset "preserves" the method, have you seen the photos of a typical page of that document as it appears now? they showed it on tv , and it really is not readable. they are trying valiantly to eventually reconstruct some data from a parchment that was erased centuries ago.

I could be wrong, if say they were trying to show a particularly bad page. Maybe there are other pages with clearer writing, but I did not see anything you could learn from.

But I admit right off I am not a historian. I should not have said that Archimedes did this this way. I should have said, I believe he did.

Actually I would be very fascinated to read anything Archimedes wrote, if it is available. The palimpset seems not to be such. Is there a source for other works? I.e. a website for actual mathematical works?

By the way, although not a historian, I have of course read Plutarch's account of the siege of Syracuse by Marcellus, and the story of Archimedes inventions used in defense of the city, and of his death. I use this type of historical data to entertain my classes, and hopefully give some life to the subject.
 
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  • #55
Here's the relevant passage concerning the computation of the ball's volume:

"Archimedes is able to perform infinite sums: he takes a sphere, for instance, and calculates its volume as the infinite sum of the circles from which it is made... This was Archimedes' breakthrough, comparable to the modern integral calculus."
 
  • #56
thanks. where does that come from?

by the way, if you understand "circle" to mean very small cylinder, then this is exactly the method I gave as his, and that saltydog illustrated with his pancakes.
 
  • #57
Mathwonk: The palimpsest is in an awful state, so I think what is going on is a race to transcribe whatever can be retrieved from it, before the manuscript disintegrates completely.
 
  • #58
mathwonk said:
thanks. where does that come from?
It comes from the link, in Dr. Naetz's comment there
This should be it:
http://www.thewalters.org/archimedes/frame.html
 
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  • #59
oh yes, and the sentence that he able to perform infinite sums also argues to me that he used limits.

Ok I checked that link, without however finding the quote you mention. THis link does not work so well onmy browser for some reason. I want to say however that these quotes found on this site do not have the force of historical reliability.

I.e. although I am not a historian I am more careful than the trnascribers of these statements. They quote as fact, statements which are written with considerably more caution in the original documents.

For example historians question the strict accuracy of the amazing descriptions of machines lifting ships from the water and so on, which occur merely as repeated stories in the original documents, not as strict historical fact.

Moreover the account of Archimedes death given on this website, is but one if several competing accounts. yet the website gives it as the truth.

so one should be careful about citing sentences found on some websites as correct. Many websites seem often to be much less reliable as sources of information than the original sources.

To get a better idea of Archimedes siege of Syracuse one should actually read Plutarch. And even then one is dependent on the translation, if one does not read Greek. Even then one is dependent on the accuracy of an old document which may or may not be genuine.

I.e I am not a historian, but I try to be a critical scholar.
 
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  • #60
mathwonk said:
oh yes, and the sentence that he able to perform infinite sums also argues to me that he used limits.
I "recently" saw a documentary on the palimpsest, where Netz said that some of the crucial passages Heiberg had been unable to transcribe had now yielded to modern reading devices.
the most interesting of these was precisely concerned with how Archimedes managed to compute infinite sums..

EDIT:
Unfortunately, both that program and the site are "popular" versions, I really would like a scholarly presentation of what they've found, but that is still lacking, I think.
 
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