What is the volume of a sphere?

[arildno]

i have since read some archimedes, not in detail, nd other sources. it now seems likely that eh did the volume of a pyramid first, then approximated the volume of a sphere by pyramids with vertices at the center of the sphere, then took limits and obtained the volume as 1/3 the product of the surface area and the radius, in perfect analogy with the case of a circle. then he may showed the surface area of a sphere agreed with that of the lateral area of a cylinder, finishing it off.

the works of archimedes are highly recommended, in print from dover.

Thanks for the update!

 

[coomast]

I am not going into the discussion of the spere or ball or whatever the round thing is called, I'm not a native speaking English person, I make up from the context what is exactly meant :-)
However, one thing struck me in the following post, namely the second and the last sentence:

Well, of all geniuses this world has seen, to me at least, Archimedes is still the greatest.
The gap between Archimedes and his contemporary researchers (by no means incompetent fellows) seems to have been so vast that they simply couldn't adjust themselves to his level (since he does not seem to have made lasting impact upon their thinking).
Since the Hellenic period in which Archimedes lived was the golden age of ancient technology (not only A. worked then), this fact is really remarkable.
This is quite different when comparing Newton, Gauss or Einstein to their ages; there were enough others who were able to appreciate their works.

I agree that this is true, but there are others who were also far ahead of their time. Evariste Galois for example. Don't misunderstand me, I am not seeking to open a discussion again, I completely agree with you Arildno. Taking this a bit further, what about Giordano Bruno?

 

[arildno]

Well, Abel would likely have understood some of Galois' work, if he had known it, and it didn't take more than 20-30 years after Galois' death before others recognized his importance.

However, it should be said that the manner in which Galois wrote his work, it was fairly illegible, and it had to be "cleaned up".
While there was rigour in his thinking, it was well hidden..
(Anyways, that's what I've heard about Galois)

 

[coomast]

Well, Abel would likely have understood some of Galois' work, if he had known it, and it didn't take more than 20-30 years after Galois' death before others recognized his importance.

However, it should be said that the manner in which Galois wrote his work, it was fairly illegible, and it had to be "cleaned up".
While there was rigour in his thinking, it was well hidden..
(Anyways, that's what I've heard about Galois)

This is certainly right. Galois' writing was difficult to read. He left out intermediate steps and didn't work systematically. However, if you can write papers and they are not recognized by the leading mathematical society, then you have the same "problem" as with Archimedes, you're ahead of your time, no?

Galois did read Lagrange and Abel, Cauchy rejected some of his papers, Poisson and Lacroix didn't come back on a memoir he wrote to them...

In Archimedes' time there were not so many people studying "science" compared to the time in which Galois lived so the chance of being understood was smaller, or is this incorrect? In Archimedes" case he was well recognized, fortunately.

Anyway, my admiration of both (and a lot of others) is the same.

An interesting book on Galois is "Galois Theory" written by Ian Stewart, ISBN 0 412 10800 3, with a small introduction on his life.

The remark on Giordano Bruno is a bit off post here. I only wanted to mention that things can go very bad if one is not understood. But in this case it is more related to religion and politics I believe.

 

[arildno]

Sure there were fewer scientists back in Archimedes' time, but that makes his insights all the more remarkable.

Consider how mathematicians work: They chat with each other all the time of various topics.
And so do all other scientists as well.

This is a positive feedback loop that spurs every one of them onto new research fields, and abandon worthless projects others have made them realize were worthless.

The lone genius is a very rare entity, mostly, gifted individuals without a social network of peers will degenerate into crackpots. Sad, but true..

Scientists need each other to stay on track and improve themselves.

 

[coomast]

Absolutely true, I can't agree more on this.

 

[mathwonk]

ok i have actually read more of archimedes and think i know how he found the volume of a sphere, or at least how he proved it. (he discovered it by setting up a lever and balancing the weights of different solids, knowing the centers of gravity of some of them, and deducing that of others.)

basic principles:
1) principle of parallel slices: two solids with equal areas for all plane slices parallel to a given plane, have equal volumes.
2) magnification principle: two pyramids with bases of equal area, have volumes in the same ratio as their heights.

these principles are proved by the method of approximation by blocks or cylinders, since solids with equal plane slices have equal approximating cylinders, and scaling the height merely scales the height of the approximating cylinders. then one proceeds as follows, first for pyramids and cones, then spheres.

step 1) right pyramids of height equal to base edge:
choose 2 opposite vertices on a cube, call them 1 and 2, and join them by a diagonal. choose a face having vertex 2 as a corner, and join every point of this face to vertex 1. this forms a right pyramid. the other two choices of faces having vertex 2 as corner, yield congruent pyramids, by rotation, and all three together make up the cube. thus the given right pyramid has volume 1/3 that of the cube, or 1/3 Bh, where B = area of base, and h = height.

step 2) using magnification principle, one extends the same formula to the case of arbitrary height in comparison to base edge, and using parallel slices one extends the same formula to pyramids which are not "right", but for which the angle to the vertex is arbitrary, since sliding a pyramid over at a new angle does not change the area of parallel slices.

step 3) approximating the base circle by polygons, hence approximating the cone by pyramids, gives the same formula for a cone, V = 1/3 Bh.

step 4) now circumscribe a cylinder about a sphere, and inscribe a double cone (vertex at center, bases at both top and bottom) in the same cylinder. then pythagoras shows that the area of a parallel slice of the cylinder has area equal to the sum of the parallel slices of the sphere and the cone.

Thus the volume of the cylinder equals the sum of the volumes of the cone and the sphere. in particular since the cone has 1/3 the volume of the cylinder, the sphere has 2/3 the volume of the circumscribing cylinder.

And that is how archimedes proved the volume of a sphere.

the by the argument above, viewing the sphere as a limit of pyramids with vertices at the center, he showed the surface area of the sphere, defined as the limit of the areas of the bases of the inscribed pyramids, was 3/R times the volume of the sphere, since tht is the formula for the base area of a pyramid in terms of the volume.

I.e. the volume of a sphere is 1/3 SR where S is the surface area and R is the radius.

and that's that! hooray for archimedes, who was obviously in almost complete command of the methods of purely integral calculus.

the only thing needing to be added, was the algebraic technique of antidifferentiating the algebraic formula for the area of the parallel slices and getting an algebraic formula for the moving volumes below each slice.

so as far as i know now it had nothing to do with ding up squares of integers at all, quite opposite to my original impression.

 

[mathwonk]

moreover archimedes said he could also compute that the volume of a bicylinder, intersection of two perpendicular cylinders, is 2/3 that of a circumscribing cube. his solution of this is lost, but you can guess it if you reflect that the horizontal slices of a bicylinder are intersections of horizontal slices of cylinders, i.e. intersections of rectangles, hence are squares.

thus you want to replace his prior use of a cone by some cone - like figure whose horizontal slices are squares. what do you guess? ...that's right! try it.

so this is archimedes actual work, and this i believe should be taught to every geometry student before attempting calculus. in fact harold jacobs' fine high school geometry book has this calculation of the volume of a sphere near the very end of his book.i also feel that this use of limits is not properly calculus, but that calculus is the combination of differentiation and integration, found in the fundamental theorem. i.e. i would preserve the term calculus for the use of antidifferentiation to compute the limits archimedes used to define volumes. of course this use of terminology is a matter of preference. note euler also declined to refer to the limits involved in infinite series, as calculus.

 

[arildno]

thanks mathwonk!
He becomes greater and greater, the more I get to know his work..

 

[dekyfineboy]

When the region between a and b of the function f(x) is rotated about the x-axis, the solid formed will have a volume

(pi)*(integration of f(x)^2). ----------------- 1

so we need the the formula of a circle so that we can put it into the formula

formula of a circle is given by r^2=x^2 + y^2 ---------------- 2
therefore making y the subject y^2=r^2 - x^2 ---------------- 3

put y=f(x) into the equation 1 and the formula for the volume of a sphere will be found.

 

[oscar_osa]

You are almost there,

When you integrate (r^2-x^2)dx you will have to fix the limits of integration, meaning, I can recommed from 0 to R, be careful, this is only half of the sphere, when you integrate the result is just r^2.R - R^3/3 this is iqual to R^3 - R^3/3 by just algebra this is equal to 2.R^3/3, as I told you this is just half of the sphere, double this number and you will obtain 4R^3/3, remember that pi was already out of the integration as constant. So to make the story short you have at the end 4.pi.R^3/3

I hope this helps

 

[dekyfineboy]

When the region between a and b of the function f(x) is rotated about the x-axis, the solid formed will have a volume

(pi)*(integration of f(x)^2). ----------------- 1

so we need the the formula of a circle so that we can put it into the formula

formula of a circle is given by r^2=x^2 + y^2 ---------------- 2
therefore making y the subject y^2=r^2 - x^2 ---------------- 3

put y=f(x) into the equation 1 and the formula for the volume of a sphere will be found.

note: the integration will have to have its limits as -r and r since that si the boundries of
the circle

 

[counterpoint]

Yea, I know I'm slow. Anyway, here's the volume using a triple integral. And I didn't know either what Daniel meant about the volume being zero, and in fact it took me a while to figure it out even after Cepheid explained it.

In spherical coordinates, the problem can be defined as follows:

vol=8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\int_0^r \rho^2 \sin(\phi)d\rho d\theta d\phi

Beautiful isn't it!

So:

8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2} \sin(\phi)(\frac{\rho^3}{3}){|}_0^r d\theta d\phi

and then:

\frac{4r^3 \pi}{3}\int_0^\frac{\pi}{2}sin(\phi)d\phi

or:

-\frac{4r^3\pi}{3}[0-1]=\frac{4\pi r^3}{3}

Don't you just love Calculus!

good.

 

[rickcjmac]

r is a constant because if you have a sphere, then r is a specific number. For example, a sphere of radius 2, r=2. That's kind of tough to grasp when looking at it for the first time.

Counterpoint gave a triple integral to represent the sphere in the first octant, which is only an eighth of the sphere, which is why he multiplied by eight to get the final volume. His answer was correct, but here is the volume of a sphere in triple integrals without cutting it up:

\int_0^{2 \pi }\int_0^{ \pi }\int_0^r\rho^2d\rho d\phi d\theta

\left( -\cos \phi \Big|_0^{ \pi } \right) (2\pi)\left(\dfrac{r^3}{3}\right)

\dfrac{4}{3}\pi r^3​

 

[ayushojha]

excuse me Jameson there is only one variable in the integration given by you. the "r" is not a variable but is a constant which is actually the limit for the variable "x" (upper limit in this case). To integrate you need to put a trigonometric form of x i.e x=r cos(theta). then choose the limits for "theta" and "theta" becomes the variable instead of "x".

 

[theprofessor0]

i haven't read all the replies, but there is a classic greek proof to this problem.
take a hemisphere of radius "r", a cone and cylinder of radius=height=(r)

then, volume of cylinder is pi r^3 and of the cone is 1/3*pi r^3
thus, all we need is to find the volume of the hemisphere
if 2 solids have the same area of cs fr all arbitary slices, they have equal volumes.
thus we have, the volume sphere =volume of cylinder-vol of cone. thus=2pi/3 r^3
so, volume of a sphere is 4*pi/3 r^3

 

[Those Who Squ]

BTW,in case you didn't know,the volume of a sphere is ZERO...

Daniel.

Is this because what the OP's really wants to prove is the volume of the closed ball comprising the sphere and its interior?

Forgive what's undoubtedly a simple question, but this isn't my field. I came in because I was interested in the same proof (or, at least, a proof of the same fact).

Also everyone please excuse my name. It seems to have gotten truncated and I'll have to see if the admins will let me change it.

 

[Those Who Squ]

i think one could prove it without calculus.

I haven't read through all this lengthy thread, but I know it can be done; I did it myself many years ago and have been trying to remember the details of the algebra involved. My approach involved imagining a hemisphere with n horizontal disks of equal thickness inscribed therein.

I haven't mastered the tex system yet, either, so I can't lay it all out. Essentially, though, the radius of each slice is one leg of a right triangle of which the other is kr/n where k is that slice's position counting up from the "equator" and n is the number of slices we are using. The Pythagorean theorem gives us the radius of slice k:

sqrt (r^2 - (k^2*r^2)/n^2) )

The volume of each slice is

(pi)*(r/n)*[r^2 - (k^2*r^2)/n^2) ]

so the solution for the whole hemisphere is simply to take the Riemann sum of all the slices. This involves the sum of the first n squares, which is

1/6*n*(n+1)(2n+1)

By decomposing this in the correct way I was able to arrive at a form which converged on the familiar 2/3*(pi)*r^3 value for the volume of the hemisphere, when n is allowed to increase without bound. Now I can't remember the details. I didn't use the cone and cylinder of Archimedes.

I suppose my proof is "calculus-ish", but it's definitely not formal calculus. Certainly there's no triple integral.

 

[nehal.helal]

i have proved the volume of sphere .it is similar to you .but there is difference. i have placed the sphere on x y z axis.and the origin coincides with center of sphere.if we cut the sphere cross sectionally parallel to y-axis then the radius of each cross section y and x will haver following relation
y^2+x^2=r^2
so,y^2=r^2-x^2
and let the distance between each cross be x0. then volume of hemisphere =pi[(r^2-x0^2)x0+(r^2-4x0)x0+(r^2-9x0)x0.........r/x0]
=pi[r^3-x0^3(1+4+9.......r/x0]
=pi[r^3/-x0^3[{r/x0(r/x0+1)(2r/x0+1)}/6]
=pi[r^3-x0^3[{2r^3/x0^3}/6] because x0 is extremely small
=pi[r^3-r^3/3]
=pi[2r^3/3]
so the volume of sphere will be4/3pi*r^3 double of volume of hemisphere

 

[fzngagan]

let me try i think i have a quite simple way to prove it