What is the volume of a sphere?

[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