Derivation of the area of a sphere formula.

[mprm86]

Deduce the formula for the area of a sphere with ratio R. (I already know it is 4*pi*R^2)

 

[Jameson]

The volume of a ball is \frac{4}{3}{\pi}r^3 and the surface area of a ball is 4{\pi}r^2. The surface area is the derivative of the volume.

Jameson

 

[dextercioby]

Not really.The cube...

The simplest is:
S=\iint dS=R^{2}\int_{0}^{2\pi}d\varphi \int_{0}^{\pi} d\vartheta \ \sin\vartheta

Daniel.

 

[dextercioby]

How do you prove the volume of the ball is the one that u mentioned & how do you prove that the area of the 2-sphere is the derivative (wrt radius) of the volume of the 3-ball...?

Daniel.

 

[mprm86]

Yes, I have to prove that the volume of a sphere is \frac{4}{3}{\pi}r^3, and that the surface area of a sphere is 4{\pi}r^2, without using integrals.

 

[dextercioby]

The volume of a sphere is zero...There's a giant thread here (i think in the "Calculus & Analysis" forum) bearing this meaningless name (the volume of a sphere),in which Saltydog,Mathwonk & Galileo give Archimede's rationale...

Daniel.

 

[metacristi]

The volume of a sphere can be obtained without using integrals by remarking that the volume of a cylinder of radius r and height r is equal with the volume of a cone having the radius of the base r and the height r + the volume of a semisphere of radius r. If I remember well the demonstration involved the use of the principle of Cavalieri. So the volume of a semisphere is (2/3)*π*r³. But I do not think that the area of a sphere can be obtained without calculus, at least I do not think there is an easy (and exact) method.

 

[uart]

I don't know of any method that doesn't rely on caculus, but the volume is a pretty easy "volume of rotation" integration. Just set up a hemi-sphere as the volume of rotation of the function y=+sqrt(r^2 - x^2) over x=[0..r].

& how do you prove that the area of the 2-sphere is the derivative (wrt radius) of the volume of the 3-ball...?

dV = A dr can be obtained "by inspection".

 

[tongos]

missing calculus...integrals are good to solve such.

 

[Alkatran]

Start from the equation y = sqr(x^2 - r) (r is a constant)
Integrate it to get the surface of a circle, pi*r^2
Then say you are putting a whole bunch of discs, with volume pi*r^2 * dx along the x-axis with radius y = sqr(x^2 - r) to get this function:

Integral[-1 to 1]: sqr(x^2 - r)^2 * pi = pi*Integral[-1 to 1]: x^2 - r


Of course, you'd also need the proofs that a semi circle is defined by y = sqr(x^2 - r) and all the calculus ones.

 

[Galileo]

Archimedes solved the problem of finding the area of a sphere with given radius.

I don't know how he did it, but I know it his solution counts as one of the most notable mathematical achievements.

 

[Alkatran]

Whoops, my mistake, I meant to use the perimeter of a circle formula (2pi*r) * dx, which would give you the sum of the areas of the rounded edge of all the discs (which would approach sphere's area as dx -> 0)

 

[prdmama]

http://www.zazzle.com/derivation_of_the_formula_for_the_surface_area_tshirt-235183739395232296

I don't work for zazzle or care to serve as an ad campaign for them. But this was
the clearest and most concise derivation of the area of a sphere that I found.

 

[bejugo]

This does not precisely deduce the area formula for a sphere, but it's nonetheless a very nice video representation, conceptually: http://www.rkm.com.au/animations/animation-Sphere-Surface-Area-Derivation.html