# Derivation of the area of a sphere formula.

1. Mar 12, 2005

### mprm86

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

2. Mar 12, 2005

### 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

3. Mar 12, 2005

### 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.

4. Mar 12, 2005

### 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.

5. Mar 12, 2005

### 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.

6. Mar 12, 2005

### 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.

7. Mar 17, 2005

### 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)*&pi;*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.

Last edited: Mar 17, 2005
8. Mar 17, 2005

### uart

I dont 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].

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

9. Mar 17, 2005

### tongos

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

10. Mar 22, 2005

### 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.

11. Mar 23, 2005

### 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.

12. Mar 23, 2005

### 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)

13. Feb 17, 2009

### prdmama

14. Jun 1, 2011