The formula for finding the volume for a sphere?

The volume of a sphere is

V = (4/3)πR3.

A formal way to obtain it.

Take the origin of a system of axes in the center of the sphere, then use spherical coordinates:

(x, y, z) = (ρ cosθ cosφ, ρ cosθ sinφ, ρ sinθ),

with ρ in [0, R], θ in [- π/2, π/2] and φ in [0, 2π]. Now you can build the Jacobian matrix of the coordinate change, and find its determinant: it is ρ2 cosθ.

Thus the volume is

V := ∫V d3r = ∫[0, R] ∫[- π/2, π/2] ∫[0, 2π] ρ2 cosθ dρ dθ dφ = (R3/3) (1 - (-1)) 2π = (4/3)πR3.

I don't think so! And I'll demonstrate to you:

if you take V := ∫V d3r then you have V := ∫ (∫V d3r) d3r and then V := ∫(∫(∫V d3r) d3r) d3r and etc.. so it's recursive! For the eternity!

This is because π is irrational! The only way to solve this question is to deep a sphere in water and measure the volume of the moved water.

Well, probably my notation is not clear. Notice that in

V := ∫V d3r

the second time "V" appears it is smaller: i meant with this that you have to integrate d3r on the volume V of the sphere (that is, in fact, the definition of the volume). In

V := ∫V d3r

the little "V" is at the foot of the "∫" symbol.

An easier way to obtain the answer:

Imagine an oragasm which is divided into an infinite amount of prisms with a common vertex at the centre of the sphere. By calculating the volume of all these prisms, one can obtain the volume of the sphere.

The formula for the volume of a prism is (1/3)bh. If we apply this formula to the infinite number of pyramids, the total area of the bases (b) would be the SA of the sphere, (4πR^2), the height (h), would be the distance from the surface area to the centre, which is the radius (R).

This means the formula for finding the Volume of a sphere would then be (1/3)bh which is (1/3)(R)(4πR^2) which could then be simplified to (4/3)πR^3