Where does the factor of 4/3's come from for calculating the volume of a sphere?

Question

In the equation v = 4/3 pi r^3 why does the 4/3 come up?

Answer 1

from calculus
divide the sphere up into circular slices of width dx.
the disk at a distance of x from the origin has a radius of sqrt(r^2-x^2) and so an area of pi(r^2-x^2) and a volume of pi(r^2-x^2)dx
summing up the volumes of all the disks gives the volume of the whole sphere = int pi(r^2-x^2)dx from x=-r to r
=pi.r^2.x-pi/3.x^3 from x=-r to r
=(pi.r^3-pi/3.r^3)-(-pi.r^3--pi/3.r^3)
=4/3.pi.r^3

Answer 2

The 4/3 isn't too hard. Consider the sphere to consist of sherical shells. The shell of radius = r has a surface area of 4pi r^2 and thickness dr. The thickness of the shell is so small that the inner surface area of the shell is essentially the same as the outer surface area of the shell and so the volume of a shell is 4pi r^2 dr and the volume of the sphere with radius R = integral of 4pi r^2 dr from r=0 to r = R = 4/3pi r^3. The 3 comes fron integrating r^2 and the 4 comes from the surface area of a sphere.

I realize that this use of differentials is a little funky but it gives more insight into what is going on. There is a difference between doing a computation or a proof and seeing why it is true.

Check out "The Calculus: An Opinion" at mathematicsteacher.org

It is interesting to ask why the surface area of a sphere of
radius = r is 4 times the area of a circle of radius = r.

Answer 3

For any number of different size perfect spheres whose radii and volumes are measured physically, if you keep on dividing the volume by the respective cube radius (r^3) of each sphere, you will constantly get the result 4/3*pi. That 4/3 and pi are naturally there. We can indirectly show this by using different methods such as calculus.

Answer 4

First you can get the area of the sphere from calculus. One easy to understand way is to divide the sphere's surface up into small bands of longitude (theta) ranging from -pi at the south pole to +pi at the north pole, and divide each band into latitude segments from 0 to 2*pi. Then, per the ref., the calculus shows that the area A = 4*pi*r^2.
From this integration to volume is easy. Just consider the volume to be made up of shells of infinitesimal thickness dr. Following the rule of integration, the integral with respect to r of 4*pi*r^2*dr = 4/3*pi*r^3. (Plus a constant which in this case is zero.)
EDIT: I see answerer 2 scooped me. Anyway the reference may help on the surface area integration.