If a cube can be construed as a series of squares stacked one upon the other, can a sphere be considered as a series of circles stacked one upon the other?

Question

If a cube axbxc can be said that it is a series of “c” squares axb stacked one top of another, then volume of cube is sum of areas of all the c squares

ab+ab+…. ab c times, so abc.

Similarly could a sphere of radius r can be seen as a series of circles stacked one over other each with increasing radius from 0 to r for the top and bottom halves of sphere independently.

In that case volume of sphere is the twice the sum of all the areas of those circles.

2*pi*[ r^2+(r-1)^2+……0)

In: 0

Answer 1

Kind of! The thing you’re intuiting is called “Riemann Summation.” The thing is that all the squares you’re using to build the cube have the same shape, so it doesn’t matter how thick they are. By saying you’d use C squares for a box of height C, you’re suggesting that you use all squares of thickness 1. But you could use C/2 squares of thickness 2, or 2*C squares of thickness 1/2.

Spheres don’t work that way. If you stack circles of the same size and some small thickness, you get a cylinder. So, you have to change the size of the circles from very small at the top and bottom to very big in the middle. But the boundary between those layers will look blocky.

Of course, it’ll look less blocky if you use more circles with less thickness, so the difference between any two layers is smaller. Riemann Sums, and the integral calculus you get by working with circles so thin they have only hypothetical thickness, does look at spheres as a bunch of disks stacked on top of each other, and uses that to estimate their volume.