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

The thing with cubes is that the cut section at every height has the same size and shape, allowing the “squares” to have a height – i.e. be 3-dimensional with height 1 each so you only have to add c of those “squares”.

This doesn’t work with spheres, because if you made the circles 3-dimensional before stacking them then the result would no longer be a sphere. So you’d have to keep the circles 2-dimensional, without height.

A way to resolve this is to still treat those circles as 3-dimensional but with infinitesimal (infinitely small) height, but that’d still require you to add up infinitely many of them to get the whole sphere. This would essentially compute the volume of the sphere as an integral but it’s much more complex than the shortcut you can use with cubes.

The answer is ‘yes’, but the math for those circles is a little complicated.

But yes, you’ve essentially got stacked circles increasing in radius from a point to your given radius then decreasing again until you get another point, and boom, sphere. The rate at which your radius changes is important, though!

In the weeds:

Your circles’ radiuses need to change based on a sine function iirc in order to get a sphere instead of an hourglass, stacked cones, or some kind of spheroid.

Think circles of πr_c^2 where r_c = r_s * sine(h), then you’d have to do some calculus to sum them up based on limits of h.

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.

No, because it doesn’t hold true even for the cube.

A square is a 2D object and has no thickness. So you can put however many you like above each other and would get either

– all squares in the same place with still thickness 0 or

– the squares “stacked” on one another but with a gap, so no new object is formed

Even if you’d use infinetly many squares for the second option, there would still be an infitesimal small gap.