Volume of Sphere Derivation Proof

Proof by Integration using Calculus: Here is a sphere. If you cut a slice through it at any arbitrary position z, you get a cross-sectional circular area, as shown in yellow. The radius of this circle is x. Therefore the area of the circle shaded in yellow is given by π multiplied by its radius x squared.

Two thousand years ago Archimedes found this proof to be a piece of cake, but today school children still find this difficult to understand, therefore I have written it as simply as possible. There are many other ways to show this derivation using polar coordinates and spherical coordinates with triple integrals, but I doubt very many people would be interested in that.

This is a very simple proof using calculus, and using integration. Unfortunately, it is not possible to prove it without calculus because of the summation principle, but I promise to keep this derivation clean, minimalistic, and just as Mr Smith taught me all those years ago when I was a young Archimedean whippersnapper.

The derivation of the standard formula using calculus is very simple but it all starts here with some simple assumptions, and the use of the Pythagoras Theorem. The expression shown above describes the triangle outlined in red. I have just rearranged it for the term x, which is of course the radius of the circle. Since the expression gives x², it makes the rest of the work easier.

The expression above shows the standard formula for calculating the area of a circle, where x is the radius. However, we already have an expression for x² so just plug in the whole expression for x².

If you were to take many slices through the sphere and measure their cross-sectional area, and add them up, then eventually you would get a rough estimate of the volume. The accuracy obviously depends upon the thickness of the slices. Nevertheless, the principle here is that if the slices were so thin that the error introduced by their thickness was negligible, then essentially adding up all such areas will give you the volume.

The summation symbol above simply states that we are adding the areas of all the slices taken between the range +r and -r. Once added we should get the volume V.

In this step, we simply replace the summation symbol by the integral symbol, because the integration function finds the volume. You integrate the expression for the cross-sectional area between the limits +r and -r to find the volume.

You are integrating with respect to dz, where the r² term is a constant. The integral operator is one of those neat symbols that can solve so many problems in mathematics. When you have a constant such as π throughout the expression, you can factorise and move it behind the integral symbol out of the way.

Mathematicians love to use large square brackets because it makes them look clever. :-) But in this case it simply means that you will be performing this operation twice, once by substituting z with +r, and then again by substituting z with -r, and then subtracting the two as shown below.

When mathematicians get to make two large square brackets, they jump at the chance because it makes them feel even cleverer. However, from this point on just treat them as normal brackets. All we have done here is expand out by substituting for the +r term and -r terms.

Do not forget the minus multiplied by minus makes a plus rule. There is a lot of this going on in the second bracket, as you can see, but all very simple and straightforward algebra.

In the end, it all boils down to two fractions as you can see. Keeping the π out of the action keeps things simple as you can see; otherwise, you will be wasting a lot of ink writing it all over the place.

Well, if you understood this proof by integration using calculus then congratulate yourself. This will also work for any n-dimensional sphere. I just thought I would mention that in case you find yourself in fluidic space with species 8472, it could come in very handy.