To derive the volume of a cone formula, the simplest method is to use integration calculus. The mathematical principle is to slice small discs, shaded in yellow, of thickness delta y, and radius x. If we were to slice many discs of the same thickness and summate their volume then we should get an approximate volume of the cone.
The derivation usually begins by taking one such disc of thickness delta y, at a distance y from the vertex of a right circular cone. The radius of the disc is x, however, there will be a small error, shaded in red, due to the thickness delta y. As the thickness reduces to zero then so does the error.
Proportional triangles: As you can see, there are two right-angled triangles shaded in yellow. Their sides are proportional which is useful for building the expressions shown below.
Since the sides are proportional, we can write them like this. Nothing amazing or complicated happening here as you can see. A greater explanation is not required, as the key is to notice the proportional triangles.
Here, I am simply rearranging the expression to give x, which is the radius of the disc. At this stage, you may have figured out that this expression will be of use in another formula to calculate the volume of a disc.
Here, I am using the standard formula for calculating the volume of a disc. The volume of a disc is the same as the volume of a cylinder with a short height. In this case the disc has a radius x and a height delta y. Delta y is just another way of saying that it is a very small distance.
Substituting the previously found equation for radius into the standard equation for the volume of a disc gives us this expression.
Here we have just expanded out the power term, with simple algebra.
At this stage, the integration principle involves adding the volume of the discs to give us the volume of the cone. The summation symbol simply means that if we add the volume of all such discs between y = zero, and y = h, then we should get an approximate value of the volume of a cone. This is still an approximate value due to the error caused by the thickness of the disc delta y. The animated diagram above shows this by showing many discs sliced at various points along the y-axis.
As delta y approaches zero, then so does the error, and therefore we replace the delta y by dy and perform a definite integration operation. All we have done is replace the summation symbol with the integral operator. I normally leave PI behind the integral as it is a constant term throughout the expression and keeps it out of the way. Here we are integrating with respect to y of course.
After integrating y, the result is in the large square brackets. This makes mathematicians look clever, but it just means that you will be performing the same operation twice. Once with y = h, and then again with y = 0.
Here are the two terms expanded out and within their own square brackets. When you are calculating volume by integration, you always subtract the contents of the second brackets with the contents of the first.
The contents of the second bracket becomes zero because y = 0 and everything there is multiplied by zero. In addition, h² cancels out h³ leaving a spare h.
Removing all the un-needed terms and simplifying, it boils down to this familiar term for the volume. This is the simplest proof that Mr Smith taught me back around 1986 in Carshalton College. He was the best teacher I ever had.