Bonaventura Cavalieri observed that if a set of parallel planes cutting two figures of equal height always form cross-sections of equal area, then the volumes of the solids are equal.

Using Cavalieri’s principle we are able to derive the formulas of the volumes of many 3-D shapes such as prisms, triangular pyramids, cones and spheres. However, for the purpose of keeping this article brief, we will take the following for granted as we do this:

1. The volume of a cuboid is length * height * width
2. The formula for the area of a circle
3. We can cut a triangular prism into three congruent triangular-based pyramids.

Firstly, the volume of a prism can be proved very simply. Considering the cross-section is always constant in a prism and a cuboid, we can construct a prism and a cuboid of equal height whose cross-sections are equal. Because the cross-sections are constant, they will always be equal throughout both figures and therefore, by Cavalieri’s Principle, the volumes are equal. By definition, the volume of a cuboid is height * cross-section where the cross-section is length * width. Therefore, the volume of a prism must be height * cross-section.

Now that we know this, we can prove the area of a triangular-based pyramid. As mentioned above, we will take for granted that we can cut a triangular prism into three congruent triangular-based pyramids. Therefore, we the volume of these pyramids must be exactly a third of a triangular prism. However, we cannot write the formula as ⅓ * height * cross-section as there is no constant cross-section. However, the base of the pyramid is equal to the cross-section of the prism, therefore the formula for the volume of a pyramid is ⅓bh.

Using the above information, we can prove the volume of a cone and pyramids in general. We take an n-sided polygon as the base of a pyramid and split it up into triangular sections like so:

Now, we can split this pyramid into six triangular-based pyramids whose volumes equal  ⅓bnh (where n is a number corresponding to a certain triangle) and who share an apex with the hexagonal-based pyramid. Let the base area of the hexagonal-based pyramid be bt. We know that the volume of the entire hexagonal-based pyramid must be (⅓b1h + ⅓b2h + … + ⅓b6h). We can factorise this to ⅓h(b1 + b2 + … + b6). Given that b1 + b2 + … + b6 = bt, we can express the area of this pyramid as ⅓bth. This logic will also work for a pyramid with any polygon as a base.

Therefore, the formula for the area of all pyramids is ⅓bh, not just triangular-based pyramids. Now if we add more and more sides to the n-sided polygon, we essentially get what is a circle. This would then be the base of what is essentially a cone. Therefore, a cone is in essence, a pyramid whose base area equals πr2. Thus, by substituting this value for the base area in the formula for a volume of a pyramid, we know that the volume of a cone is ⅓πr2h.

Finally, we can use this to prove the volume of a sphere. Take a look at the following diagram:

Here, we have a hemi-sphere with a base of radius r and a cylinder whose bases have radius r and a height of r. Also, the cross-sections (in red) are at a height of y. In the cylinder, we cut a cone out whose base is the same as the cylinder (as shown in the diagram). We know that the area of the base of the cylinder must be πr2 from the area of a circle. By the volume of a cylinder (using the same logic as that of a prism), given that the height is r, the volume this cylinder must be πr3. The area of the base of the cone we have taken out also has an area of πr2 and the height is ⅓r, therefore, the area of the cone is ⅓πr3. If we take this away from the area of the cylinder, we get ⅔πr3 as the volume of this figure.

Next, let’s look at the green triangle in the cylinder. The red cross-section cuts the triangle to create a smaller triangle above. This smaller triangle is similar to the green triangle by AAA similarity. Therefore, we can represent the ratio of the height and length of the two triangles as r/r = ( r- y)/l. Given that r/r = 1, we can re-arrange this to give us y = r – l. Now, we look at the cross-sections of the two figures (in red). The cross-section of the cylinder is an annulus formed by taking a circle of radius l from a circle with radius r – l. Therefore, the area of the annulus is πr2 – π(r – l)2 = π(r2 – y2).

Now let’s look at the purple triangle in the hemi-sphere. We are given the two sides: r and y. Using the Pythagorean Theorem, the other side must be  r2 – y2 which is the radius of the cross-section of the hemi-sphere. Therefore, by the area of a circle, the area of the cross-section of the hemi-sphere is π(r2 – y2). Since the value of  can be changed to give all cross-sections of the cylindrical figure and the hemi-sphere, this means that their cross-sections are always equal. As shown above, the area of the cylindrical figure is ⅔πr3, therefore, the area of a hemi-sphere is ⅔πr3 and the area of a sphere is 4/3πr3.