Bonaventura Cavalieri observed that if a set of parallel planes cutting two figures of equal height always form crosssections 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 3D 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:
 The volume of a cuboid is length * height * width
 The formula for the area of a circle
 We can cut a triangular prism into three congruent triangularbased pyramids.
Firstly, the volume of a prism can be proved very simply. Considering the crosssection is always constant in a prism and a cuboid, we can construct a prism and a cuboid of equal height whose crosssections are equal. Because the crosssections 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 * crosssection where the crosssection is length * width. Therefore, the volume of a prism must be height * crosssection.
Now that we know this, we can prove the area of a triangularbased pyramid. As mentioned above, we will take for granted that we can cut a triangular prism into three congruent triangularbased 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 * crosssection as there is no constant crosssection. However, the base of the pyramid is equal to the crosssection 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 nsided polygon as the base of a pyramid and split it up into triangular sections like so:
Now, we can split this pyramid into six triangularbased pyramids whose volumes equal ⅓b_{n}h (where n is a number corresponding to a certain triangle) and who share an apex with the hexagonalbased pyramid. Let the base area of the hexagonalbased pyramid be b_{t}. We know that the volume of the entire hexagonalbased pyramid must be (⅓b_{1}h + ⅓b_{2}h + … + ⅓b_{6}h). We can factorise this to ⅓h(b_{1} + b_{2} + … + b_{6}). Given that b_{1} + b_{2} + … + b_{6} = b_{t}, we can express the area of this pyramid as ⅓b_{t}h. 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 triangularbased pyramids. Now if we add more and more sides to the nsided 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 πr^{2}. 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 ⅓πr^{2}h.
Finally, we can use this to prove the volume of a sphere. Take a look at the following diagram:
Here, we have a hemisphere with a base of radius r and a cylinder whose bases have radius r and a height of r. Also, the crosssections (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 πr^{2} 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 πr^{3}. The area of the base of the cone we have taken out also has an area of πr^{2} and the height is ⅓r, therefore, the area of the cone is ⅓πr^{3}. If we take this away from the area of the cylinder, we get ⅔πr^{3} as the volume of this figure.
Next, let’s look at the green triangle in the cylinder. The red crosssection 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 rearrange this to give us y = r – l. Now, we look at the crosssections of the two figures (in red). The crosssection 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 πr^{2} – π(r – l)^{2} = π(r^{2} – y^{2}).
Now let’s look at the purple triangle in the hemisphere. We are given the two sides: r and y. Using the Pythagorean Theorem, the other side must be r^{2} – y^{2} which is the radius of the crosssection of the hemisphere. Therefore, by the area of a circle, the area of the crosssection of the hemisphere is π(r^{2} – y^{2}). Since the value of can be changed to give all crosssections of the cylindrical figure and the hemisphere, this means that their crosssections are always equal. As shown above, the area of the cylindrical figure is ⅔πr^{3}, therefore, the area of a hemisphere is ⅔πr^{3} and the area of a sphere is ^{4}/_{3}πr^{3}.
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Aarit Bhattacharyahttps://etonstem.com/author/aarit04

Aarit Bhattacharyahttps://etonstem.com/author/aarit04

Aarit Bhattacharyahttps://etonstem.com/author/aarit04

Aarit Bhattacharyahttps://etonstem.com/author/aarit04
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