# Volume of pyramids, cones, and spheres (Part 3)

I’m in the middle of a series of posts concerning the area of a triangle. Today, however, I want to take a one-post detour using yesterday’s post as a springboard. In yesterday’s post, we discussed a two-dimensional version of Cavalieri’s principle. From Wikipedia:

Suppose two regions in a plane are included between two parallel lines in that plane. If every line parallel to these two lines intersects both regions in line segments of equal length, then the two regions have equal areas.

In other words, if I have any kind of shape that has cross-sections that match those of the triangles above, then the shape has the same area as the triangles. Geometrically, we can think of each triangle a bunch of line segments joined together. So while the positioning of the line segments affects the shape of the region, the positioning does not affect the area of the region.

There is also a three-dimensional statement of Cavalieri’s principle, and this three-dimensional version is much more important than the above two-dimensional version. From MathWorld:

If, in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.

Pedagogically, I would recommend introducing Cavalieri’s principle with two-dimensional figures like those from yesterday’s post since cross-sections in triangles are much easier for students to visualize than cross-sections in three-dimensional regions.

This three-dimensional version of Cavalieri’s principle is needed to prove — without calculus — the volume formulas commonly taught in geometry class. Based on my interactions with students, they are commonly taught without proof, as my college students can use these formulas but have no recollection of ever seeing any kind of justification for why they are true. When I teach calculus, I show my students that the volume of a sphere can be found by integration using the volume of a solid of revolution: $V = \displaystyle \int_{-R}^R \pi \left[ \sqrt{R^2 - x^2} \right]^2 \, dx = \frac{4}{3} \pi R^3$

Without fail, my students (1) already know this formula from Geometry but (2) do not recall ever being taught why this formula is correct. Curious students also wonder (3) how the volume of a sphere (or a pyramid or a cone) can be obtained only using geometric concepts and without using calculus.

For the sake of brevity, I only give the logical flow for how these volumes can be derived for students without using calculus. I’ll refer to this excellent site for more details about each step.

• Using a simple foldable manipulative (see also this site), students can see that $V = \displaystyle \frac{1}{3} Bh$ for a certain pyramid — called a yangma — with a square base and a height that is equal to the base length.
• Enlarging the yangma will not change the ratio of the volume of the pyramid to the volume of the prism.
• Cavalieri’s principle then shows that $V = \displaystyle \frac{1}{3} Bh$ for any square pyramid.
• Cavalieri’s principle then shows that $V = \displaystyle \frac{1}{3} Bh$ for any pyramid with a non-square base or even a cone with a circular base.
• Finally, a clever use of Cavalieri’s principle — comparing a sphere to a cylinder with a cone-shaped region removed — can be used to show that the volume of a sphere is $V = \displaystyle \frac{4}{3} \pi R^3$. I note in closing that there are other ways for students to discover these formulas, like filling an empty pyramid with rice, pouring into an empty prism of equal base and height, and repeating until the prism is filled.

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