A Non-calculus Explanation for the Volume of a Bead

  

After I wrote about the "surprising constant" that a bead formed by drilling a hole through the center of a sphere had a volume that was dependent only on the height of the remaining bead. At the time I admitted I could not think of a good "non-calculus" explanation for why the volume for a given bead height was a constant. Afterward Arjen Dijksman,(see his blog) who regularly provides helpful insights, suggested that the volume could be explained using the ideas of Mamikon Mnatsakanian that I discussed here. While I was not able to see Arjen's solution, it did lead me to realize that a "non-calculus" proof existed (with some allowances). Thinking of Mamikon's annular sweep led me to realize that every cross-section through the bead perpendicular to the axis of the drilled hole would be an annulus. In fact, using R as the original sphere's radius, r as the radius of the hole, and h as the distance above the center of the sphere where the cross section was taken, the area of the annulus would be Pi times (R² - r²-h²), which is just the area of a circle. Now by Cavalieri's Principal, "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." Several candidates exist with circular cross-sections; cylinders, cones, and spheres are examples. The cross-sections of the bead went from zero at the top, to a maximum at the center, so the cylinder is excluded. We can exclude the cone since their radii decrease linearly from top to bottom and our bead does not. And since we are already tempted with the idea that the volume of the bead is the same as a bead with a radius equal to half the total height of the bead, why not try to compare those? To make the notation easier, I decided to call the height of the bead 2H (letting H² equal the value (R² - r²). So the cross sectional area of the sphere with raidus H at any height, h, above the center will be Pi(H² - h²). As we have shown above, the cross-sectional area of the bead at the same height h is Pi (R² - r²-h ²). and since H² is equal to (R² - r²) these two are equal. So we have proven that the volume of the bead is the same as the volume of a sphere with the same height;... but did we use calculus? I fear that the proof of Cavalieri's Principal is founded in the calculus, and thus we still have a "calculus-based" explanation...but it's the best I have to offer to date.

Over the last few days I came upon a way to explain the volume of a bead with an (almost) non-calculus approach. My solution was inspired by a comment posted by Arjen Dijksman. I admitted I didn't understand his suggestion to use Mamikon's visual calculus, so he did what good teachers do. He posted a very clear explanation that even I could not miss. It's a beautiful blog, pretty math... check it out.. and Thanks loads Arjen.