High school students would probably do it in a jiffy. I've been baffled for a long time now about proving the cone volume formula of PI*R^2*H/3 .. Just haven't figured it out till date (i.e. a good solution). It's amazing that the volume formula has existed for such a long time and was invented before calculus came into being. My last attempt was basically cheating on the idea of a limiting value to determine the radius of an infinitesimal height disc that constitutes a cone.
Yesterday, however, I thought I found a better answer - not nearly close to an elegant solution - because it still uses the volume of a cylinder (which is slightly easier to explain), but a calculus-free solution nevertheless. The idea came ironically from a book on calculus (one I purchased recently). The second theorem of Pappus which states that the volume of revolution of a planar lamina about an external axis is equal to the surface area of the lamina times the distance travelled by the center of gravity of the lamina, plays the central statement in the solution. ("Central" sounds like there're too many things about it, but really there's just a little more).
So, now imagine a cylinder (volume = PI*R^2*H) with radius R and height H. Also imagine a right triangular lamina with its back (the perpendicular) resting on the cylinder. The lamina's dimensions height=h=H, base width = r. If you rotate this triangle about the cylinder's central axis you'd get a section of the cone. If you reduce the radius of the internal cylinder to zero, you'd get a perfect right-circular cone.
The volume of the cone can therefore determined as the volume of the rotation volume of this circular lamina.
Incidentially :) the position of the centroid is at (r/3, h/3) from the right vertex. Thus, the distance travelled by the center of gravity is (2*PI*r/3).
And the area of the right triangle is (r * h)/2 = (r * H/2)
Therefore, the volume of rotation (or the volume of the cone) is 2*PI*r/3 * (r * H/2) = (PI * r^2 * H) /3 which is what we set out to prove... It was so simple ..
Please don't ask me to prove Pappus's theorem without calculus..
2 comments:
And how did you get the center of gravity of the cone without calculus?
Centers of gravity of the triangle is obtained by determining the intersection point of side-bisector line segments from the opposite vertices.
In the case of this simple right triangle, we can use simple cartesian geometry to determine the coordinates (relative to the right vertex) of the center of gravity.
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