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Name: Sam
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Date: 5/2/2005

Why is the volume of a cone and a pyramid (1/3)(Base Area)(Height)? How did mathematicians come up with 1/3?


The formula is based on geometry and limits.

First, the geometry:

A cone and pyramid have one thing in common, the base area is proportional to the square of the height. If you keep the angles and overall shape the same and just increase the size (like a slight zoom of a camera lens), the base area is some constant times the square of the height: A=cH^2. The "c" is a constant that will eventually divide out. Also, the volume is proportional to the cube of the height: V=kH^3. We want to find out how k must relate to c.

Consider two cones with identical shapes: one has height H, one has height H+h. "h" is extremely small. If H were 10 meters, h would be a millimeter. The base area and volume for the first are A1=cH^2 and V1=kH^3. The other's area and volume are A2=c(H+h)^2 and V2=k(H+h)^3.

If we subtract the larger smaller volume from the larger volume, we get the very thin sheet at the bottom. If h is small enough, the thin sheet will have area A1=cH^2 and height h. The thin sheet will have a volume (cH^2)h. If you don't say that h is very small, then the sheet will have a more complex volume formula. We now have V2-V1=(cH^2)h. Let's do the algebra: k(H+h)^3- kH^3=(cH^2)h kH^3+(3kH^2)h+(3kH)h^2+kh^3 - kh^3 = (cH^2)h (3kH^2)h + (3kH)h^2 + kh^3 =(cH^2)h

Now, the limits:

"h" is tiny, so the second and third terms are extremely small compared to the first. They are like adding a penny to a million dollars. They really don't count enough to keep. Limit theory lets us drop them under the circumstances. This yields: (3kH^2)c=(cH^2)h. Divide out common terms and get: 3k=c, or k=(1/3)c. Put this in the original formula for volume: V=kH^3=(1/3)cH^3=(1/3)(cH^2)H=(1/3)AH. Volume equals (1/3)Base area times height.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College

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