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Cone Volume Formula
Name: Sam
Status: educator
Age: N/A
Location: N/A
Country: N/A
Date: 5/2/2005
Question:
Why is the volume of a cone and a pyramid (1/3)(Base
Area)(Height)? How did mathematicians come up with 1/3?
Replies:
Sam,
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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Update: June 2012
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