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### Volume of a Cone

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Date: 02/19/99 at 11:35:04
From: Jodi Anderson
Subject: Volume of a Cone

I'm teaching a tenth grade geometry class, and I'm not happy with my
discussion on finding the volume of a cone. We have actually measured
the volume of a cone and compared it to the volume of a cylinder with
the same height and radius, and found that it is about 1/3 the volume
of the cylinder, but my students still want to know why. I do not
remember ever seeing a proof of this, except possibly in Calculus.
Do you have any suggestions for a proof of this formula?

Thanks.
```

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Date: 02/19/99 at 13:16:48
From: Doctor Rob
Subject: Re: Volume of a Cone

Slice the cone into n pieces with n-1 equally spaced planes parallel to
its base. Each of these pieces is a frustum of a cone, and has volume
that is greater than the volume of a cylinder whose radius is the
radius of the smaller base, and smaller than the volume of a cylinder
whose radius is the radius of the larger base. That is because the
small cylinder is contained in the frustum of the cone contained in the
large cylinder. Here is a diagram of a cross-section of the kth piece,
from the side:

A   B               C               D   E
o--o---------------o---------------o--o
| /|               |               |\ |
|/ |               |               | \|
o--o---------------o---------------o--o
F   G               H               I   J

The cross-section of the frustum is FJDB; of the larger cylinder it is
FJEA; and of the smaller cylinder it is GIDB. A segment of the axis of
the cone and the cylinders is CH, whose length is h/n, the height of
each frustum and each cylinder. For the kth piece from the top, the
radius of the smaller base BC = CD = r*(k-1)/n, and the radius of the
larger base FH = HJ = r*k/n. Then, if V(k) is the volume of the kth
frustum, we have the inequality

Pi*[r*(k-1)/n]^2*(h/n) < V(k) < Pi*[r*k/n]^2*(h/n),
(k-1)^2*Pi*r^2*h/n^3   < V(k) < k^2*Pi*r^2*h/n^3,

Now, add up these inequalities for k = 1, 2, 3, ..., n to get the
volume of the cone in the middle,

V = V(1) + ... + V(n).

Divide by Pi*r^2*h, and multiply by n^3, and you have:

0^2 + ... + (n-1)^2 < V*n^3/(Pi*r^2*h) < 1^2 + ... + n^2.

Now you have to know the sum of the first n squares is equal to
n*(n+1)*(2*n+1)/6, so the sum of the first n-1 squares is equal to
(n-1)*n*(2*n-1)/6. This can be proved by Mathematical Induction if you
don't already know this formula. So,

(n-1)*n*(2*n-1)/6 < V*n^3/(Pi*r^2*h) < n*(n+1)*(2*n+1)/6.

Now, divide by n^3, and rearrange things a little:

(1-1/n)*(1-1/[2*n])/3 < V/(Pi*r^2*h) < (1+1/n)*(1+1/[2*n])/3.

Now recall that n could be any positive integer, and the above
inequalities must be true. As n gets large, the quantity on the left
gets closer and closer to 1/3, and the quantity on the right also gets
closer and closer to 1/3. The difference between these two quantities
is just 1/n. As n gets large, this difference gets small, approaching
zero. In fact, if you want the left side to be larger than 1/3 - a, for
some real number a > 0, just pick n > 1/(2*a), and if you want the
right side to be smaller than 1/3 + a, just pick n > 1/(3*a). That
means that for any positive real number a, we have

1/3 - a < V/(Pi*r^2*h) < 1/3 + a

The only real number x that satisfies 1/3 - a < x < 1/3 + a, for every
positive real number a, is 1/3 itself. Thus

V/(Pi*r^2*h) = 1/3,
V = (1/3)*Pi*r^2*h.

By the way, a very similar proof, depending on the same formula for
the sum of the first n squares, works for pyramids, too.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry

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