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The volume of a cone, without calculus
The volume V of a cone with base area A and height h is well known to be given by V = 13 Ah. The factor 13 arises from the integration of x^2 with respect to x. The object of this note is to start by supposing V = cAh, and to show–without calculus–that c = 13. Using the cone formula, we’ll also deduce the volume and the surface area of a sphere of radius R. Consider the frustum of height h, top area a, and base area A, cut from a cone of height e + h (e is for “extra”) and base area A. The volume of the frustum is
V = cA(e + h) − cae.
Now, the area of a cross-section of the cone is proportional to the square of its distance from the vertex, so √ a e =
A
e + h.
It follows that
e =
√ a A −
a
h, e + h =
√ A
A −
a
h
and the volume of the frustum is
V = cA
√ A
A −
a
h
− ca
√ a A −
a
h
= c
A
A − a
√ a A −
a
h = c(A +
Aa + a)h.
Now consider what happens as a tends to A. The frustum becomes a cylinder, and we find that V = 3cAh. But we know that, for a cylinder, V = Ah, so c = 13 , and we conclude that the volume of a cone is
V =
3 Ah. Typeset by AMS-TEX 1
2
As a bonus, we obtain the volume of a frustum:
V =
3 (A^ +^
Aa + a)h.
We conclude with two simple applications of the formula.
The volume of a sphere FIGURE 1 shows a sphere radius R, together with a cylinder of radius R and length 2R; cones are drilled out from each end of the cylinder to its center. If we slice each object at a distance x from its center, the area of the slice is, in each case, π(R^2 − x^2 ). Thus the two solids have the same volume, and we conclude that
V = πR^2 · 2 R − 2 · 1 3 · πR^2 · R =^4 3 πR^3.
The surface area of a sphere Given a sphere, we divide the surface into very many small (flat) pieces of area Ai, i = 1, · · · , n. We join each to the center, forming sharp cones. The volume of a typical cone is V = 13 AiR, and the total volume of all the cones is
V =
3 R
∑^ n i=
Ai =
3 RS,
where S is the surface area of the sphere. Thus 13 RS = 43 πR^3 , and so
S = 4πR^2.
