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Using Simple Geometry |

Previously, using simple geometry, we showed that the ratio of the circumference of a circle to its radius is independent of its size, and found a value for π accurate to about 10 decimal places. On this page, we'll carry that farther, and find the area of a circle and the area and volume of a sphere, using simple geometric arguments.

These formulas are, of course, trivial to find using calculus. However, the formulas were well known long before calculus was discovered; in fact, little more than the Pythagorean theorem is needed to derive them. Deriving them using calculus seems needlessly roundabout.

Figure
1: Area of a polygon,with N sides and perimeter P, with inscribed circle: |

In figure 1 we've shown a circle inscribed in a polygon. (It happens to be a hexagon but that's not important to the argument.) It has

Since there are

A circle is very much like a polygon with an arbitrarily large number of sides; as such, the area of a circle must also be the product of its radius and its perimeter, divided by 2. Since the perimeter of a circle is

As stated, this argument isn't terribly rigorous. By inscribing a second polygon within the circle to provide a lower bound on the area, and using the formal definition of a limit rather than just saying a circle is

Alternatively, we can visualize the same argument (with even less rigor) by cutting a circle in two, and cutting each half in pie wedges, as we've shown in figure 2.

Figure
2: Slicing a circle,
reassembling into a rectangle: |

Figure
3: Reassembled circle |

Figure
4: Sphere inscribed in cube: |

What we'll actually do here is find the difference between the volume of a cube and the volume of a sphere. We'll do this by

Figure
5: Half-sphere in half-cube: |

Take two objects, each the same length, each oriented parallel to theThe Slicing Principle:zaxis. Pass a plane through both of them, perpendicular to thezaxis. If theareaof the cut is the same for each object,no matter where the plane is located along the z axis, then the two objects must have the same total volume.

Let's make this concrete. Consider a pyramid. Cut it into thin horizontal "plates". The total volume of the pyramid is the sum of the volumes of all of the plates. If we allow the plates to slip across each other, so that the pyramid leans to one side or the other, the total volume is not affected. Similarly, if we change theshapeof each platewithoutchanging its thickness or volume (perhaps by making it circular rather than square), then the total volume of the pyramid -- and its height -- will not be affected, even though its new outline may be quite different. We can rephrase the condition above: Instead of slicing through each object with a plane at a particular location on thezaxis, we shall cut each object into plates, with each plate perpendicular to thezaxis. The volume of each object is the sum of the volumes of the plates it's been sliced into. If wepair upthe plates between the two objects based on their location along thezaxis, and if, in each pair of plates, the two plates have the same volume, then the two objects must have the same total volume.

Figure
6: Cone and block, total volume equal tovolume of half-cube minus volume of half-sphere: |

To start, note that the half-cube in figure 5 is oriented with the base lying on the

(1)

The cross section of the sphere is a circle, with radius which depends on

(2)

Figure
7: Half-sphere in half-cube,sliced by plane, to obtain cross section: |

(3)

And the net area of the region

(4)

Next we'll slice through the block and cone in figure 6 with the same plane, at the same height above the

(5)

Figure
8: Cross section through half-cube,with half-sphere cut out; cut z unitsfrom cube base: |

(6)

The sum of (5) and (6) is:

(7)

This is equal to (4), the area of the slice through the cube, minus the area of the slice through the sphere, as shown in figure 7. Consequently, we can conclude that the volume of the half-cube

The volume of the block is, of course,

(8)

We determined,

Figure
9: Cross sections through block and cone,showing total area through the two of them: |

(9)

And the difference between the volume of the half-cube and the volume of the hemisphere is:

(10)

But we know the volume of a half-cube, with height

(11)

And the volume of the sphere is, of course, twice the volume of the hemisphere:

(12) |

We imagine the sphere inscribed in a polyhedron. We don't require the polyhedron be regular; consequently, we can have as many faces on it as we like. We imagine a polyhedron of many tiny faces. By drawing lines from each vertex to the center of the polyhedron, and then cutting up the polyhedron along the surfaces extending between each edge and the two lines drawn from that edge's endpoints to the center, we can divide it up into pyramids. Each pyramid is built on one face, and has its point at the center. (Of course, the pyramids are not regular 4-sided pyramids, but we don't care.)

The volume of each pyramid is r/3 times the area of the face on which it's built.

The total area of all the pyramid bases is equal to the total surface area of the polyhedron. The total volume of the polyhedron is the sum of the volumes of all the pyramids. Consequently, the total volume of the polyhedron must be

(13)

where

By choosing a polyhedron with many facets, we can make the its volume as close as we like to the volume of the inscribed sphere. Similarly, its surface area can be made as close as we like to the surface area of the inscribed sphere. And so we must have, for the sphere as well as each polyhedron,

(14)

where

(15) |

Page created on 2/26/2008