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Surface Area of a Sphere

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Date: 10/03/97 at 09:21:54
From: Stephanie Kaszuba
Subject: Surface Area of a Sphere

I would like help in assisting my daughter in answering the question,
"How is the surface area of a sphere calculated, and why?".

Bob Kaszuba
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Date: 10/03/97 at 11:29:10
From: Doctor Rob
Subject: Re: Surface Area of a Sphere

Lengths, areas, and volumes of curved objects are hard to compute
with the tools of ordinary geometry.  That is one reason that the
calculus was invented. Without resorting to the calculus, however,
one could establish some of the formulas by the following sort of
reasoning.

Construct two polyhedral objects, one contained in the curved object,
and one containing it. The surface area of the curved object must
exceed that of the contained object, and be exceeded by that of the
containing object.

Measure the surface area of the polyhedral objects. Construct two
more complicated polyhedral objects, whose surface area lies between
those ofthe previously constructed ones and the curved object.
Measure their surface areas.  Repeat this until the approximation
is sufficiently good for your purposes.

If you can do this symbolically, you may be able to get formulas
for the surface areas of the sequences of contained and containing
polygonal or polyhedral objects. Then you may be able to prove that
the difference between the surface areas of the corresponding
containing and contained objects gets arbitrarily small as you step
through better and better approximating objects. In fact, you may
be able to determine the limit that both sequences approach, which
is the surface area of the curved object.

In the case of a sphere, you could start with an octahedron,
composed of eight equilateral triangles, formed by putting two
square pyramids base-to-base. If the radius of the sphere is r,
then the length of each side of the triangles of the inscribed
octahedron will be r*Sqrt[2], and their areas will be r^2*Sqrt[3]/2,
for a total surface area of 4*r^2*Sqrt[3]. For the circumscribed
octahedron, the length of each side of the triangles will be
r*Sqrt[6], and their areas will be 3*r^2*Sqrt[3]/2, for a total
surface area of 12*r^2*Sqrt[3].

Now on each face of the inscribed octahedron build a triangular
pyramid, the height of which is just enough to make the vertex be on
the surface of the sphere, and whose lateral sides are isosceles
triangles. The surface area of this new polyhedron with 24 congruent
isosceles triangle faces will be larger than that of the octahedron,
but still smaller than that of the sphere. Using solid geometry, one
can figure out the length of the new edges to be 2*r*Sqrt[3]/3. This
allows one to determine the areas of each of the 24 isosceles
triangles, and the total surface area of this polyhedron.

A construction on the circumscribing octahedron which chops off the
six corners with planes tangent to the sphere will create a polyhedron
with 14 faces, eight squares, six equilateral triangles, called
a "cuboctahedron," which circumscribes the sphere and whose surface
area is less that that of the octahedron but more that that of the
sphere. Its surface area can be computed in a similar way.

Repeating this process of approximating the sphere by inscribing and
circumscribing polyhedra with more and more faces, by adding pyramids
inside and chopping off pyramids outside, will give better and better
approximations to the surface area of the sphere.

This is the kind of approach used by Archimedes. Among other things,
he used it to show that the area of a circle is the square of the
radius times the constant we call Pi. His approach leads to the
following facts.

Let x(0) = 0, and for every n > 1, let x(n) = 2 + Sqrt[x(n-1)].
Then as n grows without bound, the expression Sqrt[4-x(n)]*2^n
approaches Pi as a limit. This is derived from using inscribed regular
polygons with 2^n sides. Archimedes used this kind of argument (but
not exactly this) to show that 3+10/71 < Pi < 3+1/7.

Using the calculus to find this surface area involves using spherical
coordinates centered at the center of the sphere. The surface area is
given by

2*Pi        Pi
S = Integral    Integral   r^2 sin(phi) d(phi) d(theta),
0           0

= 2*Pi*r^2*[cos(0) - cos(Pi)],

S = 4*Pi*r^2.

This is much quicker, and it gives an exact answer, not an approximate
one, but you have to know enough of the calculus to understand why
this double integral equals the surface area, and to evaluate it.

I hope this helps.  If this doesn't answer the question, write again.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
College Calculus
College Higher-Dimensional Geometry
High School Calculus
High School Higher-Dimensional Geometry

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