How do you get the answer for the surface area figures using pi?

Answer 1

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.