Derivation of Geometric Formulas

Date: 5/29/96 at 18:35:16
From: Anonymous
Subject: Derivation of Geometric Formulas

Could you please answer these questions? Your help would be very much 
appreciated.

How were the formulas for surface area, total surface area, and volume 
of a sphere derived? How were the formulas for the volume of a pyramid 
and cone derived?

Please give an explanation of the answers to the above questions and 
include equations if there are any. Thank you.

SwimDuck

Date: 5/31/96 at 11:45:38
From: Doctor Anthony
Subject: Re:Derivation of Geometric Formulas

In the case of the surface area of a sphere, consider an elementary 
element of arc at position (a,theta) (in polar coordinates) rotated 
through 360 degrees about the x axis, thereby forming a hoop of radius 
a.sin(theta) and surface area a.d(theta).2.pi.a.sin(theta) = 2.pi.a^
2.sin(theta).d(theta)

To get the total surface area we integrate this between theta = 0 and 
pi:

Surface area = 2.pi.a^2.INT[sin(theta).d(theta)] between 0 and pi.
             = 2.pi.a^2.[-cos(theta)] between 0 and pi
             = -2.pi.a^2[-1 - 1]
             = 4.pi.a^2

For volume consider the circle x^2 + y^2 = a^2, and take a slice of 
thickness dx, and radius y (= sqrt(a^2-x^2)).  Rotate this about the 
x axis to get a disk of volume pi.y^2.dx.  Integrate between 
x = -a and +a:

Volume = pi.INT[(a^2 - x^2).dx]
       = pi.[a^2.x - (1/3)x^3]  between -a and +a
       = pi.[a^3 - (1/3)a^3 + a^3 - (1/3)a^3]
       = pi.[2a^3 - (2/3)a^3]
       = pi.[(4/3)a^3]

For volume of cone of height h and base radius a, consider the line 
y = (a/h)x. By rotating this about the x axis you will generate the 
cone. The volume of an elementary disk of thickness dx and radius y is 
pi.y^2.dx = pi.(a^2/h^2)x^2.dx  Integrate between 0 and h:

Volume = pi.INT[(a^2/h^2)x^3/3]  between 0 and h
       = pi.(a^2/h^2)(h^3/3)
       = (1/3).pi.a^2.h
     
In the case of a pyramid, you take slices, parallel to the base, of 
thickness dx and area calculated by ratio of similar rectangles or 
square with the shape of the base.  You get the same formula as for 
the cone, namely(1/3).area of base times perpendicular height. 

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/