Surface Area of a Sphere

Topics: Volume, Sphere, Area Pages: 2 (580 words) Published: April 1, 2013
|  |Subject: Surface area of a sphere |

A connection which could be illustrated, and could be understood by students who know the perimeter of a circle, runs as follows: Put the sphere of radius R inside a cylinder, with the cylinder just touching the equator, and cut off at the height of the top and bottom of the sphere. (A cutaway view is in the diagram.) [pic]

What is the area of the curved part of the cylinder? 2 Pi R x 2R = 4 Pi R2. This is found by slicing the cylinder surface and rolling it out as a rectangle. Now, it is NOT an accident that the cylinder surface is EXACTLY the area of the sphere. Take in small horizontal slice through the diagram. (I have colored one such slice orange.) This cuts a rectangle out of the rolled out cylinder and slightly distorted rectangle out of the sphere. (If the slice is very thin then the distortion is "slight".) In the cross-sectional view below hc is the height of the slice on the cylinder, hs is the length of the arc on the sphere cut out by the slice, r is the radius of the distorted rectangle on the sphere and R is the radius of the sphere. [pic]

The area of the orange rectangle on the cylinder is 2 Pi R hc and the area of the distorted orange rectangle on the sphere is approximately 2 Pi r hs. These two areas are approximately equal (the proof is outlined below). Since the cylinder and the sphere can be decomposed into these rectangular strips, the area of the sphere and the area of the cylinder are approximately equal. This argument is in the spirit of how the Greeks compared slices to show that areas and volumes are the same. It is not only interesting reasoning in proportion, etc., but it is a lesson in History. In the same spirit, you can compare the VOLUME of a hemi-sphere with that of the cylinder with an inverted cone removed. (I think of the hemisphere with the equator at the bottom and the cone...
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