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Surface Area of an Oblate Spheroid

Date: 06/02/99 at 08:53:14
From: Lisa
Subject: Finding the surface area of an oblate spheroid

Dr. Math,

Could you please give an example problem on solving for the surface 
area of an oblate spheroid using the inverse hyperbolic sine function? 


Date: 06/02/99 at 11:17:07
From: Doctor Rob
Subject: Re: Finding the surface area of an oblate spheroid

Here is an example. The numbers 3 and 5 were chosen because 5^2 - 3^2 
= 4^2, so that e would be a rational number.

An oblate spheroid is formed by rotating an ellipse with semi-axes 
3 cm and 5 cm about its minor axis. Find its surface area.

The semi-axes of the spheroid are 3, 5, and 5. The formula for the 
surface area of an oblate spheroid with axes a, b, and b, and a < b, 

   S = 2*pi*b*(b+a*arcsinh[b*e/a]/[b*e/a]),

where e = sqrt(b^2-a^2)/b. In this case, a = 3 and b = 5, so the value 
of e = sqrt(5^2-3^2)/5 = 4/5, and b*e/a = 4/3, so

   S = 2*pi*5*(5+3*arcsinh[4/3]/[4/3])
     = 10*pi*(5+9*arcsinh[4/3]/4)
     = 50*pi + (45/2)*pi*arcsinh(4/3)

Since arcsinh(4/3) = 1.09861, to six significant figures, and pi = 
3.14159, also to six significant figures, the answer is

   S = 234.736 cm^2,

to six significant figures.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Trigonometry

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