Surface Area of an Oblate SpheroidDate: 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? Thanks. 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, is 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 http://mathforum.org/dr.math/ |
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