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Surface Area of an Ellipsoid

Date: 02/11/97 at 16:41:40
From: Tony Rojas
Subject: Ellipsoid Question

Dear Dr. Math:
Is there any formula or computer program that can calculate the 
surface area of an ellipsoid: (X/A)^2 + (Y/B)^2 + (Z/C)^2 = 1, 
given A, B, and C?
I would greatly appreciate an answer.
Thank you,
Tony Rojas

Date: 02/17/97 at 08:02:23
From: Doctor Mitteldorf
Subject: Re: Ellipsoid Question

Dear Tony,

Here's what I've been able to find:

There are formulas where two of the three axes of the ellipsoid are 
equal.  In the "prolate" (cigar-shaped) case, the axis a is longer than 
b = c. The area is: 2*pi*a^2 + pi*b^2/e*ln((1+e)/(1-e))

In the "oblate" (flying saucer) case, the axes a = c are longer than 
b. The area is: 2*pi*b^2 + 2*pi*a*b/e * arcsin(e)
In both of these formulae, e is the eccentricity of the ellipse, 
defined by sqrt(1-b^2/a^2).


To solve the area for the general case, where all three axes are 
different, you need a little vector calculus.  The computation is not 
difficult in principle, but it yields an integral which I don't know 
how to do, and for which you may need a "special function" - an 
elliptic integral or Bessel function.  (This is a guess on my part.)

Here's one way to set up the integral:  First project the ellipsoid 
onto the xy plane.  The area of the image on the plane (an ellipse) 
can be written as a double integral:

        Integral dx from -a to +a
        Integral dy from -b*sqrt(1-x^2/a^2) to +b*sqrt(1-x^2/a^2)

If you're looking for the area of the plane projection (the xy 
ellipse), then the integrand is just unity, and the integral 

But if you want to find the area of one dome of the ellipsoid, you can 
use as an integrand the ratio of a surface element of the curved dome 
to the corresponding surface element on the xy plane underneath.  This 
ratio is just the reciprocal of the cosine of the angle between the 
plane and the local tangent to the dome.  This cosine is also the same 
as the cosine of the angle between the normal to the dome at any given 
point and the z axis.  You can compute this, in turn, by taking the z 
component of the gradient of the function: 

  x^2/a^2 + y^2/b^2 + z^2/c^2

This function has the ellipse as a contour surface.  The z component 
of the gradient is:

           ( x^2/a^4 + y^2/b^4 + z^2/c^4)

In order to actually do the integral, you must eliminate z from the
above in favor of x and y:

             let z = c^2 * (1 - x^2/a^2 - y^2/b^2)

When you do this, the integral takes on a form which is, I believe, an 
elliptic integral - in any case, it's an integral I don't know how to 
do.  If you've got a numerical integration package, it shouldn't be 
hard to set it up to do the integral; if you find someone who knows 
more about Bessel functions than I do, it may be possible to avoid 

-Doctor Mitteldorf,  The Math Forum
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Associated Topics:
College Calculus
College Higher-Dimensional Geometry

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