Surface Area of an EllipsoidDate: 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 simplifies. 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: z/c^2 ------------------------------ ( 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 this. -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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