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Surface Area and Volume of an Ellipsoid

Updated on August 17, 2013
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TR Smith is a product designer and former teacher who uses math in her work every day.

An ellipsoid is a sphere-like solid shape that is the 3-d analogue of an ellipse. Whereas a sphere has a single radius, an ellipsoid has three semi-axes: half-height, half-length, and half-width. If A, B, and C represent each semi-axis, then the volume and surface area of an ellipsoid are functions of A, B, and C.


Volume of an Ellipsoid

The volume of an ellipsoid is easy to learn and remember because it is so similar to that of a sphere. Recall that a sphere's volume is (4*pi/3)r^3. If you replace each of the three r's with A, B, and C, you get the formula for ellipsoid volume:

V = (4*pi/3)ABC


Ellipsoid Surface Area Equations

The formula for the volume of an ellipsoid is much more complicated than the analogous formula for a sphere. A sphere's surfaces area is a simple quadratic function in r, S(r) = 4*pi*r^2, however no such quadratic function in A, B, and C exists for computing the surface area of an ellipsoid.

Complicating the matter is that the surface area formula is different depending on the relative sizes of A, B, and C. If all three axis lengths are distinct there is actually no closed form expression involving elementary functions, but you can express the surface area in terms of elliptic integrals.

If the ellipsoid is oblate or prolate, cases in which two axis lengths are equal, then there there are closed-form expressions. These types of ellipsoids are called spheroids because they can be obtained as solids of revolution by spinning an ellipse about one of its axes.


Oblate Ellipsoid (Oblate Spheroid)

An oblate ellipsoid is one for which the semi-axis lengths are A, A, and B, where B is less than A. This type is spheroid is flattened like an M&M candy. The formula for the surface area of an oblate spheroid is

S = 2*pi*[1 + (1-m^2)arctanh(m)/m]*A^2

where arctanh is the inverse hyperbolic tangent function given by

arctanh(x) = 0.5*Ln[(x+1)/(1-x)]

and m = sqrt(1 - (B/A)^2).


Prolate Ellipsoid (Prolate Spheroid)

A prolate spheroid is one for which the axis lengths are A, B, and B, where B is less than A. This type of spheroid is elongated like a rugby ball. The formula for the surface area of a prolate ellipsoid is

S = 2*pi*[1 + A*arcsin(m)/(mB)]*B^2

where arcsin is the inverse sine function and m = sqrt(1 - (B/A)^2).


General Ellipsoid

If all three axis lengths A, B, and C are unequal, with A greater than B and B greater than C, then the formula for the surface area is

In the formula, E(φ, k) is the incomplete elliptic integral of the second kind and F(φ, k) is the incomplete elliptic integral of the first kind. Neither can be evaluated in terms of simpler known functions, but they can be computed with convergent algorithms.

You can approximate the surface area of an ellipsoid with the equation

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    • FitnezzJim profile image

      FitnezzJim 3 years ago from Fredericksburg, Virginia

      For the general ellipsoid case, with sem-axis A greateer than B greater than C, is it true to say an upper bound on the surface area is S(r) = 4*pi*A^2 and a lower bound is S(r) = 2*pi*BC? Is it more restrictive than that?

    • calculus-geometry profile image
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      TR Smith 3 years ago from Eastern Europe

      Hi Jim,

      Thanks for the great question. You're right, you can give upper and lower bounds on the surface area with the formulas

      S_Upper = 4*pi*A^2

      S_Lower = 4*pi*BC

      ( I assume you meant a coefficient of 4 in that second equation.)

      You can get even tighter bounds making similar modifications to the spherical surface area formula 4*pi*R^2. If you replace R^2 with the arithmetic mean of the squared semi-axis lengths, you get this upper bound function that is less than the previous upper bound:

      S_UpperNew = 4*pi*[(A^2 + B^2 + C^2)/3]

      And if you replace R^2 with the squared geometric mean of the semi-axis lengths, you get a new lower bound that is greater than the previous lower bound:

      S_LowerNew = 4*pi*(ABC)^(2/3)

      You can get even more restrictive bounds by expanding the elliptic integral surface area formula into an asymptotic series, and then truncating/rounding some terms. But that gets ugly quickly. The approximation formula given at the end of the article has a relative error of no more than 1.061%, so that's usable for practical purposes.

    • FitnezzJim profile image

      FitnezzJim 3 years ago from Fredericksburg, Virginia

      I meant the two. Consider, if that third axis is significantly smaller than the other two, then what you are looking at is the area of the topside and bottom side of an ellipse, or twice the area of one - pi*BC (or would it be pi*AB?). That would make a good test question, would it not? Prove or disprove the following conjecture ...

    • calculus-geometry profile image
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      TR Smith 3 years ago from Eastern Europe

      Oh I see what you mean. You're right, in an extreme case where one axis is much less than the other two you can approximate the surface area as the area of two ellipses, which would by 2*pi times the product of the other half axis lengths. That underestimate would give you a better lower bound than the geometric average.

    • calculus-geometry profile image
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      TR Smith 3 years ago from Eastern Europe

      For an arbitrary ellipsoid, with no extra info about the relative lengths of the three axes, these bounds are tighter than the ones I gave in my first comment --

      S_lower = 4*pi*(AB + BC + AC)/3

      S_upper = 4*pi*sqrt[( (AB)^2 + (BC)^2 + (AC)^2 )/3]

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