Surface Area of Ellipsoids

[SNNAP Board Admin]

I recent ran into the problem of calculating the surface of an ellipsoid (i.e., the surface of cell body in shape of a football). As it turns out, this is a rather difficult problem. The soluation that I found on the web is presented below:

There are three types of ellipsoids, two symmetric with a circular
cross-section (i.e., two of the major axes equal) and a third asymmetric
with an elliptical cross-section (all three axes unequal). One of the
symmetric types is oblate (an M&M shape) with the two equal axes larger
than the third axis. The other symmetric type is prolate (a football
shape) with the two equal axes smaller than the third axis.

SURFACE AREA OF AN OBLATE ELLIPSOID (M&M SHAPE)
Analytical solutions for the surface area of the two symmetric
ellipsoids can be found. First consider the M&M shape (two major
axes = a and one minor axis = b, with a > b). Put the a axis on the y
coordinate axis, the smaller b axis on the x coordinate axis, and then
rotate the 2-D hemielliptical curve around the x axis to get the 3-D
surface. To find the area of the surface created from a curve in this
way, evaluate the following integral,

/ pi/2
| 2 2
A = | 2 pi y(p) sqrt(( x'(p) ) + ( y'(p) ) ) dp ,
|
/ - pi/2

over the angle p (measured from the y axis). Here, x and y are the
coordinates of a point on the curve (functions of the angle p measured
from the y axis), and x' and y' are the first derivatives with respect
to p. For the M&M ellipse, y = a cos(p), x = b sin (p), y' = -a sin(p)
and x' = b cos(p). With these substitutions, the equation for the area
becomes:

/ pi/2
| 2 2
A = | 2 pi a COS (p) sqrt((b COS (p)) + (a SIN (p)) ) dp .
|
/ - pi/2

I used a symbolic algebra program to find the solution:

2
A = 2 a pi -

2 2 2 2 2 2
a b pi LN (a - sqrt(a - b )) a b pi LN (sqrt(a - b ) + a)
-------------------------------- + --------------------------------
2 2 2 2

sqrt(a - b ) sqrt(a - b )

This can be rewritten in simpler form as:

2 2 2
2 pi b 1 + E sqrt(a - b )
A = 2 a pi + ------- LN ( ------- ) , where E = --------------- .
E 1 - E a

For a hemiellipsoid, cut the M&M in half through the plane of the
circular cross-section, and integrate between 0 and pi/2 (instead of
-pi/2 and pi/2). In this case, the area becomes:

2
2 pi b a(1 + E)
A = a pi + ------- LN ( ---------- ) .
E b

You can show that the area for the hemiellipsoid is equal to one half
the area of the full ellipsoid, as expected.

The surface area of a section of an ellipsoid smaller than a
hemiellipsoid is more complicated. Integrate between t and pi/2, where
t is the longitude of the circular edge of the section. In this case,
the area becomes:

A =

2 2 2 2 2 2 2
a b pi LN (sqrt((a - b ) SIN (t) + b ) + sqrt(a - b ) SIN (t))
- ------------------------------------------------------------------- +
2 2
sqrt(a - b )

2 2 2
a b pi LN (sqrt(a - b ) + a)
-------------------------------- -
2 2
sqrt(a - b )

2 2 2 2 2
a pi SIN (t) sqrt((a - b ) SIN (t) + b ) + a pi .

SURFACE AREA OF A PROLATE ELLIPSOID (FOOTBALL SHAPE)
In the case of the surface of a football ellipse, (two minor axes
b and one major axis a, again with a > b), put the b axis on the y
coordinate axis, the larger a axis on the x coordinate axis, and rotate
the 2-D hemielliptical curve around the x axis to get the 3-D surface.
In this case the curve is defined by y = b cos(p), x = a sin (p),
y' = -b sin(p) and x' = a cos(p). The equation for the area becomes:

/ pi/2
| 2 2
A = | 2 pi b COS (p) sqrt((a COS (p)) + (b SIN (p)) ) dp .
|
/ - pi/2

My symbolic algebra program gave me the following solution:

/ 2 2 \
2 | sqrt(a - b ) |
2 a b pi ATAN |----------------|
\ b / 2
A = ----------------------------------- + 2 b pi .
2 2
sqrt(a - b )

I was able to show that this is equal to the following, which is in the
form that several of my respondants gave me (and using E defined above):

2 a b pi ASIN (E) 2
A = ------------------- + 2 b pi .
E

For a hemiellipsoid, the answer is simply one half of this, or

a b pi ASIN (E) 2
A = ------------------ + b pi .
E

For a section of a hemiellipse (t < p < pi/2),

a b pi ASIN ( E SIN (t) ) a b pi ASIN (E)
A = - --------------------------- + ----------------- -
E E

2 2 2 2 2
b pi SIN (t) sqrt((b - a ) SIN (t) + a ) + b pi .

SURFACE AREA OF AN ASYMMETRIC ELLIPSOID
Here the axes a, b, c are all unequal, with a > b > c. The
surface area of the whole ellipsoid is given by:

2 2 pi a b / 2 2 \
A = 2 pi c + ---------- * | cos (t) F(t,k) + sin (t) E(t,k) |
SIN (t) \ /

2 2 2 2 2 2
where k = SQRT( a ( b - c )/ b ( a - c ) ), cos (t) = c/a, and
F(t,k) and E(t,k) are elliptic integrals of the first and second kind,
respectively. This is from Kwok LS, "The surface area of an ellipsoid
revisited," letter to the editor, J. theor. Biol. (1989) 139, 573-574.
The letter corrects some errors in an earlier study (Kwok LS,
"Calculation and application of the anterior surface area of a model
human cornea," J. theor. Biol, (1984) 108, 295-313), but itself has a
few errors that I believe I have corrected. F(t,k) and E(t,k) have no
analytic solutions, but are tabulated, and can also be evaluated
numerically. A reference for the latter that includes a short BASIC
program (and a couple of errors) is Utsunomiya T et al., "Doppler color
flow 'proximal isovelocity surface area' method for estimating volume
flow rate: effects of orifice shape and machine factors," J. Am. Coll.
Cardiol. (1991) 17, 1103-1111. I think Kwok's original 1984 paper has
the equation for a partial ellipsoid, but since I was unable to
understand it, I am not completely sure.

This solution was provided by:

************************************************************************
* Sandy F.C. Stewart, PhD *
* Hydrodynamics & Acoustics Branch *
* Food & Drug Administration *
* Rockville, MD 20851 *
* *
* 301-443-6113 * fax [phone] * sxs @ fdadr . cdrh . fda . gov *
************************************************************************

Doug

[SNNAP Board Admin]

There are three types of ellipsoids, two symmetric with a circular
cross-section (i.e., two of the major axes equal) and a third asymmetric
with an elliptical cross-section (all three axes unequal). One of the
symmetric types is oblate (an M&M shape) with the two equal axes larger
than the third axis. The other symmetric type is prolate (a football
shape) with the two equal axes smaller than the third axis.

SURFACE AREA OF AN OBLATE ELLIPSOID (M&M SHAPE)
Analytical solutions for the surface area of the two symmetric
ellipsoids can be found. First consider the M&M shape (two major
axes = a and one minor axis = b, with a > b). Put the a axis on the y
coordinate axis, the smaller b axis on the x coordinate axis, and then
rotate the 2-D hemielliptical curve around the x axis to get the 3-D
surface. To find the area of the surface created from a curve in this
way, evaluate the following integral,

/ pi/2
| 2 2
A = | 2 pi y(p) sqrt(( x'(p) ) + ( y'(p) ) ) dp ,
|
/ - pi/2

over the angle p (measured from the y axis). Here, x and y are the
coordinates of a point on the curve (functions of the angle p measured
from the y axis), and x' and y' are the first derivatives with respect
to p. For the M&M ellipse, y = a cos(p), x = b sin (p), y' = -a sin(p)
and x' = b cos(p). With these substitutions, the equation for the area
becomes:

/ pi/2
| 2 2
A = | 2 pi a COS (p) sqrt((b COS (p)) + (a SIN (p)) ) dp .
|
/ - pi/2

I used a symbolic algebra program to find the solution:

2
A = 2 a pi -


2 2 2 2 2 2
a b pi LN (a - sqrt(a - b )) a b pi LN (sqrt(a - b ) + a)
-------------------------------- + --------------------------------
2 2 2 2

sqrt(a - b ) sqrt(a - b )


This can be rewritten in simpler form as:

2 2 2
2 pi b 1 + E sqrt(a - b )
A = 2 a pi + ------- LN ( ------- ) , where E = --------------- .
E 1 - E a

For a hemiellipsoid, cut the M&M in half through the plane of the
circular cross-section, and integrate between 0 and pi/2 (instead of
-pi/2 and pi/2). In this case, the area becomes:

2
2 pi b a(1 + E)
A = a pi + ------- LN ( ---------- ) .
E b

You can show that the area for the hemiellipsoid is equal to one half
the area of the full ellipsoid, as expected.

The surface area of a section of an ellipsoid smaller than a
hemiellipsoid is more complicated. Integrate between t and pi/2, where
t is the longitude of the circular edge of the section. In this case,
the area becomes:

A =

2 2 2 2 2 2 2
a b pi LN (sqrt((a - b ) SIN (t) + b ) + sqrt(a - b ) SIN (t))
- ------------------------------------------------------------------- +
2 2
sqrt(a - b )


2 2 2
a b pi LN (sqrt(a - b ) + a)
-------------------------------- -
2 2
sqrt(a - b )


2 2 2 2 2
a pi SIN (t) sqrt((a - b ) SIN (t) + b ) + a pi .


SURFACE AREA OF A PROLATE ELLIPSOID (FOOTBALL SHAPE)
In the case of the surface of a football ellipse, (two minor axes
b and one major axis a, again with a > b), put the b axis on the y
coordinate axis, the larger a axis on the x coordinate axis, and rotate
the 2-D hemielliptical curve around the x axis to get the 3-D surface.
In this case the curve is defined by y = b cos(p), x = a sin (p),
y' = -b sin(p) and x' = a cos(p). The equation for the area becomes:

/ pi/2
| 2 2
A = | 2 pi b COS (p) sqrt((a COS (p)) + (b SIN (p)) ) dp .
|
/ - pi/2

My symbolic algebra program gave me the following solution:

/ 2 2 \
2 | sqrt(a - b ) |
2 a b pi ATAN |----------------|
\ b / 2
A = ----------------------------------- + 2 b pi .
2 2
sqrt(a - b )

I was able to show that this is equal to the following, which is in the
form that several of my respondants gave me (and using E defined above):


2 a b pi ASIN (E) 2
A = ------------------- + 2 b pi .
E


For a hemiellipsoid, the answer is simply one half of this, or


a b pi ASIN (E) 2
A = ------------------ + b pi .
E


For a section of a hemiellipse (t < p < pi/2),



a b pi ASIN ( E SIN (t) ) a b pi ASIN (E)
A = - --------------------------- + ----------------- -
E E


2 2 2 2 2
b pi SIN (t) sqrt((b - a ) SIN (t) + a ) + b pi .


SURFACE AREA OF AN ASYMMETRIC ELLIPSOID
Here the axes a, b, c are all unequal, with a > b > c. The
surface area of the whole ellipsoid is given by:


2 2 pi a b / 2 2 \
A = 2 pi c + ---------- * | cos (t) F(t,k) + sin (t) E(t,k) |
SIN (t) \ /


2 2 2 2 2 2
where k = SQRT( a ( b - c )/ b ( a - c ) ), cos (t) = c/a, and
F(t,k) and E(t,k) are elliptic integrals of the first and second kind,
respectively. This is from Kwok LS, "The surface area of an ellipsoid
revisited," letter to the editor, J. theor. Biol. (1989) 139, 573-574.
The letter corrects some errors in an earlier study (Kwok LS,
"Calculation and application of the anterior surface area of a model
human cornea," J. theor. Biol, (1984) 108, 295-313), but itself has a
few errors that I believe I have corrected. F(t,k) and E(t,k) have no
analytic solutions, but are tabulated, and can also be evaluated
numerically. A reference for the latter that includes a short BASIC
program (and a couple of errors) is Utsunomiya T et al., "Doppler color
flow 'proximal isovelocity surface area' method for estimating volume
flow rate: effects of orifice shape and machine factors," J. Am. Coll.
Cardiol. (1991) 17, 1103-1111. I think Kwok's original 1984 paper has
the equation for a partial ellipsoid, but since I was unable to
understand it, I am not completely sure.

NUMERICAL METHOD FOR SPHERE, OBLATE, PROLATE, OR ASYMMETRIC ELLIPSOID
As a final resort, I came up with a computer program that breaks
up the surface of the ellipsoid into triangles, and computes the area by
summing the area contributions from the individual triangles. Using
100 x 100 x 2 triangles, I can get answers within 10e-4 of the equations
for spheres, oblate and prolate ellipsoids, and within 10e-2 of the
BASIC program given in Utsunomiya's paper. On a 150 MHz Pentium machine,
this takes about 3 seconds to evaluate. The algorithm is too long to
list here, but I will be happy to send the code (implemented in Turbo
Pascal) to anyone who wants it.

Thanks again,

--Sandy Stewart



************************************************************************
* Sandy F.C. Stewart, PhD *
* Hydrodynamics & Acoustics Branch *
* Food & Drug Administration *
* Rockville, MD 20851 *
* *
* 301-443-6113 * fax [phone] * [e-mail] *
************************************************************************

[SNNAP Board Admin]

I see that the equations didn't turn out correctly. You can find the original message @

isb.ri.ccf.org/biomch-l/archives/biomch-l-1996-11/00112.html

[SNNAP Board Admin]

The new version allows you to specify how you want to format the message body. If you need to preserve spaces (like equations), you need to choose "Display as text" option. Once the new script is installed on server, you may use that option.

cai