Post by SNNAP Board Admin on Nov 30, 2004 17:37:21 GMT 5
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
crosssection (i.e., two of the major axes equal) and a third asymmetric
with an elliptical crosssection (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 2D hemielliptical curve around the x axis to get the 3D
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 crosssection, 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 2D hemielliptical curve around the x axis to get the 3D 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, 573574.
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, 295313), 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, 11031111. 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 *
* *
* 3014436113 * fax 3014431343 * sxs @ fdadr . cdrh . fda . gov *
************************************************************************
Doug
There are three types of ellipsoids, two symmetric with a circular
crosssection (i.e., two of the major axes equal) and a third asymmetric
with an elliptical crosssection (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 2D hemielliptical curve around the x axis to get the 3D
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 crosssection, 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 2D hemielliptical curve around the x axis to get the 3D 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, 573574.
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, 295313), 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, 11031111. 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 *
* *
* 3014436113 * fax 3014431343 * sxs @ fdadr . cdrh . fda . gov *
************************************************************************
Doug