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I'm a biology PhD student looking for some help on how to calculate (or estimate) the surface area of an ellipsoid truncated parallel to the long axis. Any help would be greatly appreciated.

Thanks,

Murphy24

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- #1

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I'm a biology PhD student looking for some help on how to calculate (or estimate) the surface area of an ellipsoid truncated parallel to the long axis. Any help would be greatly appreciated.

Thanks,

Murphy24

- #2

Mark44

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You would probably use a double integral to solve this type of problem.

- #3

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The flat surface of the truncated part is relatively easy. It is an ellipse with area A = pi*a * b where a and b are the two semi-diameters. (If a=b then you have a circle a=b=r and A=pi r^2).

I'm a biology PhD student looking for some help on how to calculate (or estimate) the surface area of an ellipsoid truncated parallel to the long axis. Any help would be greatly appreciated.

Thanks,

Murphy24

The other part is a rather nasty integral, an elliptic integral without closed form solution (except to invent a new function whose value is that area).

There is a formula in terms of standard elliptic integrals and there is an approximation formula. Check out: http://en.wikipedia.org/wiki/Ellipsoid" [Broken].

EDIT: Ahhh... that only gives the total ellipsoidal area. To get the truncated area or the area of the piece to subtract out. Let me see what I can come up with. Possibly I can work out a way to numerically estimate via software.

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- #4

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Below is a python script I wrote which will calculate the value you want. You can download the python interpreter from www.python.org if you don't already have it installed on your computer. (Open the file in IDLE and run.)

It's a slow brute force approach but it will work. The ellipsoid is assumed to be truncated at a certain height above the*center* which is given by z0. Note indention is critical in python as the interpreter uses it to group blocks of code. I get about 5 decimal place accuracy on the area of a sphere of radius 2. The equation for the ellipsoid is:

[tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =1[/tex]

[tex] z < z_0[/tex]

As you change a,b, and c, and z0 be sure you enter them as decimals as otherwise python will do integer division truncating the fractions.

EDIT: BTW This code only gives the area of the ellipsoid surface. You'll need to add in the flat ellipse.

EDIT2: You can lower the number of steps for a quicker, less accurate answer.

It's a slow brute force approach but it will work. The ellipsoid is assumed to be truncated at a certain height above the

[tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =1[/tex]

[tex] z < z_0[/tex]

As you change a,b, and c, and z0 be sure you enter them as decimals as otherwise python will do integer division truncating the fractions.

Code:

```
from math import *
z0 = 2.0 #Height at which ellipsoid is truncated.
a = 2.0
b = 2.0
c = 2.0
phi0 = acos(z0/c)
steps = 3000 #Gives about 5 decimal place accuracy in a few minutes.
def Jacob(phi,theta,A,B,C):
return abs(sin(phi))*sqrt( C*C*sin(phi)*sin(phi)*(B*B*cos(theta)*cos(theta)+ A*A*sin(theta)*sin(theta))+A*A*B*B*cos(phi)*cos(phi))
Int=0.0
dphi = (pi-phi0)/steps
dtheta = 2*pi/steps
for p in xrange(steps):
for t in xrange(steps):
phi = phi0 + p*dphi
theta = t*dtheta
Int += Jacob(phi,theta,a,b,c)*dtheta*dphi
print "The Area is:", Int
```

EDIT2: You can lower the number of steps for a quicker, less accurate answer.

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I found this site yesterday and modified the equation for the surface area for a spherical cap to calculate the surface area of the ellipsoidal cap:

http://mathworld.wolfram.com/SphericalCap.html

So this site says that the surface area of a spherical cap = pi*(a^2+h^2). I suspect this is just an estimation, but its probably accurate enough for my purposes. I am aiming to estimate the surface area of a membrane that partially covers the top of reptile eggs, so it doesn't need to be a perfect estimate, just one that doesn't introduce too much error.

I reasoned that just as the area of an ellipse is pi*(a*b) whereas the area of a circle is pi*(r^2), that maybe I could modify the formula for the area of a spherical cap from pi*(a^2+h^2) to one that incorporates both radii of the base of the truncated ellipsoid: pi*(a*b+h^2).

First of all, is the site that I am referencing reliable at all? And secondly, is my modification of the equation for a spherical cap a valid one?

Thanks again,

Murphy24

- #6

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The formula they give is exact for spherical caps. It should be pretty darn close for your purposes (where a and b are relatively close to each other). Close enough that your worry is more that the egg isn't a perfect ellipsoid than that your formula isn't giving the perfectly exact area.

I found this site yesterday and modified the equation for the surface area for a spherical cap to calculate the surface area of the ellipsoidal cap:

http://mathworld.wolfram.com/SphericalCap.html

So this site says that the surface area of a spherical cap = pi*(a^2+h^2). I suspect this is just an estimation, but its probably accurate enough for my purposes. I am aiming to estimate the surface area of a membrane that partially covers the top of reptile eggs, so it doesn't need to be a perfect estimate, just one that doesn't introduce too much error.

I reasoned that just as the area of an ellipse is pi*(a*b) whereas the area of a circle is pi*(r^2), that maybe I could modify the formula for the area of a spherical cap from pi*(a^2+h^2) to one that incorporates both radii of the base of the truncated ellipsoid: pi*(a*b+h^2).

First of all, is the site that I am referencing reliable at all? And secondly, is my modification of the equation for a spherical cap a valid one?

Thanks again,

Murphy24

Good luck.

- #7

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Thanks mate :)

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