Deriving Expression for Volume of Ellipsoid

In summary: In the end, you'll get a nice simple answer that you can integrate with respect to theta. In summary, the conversation discussed deriving an expression for the volume of an ellipsoid defined by \frac{x^2}{a^2} +\frac{y^2}{b^2} +\frac{z^2}{c^2} \leq 1 and using a change of variables and spherical coordinates to set up the integral. The final answer involved integrating with respect to u and v using trig substitutions to get the volume of the ellipsoid as 4/3 * Pi * abc.
  • #1
tandoorichicken
245
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I've been asked to derive an expression for the volume of an ellipsoid. I know what the expression is, I just don't know how to get there from the information given. All that is given is that it is defined by

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

a standard inequality/definition for an ellisoid. Not quit sure how to go about this.
 
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  • #2
tandoorichicken said:
I've been asked to derive an expression for the volume of an ellipsoid. I know what the expression is, I just don't know how to get there from the information given. All that is given is that it is defined by

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

a standard inequality/definition for an ellisoid. Not quit sure how to go about this.
Have you tried anything? Like first drawing a picture. Certainly it is straight forward to set up the integral. The ellipsoid's widest girth comes in the xy-plane where it is an ellipse:
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1[/tex]
That is is covered by taking x from -a to a and, for each x, y from [itex]-b\sqrt{1-\frac{x^2}{a^2}[/itex] to [itex]b\sqrt{1-\frac{x^2}{a^2}}[/itex]. For each (x,y), z ranges from [itex]-c\sqrt{1-\frac{x^2}{a^2}- \frac{y^2}{b^2}}[/itex]. The volume is
[tex]\int_{x=-a}^a\int_{y=-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}}\int_{z=-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}dzdydx[/tex]
or
[tex]2c\int_{x=-a}^a\int_{y=-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}}\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}dydx[/tex]
 
  • #3
Use spherical coordinates, with a twist.
 
  • #4
I agree with the previous poster, use spherical coordinates. You have (x²/a²) + (y²/b²) + (z²/c²) = 1. Spherical coordinates woudldn't work too well with that equation the way it is. It would be much better if your ellipse was transformed into a sphere with an equation like u^2 + v^2 + w^2 = 1. This suggests a change of variables. Consider u^2 = (x²/a²) and likewise for the other components for the three obvious substitutions.
 
  • #5
if anyone could further evaluate this problem, i would greatly appreciate it. I followed the parameterization suggestion and set u=x/a, v=y/b, and w=z/c and switched the bounds on the integral. However, after integrating once I run into the problem of

2C * sqrt(1-u^2-v^2)

When I integrate that, I am unable to do it by hand and must do it on a computer system, which gives me a very long answer with two arctangents. I am pretty sure that will not work out to give me 4/3Piabc.

Thanks in advance.
 
  • #6
That's not difficult to integrate. It is essentially [itex]\sqrt{a^2- u^2}[/itex] for which the standard substitution is the trig substitution [itex]u= a sin(\theta)[/itex]. Here that would be [itex]u= \sqrt{1- v^2}sin(\theta)[/itex]. Integrating that, with respect to u, will give something involving [itex]\sqrt{1- v^2}[/itex] which you do with another sin substitution.
 
  • #7
I'm sorry but I don't understand. The formula listed in my previous post...

2C * sqrt(1-u^2-v^2)

is the resulting formula after having already integrated with respect to u.
You said "integrating that, with respect to u," but I need to integrate that with respect to v.

Also, if I introduce a sin(theta) into the equation, I don't see a point in the future at which that constant will disappear, and thus my answer would have a trig function in it, although the correct answer should be
4/3 * Pi * abc. Correct me if I'm wrong, I'm just not following your suggested method.
 
  • #8
Ok, so just do 2C*sqrt[ (1-u^2)-v^2] and integrate with respect to v. By making the trig substitution, you're actually integrating with respect to sin(theta) instead, so it goes away then
 

1. How do you derive the expression for the volume of an ellipsoid?

To derive the expression for the volume of an ellipsoid, we use the formula V = (4/3)πabc, where a, b, and c are the semi-axes of the ellipsoid. This formula is derived using the integral of the volume element in three-dimensional space.

2. What is the difference between an ellipsoid and a sphere?

An ellipsoid is a three-dimensional geometric shape that resembles a stretched or flattened sphere. It has three unequal semi-axes, making it a more general shape than a sphere, which has three equal semi-axes.

3. Can the expression for volume of an ellipsoid be simplified?

Yes, the expression for volume of an ellipsoid can be simplified using the eccentricity of the ellipsoid. The eccentricity is defined as the ratio of the distance between the foci to the length of the major semi-axis. The simplified form of the expression is V = (4/3)πabc(1 - e^2), where e is the eccentricity.

4. How is the volume of an ellipsoid used in real life?

The volume of an ellipsoid is used in various real-life applications, such as calculating the volume of planets and other celestial bodies, designing containers and packaging, and determining the size of organs in medical imaging. It is also used in geodesy and surveying for measuring the shape of the Earth.

5. Can the expression for volume of an ellipsoid be extended to higher dimensions?

Yes, the expression for volume of an ellipsoid can be extended to higher dimensions. In four-dimensional space, the formula becomes V = (π^2)/2 * a * b * c * d, where a, b, c, and d are the semi-axes of the four-dimensional ellipsoid. This can be further extended to any number of dimensions.

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