Integration for the Volume of an Ellipsoid

In summary, the conversation discusses finding the volume of an ellipsoid and the best approach for solving the problem. The suggestion is made to rotate the x-y plane to a different coordinate system to make it easier to find the points of intersection. It is also noted that due to symmetry, the volume can be found in one octant and multiplied by 8. The article on rotation of axes is recommended as a resource for finding the appropriate angle.
  • #1
Daniel Sellers
117
17

Homework Statement


Let E be the ellipsoid x^2 + 2xy +5y^ +4z^2 = 1
Find the Volume of E

Homework Equations


None, just various integration methods.

The Attempt at a Solution


I know we're not supposed to say 'I don't know where to start' but with this one I really don't. If the best approach is a coordinate change, how would I go about searching for the correct transformation to map this to a sphere?

I tried to solve a similar problem concerning an ellipse (not ellipsoid) and failed to find a substitution. Someone showed me how to solve that problem in cartesian coordinates but that method won't work here.

Can anyone with experience with this type of problem at least tell me whether it's best to search for a substitution or attempt the problem in cartesian? And what my first step in either direction ought to be?
 
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  • #2
Daniel Sellers said:

Homework Statement


Let E be the ellipsoid x^2 + 2xy +5y^ +4z^2 = 1
Find the Volume of E

Homework Equations


None, just various integration methods.

The Attempt at a Solution


I know we're not supposed to say 'I don't know where to start' but with this one I really don't. If the best approach is a coordinate change, how would I go about searching for the correct transformation to map this to a sphere?

I tried to solve a similar problem concerning an ellipse (not ellipsoid) and failed to find a substitution. Someone showed me how to solve that problem in cartesian coordinates but that method won't work here.

Can anyone with experience with this type of problem at least tell me whether it's best to search for a substitution or attempt the problem in cartesian? And what my first step in either direction ought to be?
Some advice I gave in another thread that wasn't helpful might actually be helpful here. If you rotate the x-y plane to a different, X-Y, coordinate system, it should be easier to find the points at which the ellipsoid intersects the various planes. The z-axis can stay unchanged -- just the x-y plane.

See https://en.wikibooks.org/wiki/Conic_Sections/Rotation_of_Axes.

Also, due to symmetry, you can find the volume in one octant of your transformed solid, then multiply that value by 8 to get the volume of the whold object.
 
  • #3
Mark44 said:
Some advice I gave in another thread that wasn't helpful might actually be helpful here. If you rotate the x-y plane to a different, X-Y, coordinate system, it should be easier to find the points at which the ellipsoid intersects the various planes. The z-axis can stay unchanged -- just the x-y plane.

See https://en.wikibooks.org/wiki/Conic_Sections/Rotation_of_Axes.

Also, due to symmetry, you can find the volume in one octant of your transformed solid, then multiply that value by 8 to get the volume of the whold object.

Unfortunately that doesn't seem to help, or maybe it would if I had a lot more practice with this type of problem. Applying that transformation results in a mess of trig functions which isn't any easier to interpret or integrate. This article doesn't explain how to find the angle I need to rotate through either.

I am trying to find a coordinate change which will map this ellipsoid to a sphere, but maybe that's not possible here?
 
  • #4
If I set z = 0 and solve for x in terms of y I can see that the major axis of the ellipse in the xy plane is the line x = -y which would need to rotated by pi/4 to align with a coordinate axis. Applying the transformation you recommended above and plugging in pi/4 results in 4X^2 + 2Y^2 + 4XY + 4z^2 which is slightly less ugly but still presents the same problem... and still seems to be a tilted ellipse.

If I use -pi/4 the numbers are a little different but the XY term is still there just with a -6 coefficient instead of 4
 
  • #5
Daniel Sellers said:
This article doesn't explain how to find the angle I need to rotate through either.
Sure it does, in the first section under Graphing.
##\tan(2\theta) = \frac B {A - C}##. In your equation, A = 1, B = 2, and C = 5. When I put these numbers in I get a value for ##\tan(2\theta)## of -1/2.

Daniel Sellers said:
I am trying to find a coordinate change which will map this ellipsoid to a sphere, but maybe that's not possible here?
You'll have to take care of the rotation first. Either way, I don't recommend turning the ellipsoid into a sphere.

Daniel Sellers said:
If I set z = 0 and solve for x in terms of y I can see that the major axis of the ellipse in the xy plane is the line x = -y
I don't think so. That would be a rotation of -45°, which is too much.

Daniel Sellers said:
If I set z = 0 and solve for x in terms of y I can see that the major axis of the ellipse in the xy plane is the line x = -y which would need to rotated by pi/4 to align with a coordinate axis. Applying the transformation you recommended above and plugging in pi/4 results in 4X^2 + 2Y^2 + 4XY + 4z^2 which is slightly less ugly but still presents the same problem... and still seems to be a tilted ellipse.
If you rotate by the right amount, the mixed term (xy term) goes away. The fact that you still have an XY term is a sign that your rotation angle isn't right.
 
  • #6
Mark44 said:
Sure it does, in the first section under Graphing.
##\tan(2\theta) = \frac B {A - C}##. In your equation, A = 1, B = 2, and C = 5. When I put these numbers in I get a value for ##\tan(2\theta)## of -1/2.

You'll have to take care of the rotation first. Either way, I don't recommend turning the ellipsoid into a sphere.

I don't think so. That would be a rotation of -45°, which is too much.

If you rotate by the right amount, the mixed term (xy term) goes away. The fact that you still have an XY term is a sign that your rotation angle isn't right.

Alright, that's helpful. I did ask for a starting place after all. The angle of rotation should be given by (1/2)Arctan(-1/2) then, correct? Am I going to need to compute a Jacobian for this transformation?
 
  • #7
Daniel Sellers said:
Alright, that's helpful. I did ask for a starting place after all. The angle of rotation should be given by (1/2)Arctan(-1/2) then, correct? Am I going to need to compute a Jacobian for this transformation?
I don't think so, because it's only a rotation -- the lengths of things aren't going to change. I'm not positive in saying that the Jacobian isn't needed, though.
The Jacobian does play a role in converting from Cartesian to polar form -- that's where the r comes from in going from ##dxdy## to ##rdrd\theta##.
 
  • #8
Mark44 said:
I don't think so, because it's only a rotation -- the lengths of things aren't going to change. I'm not positive in saying that the Jacobian isn't needed, though.
The Jacobian does play a role in converting from Cartesian to polar form -- that's where the r comes from in going from ##dxdy## to ##rdrd\theta##.

You, sir, are a gentleman and a scholar. By applying that rotation I was able to eliminate the XY term and then do a simple (u,v,w) transformation, calculate it's Jacobian and then perform a simple integration over a sphere of radius 1. It worked out to a tidy pi/3 which was correct.

If I had to integrate a function over this volume, and hence was unable to perform a rotation, I still have no idea how I would do that. Thankfully that seems to be slightly beyond the scope of my class as we have moved onto to vector fields and gradients.

Thanks a ton

EDIT: Actually do know that I would have to find a direct substitution like I was trying to in the first place, but I don't know how to do THAT so, six of one...
 

What is an ellipsoid?

An ellipsoid is a three-dimensional geometric shape that resembles a stretched-out sphere. It is defined by three radii, known as the semi-major axis, semi-minor axis, and polar axis.

What is integration for the volume of an ellipsoid?

Integration for the volume of an ellipsoid is a mathematical process used to find the total space enclosed by the surface of an ellipsoid. It involves using calculus to calculate the volume under the surface of the ellipsoid.

What is the formula for finding the volume of an ellipsoid?

The formula for finding the volume of an ellipsoid is V = (4/3)πabc, where a, b, and c are the semi-major, semi-minor, and polar axes, respectively. This formula is derived from the integration process for finding the volume of an ellipsoid.

What are the applications of finding the volume of an ellipsoid?

Finding the volume of an ellipsoid has many practical applications, such as in engineering, physics, and astronomy. It is used to calculate the volume of objects with ellipsoidal shapes, such as planets and satellites, and is also used in the design of structures and machinery.

Are there any other methods for finding the volume of an ellipsoid?

Yes, there are other methods for finding the volume of an ellipsoid, such as using geometric formulas or approximations. However, integration is the most accurate method and is widely used in scientific and engineering fields.

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