Outward flux through an ellipsoid

In summary, the problem involves computing the outward flux of F=r/|r|^3 through the ellipsoid 4x^2+9y^2+6z^2=36. The divergence theorem cannot be used on the region inside S due to F not being continuous at 0. The bounds for the sphere of radius 100 are \rho from 0 to 100, \theta from 0 to 2\pi, and \phi from 0 to \pi. For the ellipse, the bounds on \theta and \phi are the same, but the bounds on \rho will be functions of \theta and \phi. Alternatively, the flux can be found by integrating over the surface with a parameterization
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Homework Statement


Let r(x,y,z)=<x,y,z>. Compute the outward flux of F=r/|r|^3 through the ellipsoid 4x^2+9y^2+6z^2=36.


Homework Equations





The Attempt at a Solution


I know that I can't use the divergence theorem on the region inside S because F isn't continuous at 0. But I can do it on the region between S and say, a sphere of radius 100 right? An how would I set that up? I'm having some trouble figuring out the bounds between the ellipsoid and sphere.
 
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The bounds depend on the coordinate system you are using. Using spherical coordinates, the bound for the sphere of radius 100 are, of course, [itex]\rho from 0 to 100[/itex], [itex]\theta[/itex] from 0 to [itex]2\pi[/itex], and [itex]\phi[/itex] from 0 to [itex]\pi[/itex]. For the ellipse the bounds on [itex]\theta[/itex] and [itex]\phi[/itex] are the same but the bounds on [itex]\rho[/itex] will be functions of [itex]\theta[/itex] and [itex]\phi[/itex]. And, of course, you will still need to find the flux through the sphere directly.

Perhaps it would be simpler just to find the flux through the ellipsoid by integrating over the surface. [itex]x= 3 sin(\phi)cos(\theta)[/itex], [itex]y= 2 sin(\phi)sin(\theta)[/itex] and [itex]z= \sqrt{6}cos(\phi)[/itex] with [itex]\theta[/itex] from 0 to [itex]2\pi[/itex] and [itex]\phi[/itex] from 0 to [itex]\pi[/itex] is an appropriate parameterization for the surface of the ellipsoid.
 

1. What is outward flux through an ellipsoid?

The outward flux through an ellipsoid is a measure of the flow of a vector field through the surface of the ellipsoid. It is a fundamental concept in the study of electromagnetism and fluid mechanics.

2. How is outward flux through an ellipsoid calculated?

The outward flux through an ellipsoid is calculated by integrating the dot product of the vector field and the outward unit normal vector over the surface of the ellipsoid. This can be done using the Gauss's law for flux or using the divergence theorem.

3. What is the significance of outward flux through an ellipsoid?

Outward flux through an ellipsoid is an important concept in understanding the behavior of electric and magnetic fields, as well as fluid flow. It helps in determining the strength and direction of the flow through a given surface, and can be used to solve various physical problems.

4. Can the outward flux through an ellipsoid be negative?

Yes, the outward flux through an ellipsoid can be negative. This indicates that the vector field is flowing inwards through the surface of the ellipsoid, rather than outwards. This can occur in situations where there is a sink or a source within the ellipsoid.

5. How does the shape of the ellipsoid affect the outward flux?

The shape of the ellipsoid has a significant impact on the outward flux. A more elongated ellipsoid will have a higher flux compared to a more spherical ellipsoid with the same volume. This is because the outward unit normal vector varies with the surface area of the ellipsoid, and a more elongated shape will have a larger surface area for the same volume.

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