Outward flux through an ellipsoid

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SUMMARY

The discussion focuses on computing the outward flux of the vector field F = r/|r|^3 through the ellipsoid defined by the equation 4x² + 9y² + 6z² = 36. Participants clarify that the divergence theorem cannot be applied due to the discontinuity of F at the origin. Instead, they suggest using spherical coordinates to establish bounds for a surrounding sphere of radius 100, while also proposing a direct surface integral approach for the ellipsoid using the parameterization x = 3sin(φ)cos(θ), y = 2sin(φ)sin(θ), and z = √6cos(φ).

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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.


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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, \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. 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. x= 3 sin(\phi)cos(\theta), y= 2 sin(\phi)sin(\theta) and z= \sqrt{6}cos(\phi) with \theta from 0 to 2\pi and \phi from 0 to \pi is an appropriate parameterization for the surface of the ellipsoid.
 

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