How do you evaluate the vector calculus question over an ellipsoid?

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SUMMARY

The evaluation of the vector calculus integral \(\oint\nabla(1/r) \cdot n \, dS\) over the ellipsoid defined by \(x^2/4 + y^2/9 + z^2/25 = 1\) leads to a definitive result of \(-4\pi\). While some participants argued that the answer is 0 due to the harmonic nature of \(1/r\) in simply connected regions, it is established that \(1/r\) is not harmonic at the origin. Thus, the correct evaluation incorporates the singularity at the origin, confirming the result of \(-4\pi\).

PREREQUISITES
  • Understanding of vector calculus, specifically divergence and gradient operations.
  • Familiarity with harmonic functions and their properties.
  • Knowledge of ellipsoids and their mathematical representation.
  • Proficiency in evaluating surface integrals in three-dimensional space.
NEXT STEPS
  • Study the properties of harmonic functions and their singularities.
  • Learn about the divergence theorem and its applications in vector calculus.
  • Explore surface integrals over various geometric shapes, including ellipsoids.
  • Investigate the implications of singularities in vector fields and their effects on integral evaluations.
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Students and professionals in mathematics, particularly those focusing on vector calculus, as well as educators seeking to clarify concepts related to harmonic functions and surface integrals.

BrandonATC
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The question is

Let r=r(x,y,z) where it is the distance from a point O. Evaluate

[tex]\oint[/tex][tex]\nabla[/tex](1/r)*ndS

(where * is the dot product)

over the ellipsoid

x^2/4 +y^2/9+z^2/25=1

I thought the answer was 0 since the ellipsoid is a simply connected region in R^3 and the div(grad(1/r))= 0 since 1/r is harmonic.

But some people in the class said the answer was -4pi

Thanks
 
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Welcome to PF!

Hi BrandonATC! Welcome to PF! :smile:

(have a pi: π and a del: ∇ and try using the X2 icon just above the Reply box :wink:)
BrandonATC said:
… I thought the answer was 0 since the ellipsoid is a simply connected region in R^3 and the div(grad(1/r))= 0 since 1/r is harmonic.

No, it's not harmonic at the origin. :wink:
 
thank you, just wondering how do you show the answer is -4pi?
 

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