Integral of a closed surface over a general region

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SUMMARY

The discussion centers on demonstrating the relationship between the surface integral of a vector field and the volume enclosed by that surface, specifically showing that \(\frac{1}{3}\oint\oint_{S}\vec{r} \cdot d\vec{s} = V\), where \(V\) is the volume enclosed by the closed surface \(S = \partial V\). The user initially applied Gauss' theorem but struggled to progress from the expression \(\frac{1}{3}(\int\int\int_{V}(r_{x}+r_{y}+r_{z})dxdydz)\). A conceptual understanding was provided, illustrating that the integral represents the sum of pyramidal volumes formed by the surface elements, confirming that the integral equals the volume enclosed by the surface.

PREREQUISITES
  • Understanding of surface integrals and vector fields
  • Familiarity with Gauss' theorem
  • Knowledge of triple integrals in multivariable calculus
  • Concept of volume calculation using geometric shapes
NEXT STEPS
  • Study the application of Gauss' theorem in vector calculus
  • Explore surface integrals and their geometric interpretations
  • Learn about the divergence theorem and its implications
  • Investigate the relationship between surface area and volume in three-dimensional geometry
USEFUL FOR

Students and professionals in mathematics, particularly those studying vector calculus, surface integrals, and geometric interpretations of integrals. This discussion is beneficial for anyone looking to deepen their understanding of the relationship between surface integrals and volume calculations.

Kwandae
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I have been working on this problem for a few hours and am completely stuck. It seems like a simple problem to me but when I attempt it I get nowhere. The problem is:

Show that

\frac{1}{3}\oint\oint_{S}\vec{r} \cdot d\vec{s} = V

where V is the volume enclosed by the closed surface S= \partial V

I have tried to use Gauss' theorem to get as far as

\frac{1}{3}(\int\int\int_{V}(r_{x}+r_{y}+r_{z})dxdydz)

But am completely stuck on what to do from this point or even if I started this correctly. It's been about 2 years since I have done any surface integrals so I was hoping if someone here could maybe give me a helping push in the right direction
 
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I see no one has commented for several hours, so I'll at least indicate I'm interested.
I just think of ds a square element of surface and r a radius from it to some center. The dot product takes only the component of r that is perpendicular to ds. If we imagine more lines from the center to the vertices of the square ds, we'll have a pyramid. It's volume will be 1/3*r.ds. If we continue to picture an infinite number of these pyramids to an infinite number of ds elements making up the whole surface area, we will see that all the pyramids make up exactly the volume enclosed by the surface. So the integral is indeed equal to the volume.

If you need a more mathematical analysis, I suggest you take it over to the math forum.
 

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