Deriving Vector Area of a Surface S Using the Cross Product

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SUMMARY

The discussion centers on deriving the vector area of a surface \( S \) using the cross product. The key equation presented is \( \textbf{a} = \frac{1}{2} \oint{\textbf{r} \times d\textbf{l}} \). Participants highlight the use of Stokes' Theorem and the cyclicity of the triple product as essential techniques for this derivation. The conversation also explores alternative physical and geometrical interpretations of the result, indicating a desire for a deeper understanding beyond mathematical manipulation.

PREREQUISITES
  • Understanding of vector calculus, specifically Stokes' Theorem
  • Familiarity with cross product operations in three-dimensional space
  • Knowledge of line integrals and surface integrals
  • Basic concepts from electromagnetism as referenced in Griffiths' textbook
NEXT STEPS
  • Study Stokes' Theorem in detail to understand its applications in vector calculus
  • Learn about the properties of the cross product and its geometric interpretations
  • Explore line integrals and surface integrals in the context of electromagnetism
  • Review Griffiths' textbook on electromagnetism for additional insights on vector areas
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as mathematicians and engineers interested in vector calculus and its applications in physical contexts.

rbwang1225
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Define

[itex]\textbf{a}\equiv\int_{S}{d\textbf{a}}[/itex]

How do I show that [itex]\textbf{a}=\frac{1}{2}\oint{\textbf{r}\times d\textbf{l}}[/itex]

Actually, this is the problem of the EM book of Griffiths, but I don't understand his hint.

Any help would be appreciated.
 
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One trick is to dot the latter expression with an arbitrary constant vector, and then use cyclicity of the triple product. You should be able to cast it into a form where it is easy to use Stokes' Theorem.
 
O.K., I got the result.

But is there any other more physical or geometrical way to derive that?

Thanks!
 

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