Surface level and surface integral

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SUMMARY

The discussion centers on the mathematical concepts of surface levels and surface integrals, specifically how to extend the gradient theorem from line integrals to surface integrals. The user defines a level surface as f(x,y,z)=k and proposes that the associated surface integral can be expressed as ∇f(S(t,s))·(S't(t,s)×S's(t,s))=k. The conversation also explores the relationship between level curves in 2D and level surfaces in 3D, concluding that integral curves crossing a surface are governed by the conditions ∇f·dS/dt=0 and ∇f·dS/ds=0.

PREREQUISITES
  • Understanding of vector calculus, specifically gradient and surface integrals.
  • Familiarity with level curves and level surfaces in multivariable functions.
  • Knowledge of the gradient theorem and its application in line integrals.
  • Basic proficiency in parametric equations for curves and surfaces.
NEXT STEPS
  • Study the application of the gradient theorem to surface integrals in vector calculus.
  • Explore the concept of level surfaces in multivariable calculus.
  • Learn about the cross product and its role in vector calculus.
  • Investigate the implications of integral curves in the context of differential geometry.
USEFUL FOR

Mathematicians, physics students, and anyone studying multivariable calculus or vector calculus who seeks to deepen their understanding of surface integrals and gradient fields.

Jhenrique
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I have some questions, all associated. So, first, if a curve level is defined as:
[tex]f(x,y)=k[/tex]
or vectorially as:
[tex]f(c(t))=k[/tex]
and its curve integral associated as:
[tex]\bigtriangledown f(c(t))\cdot c'_{t}(t)=k[/tex]

Then, how is the equation of a surface integral associated to surface level:
[tex]f(x,y,z)=k[/tex]
[tex]f(S(t,s))=k[/tex]

Would be this?
[tex]\bigtriangledown f(S(t,s))\cdot (S'_{t}(t,s)\times S'_{s}(t,s))=k[/tex]

And more, all this above make I think if is possible to extend the gradient's theorem (that is specific to line integral):
[tex]\int_{t_0}^{t_1} \bigtriangledown f\cdot \hat{t}\;ds=\Delta f[/tex]
to surface integral...?
 
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Last edited:
I think I found the answer for my question. In 2D, we have a level curve given by the equation f(s(t))=k, and the integral curves that cross it is given by ∇f·ds/dt=0. In 3D, we have a level surface given by the equation f(S(t,s))=k, and the integral curves that cross this surface are probably given by two equations ∇f·dS/dt=0 and ∇f·dS/ds=0. Someone understood what I wanted say!?
 

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