Stokes theorem and line integral

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SUMMARY

The discussion centers on proving the equation 2A = ∮ ⃗{r} × d⃗{r} using Stokes' theorem. The participants explore different formulations of Stokes' theorem, particularly in relation to vector fields and planar surfaces in the xy-plane. The correct approach involves defining the vector field F = (-y/2, x/2, 0) and calculating its curl, resulting in ∇ × ⃗{F} = (0, 0, 2). This leads to the conclusion that the line integral equals the double integral over the surface area, confirming the initial equation.

PREREQUISITES
  • Understanding of Stokes' theorem and its applications in vector calculus.
  • Familiarity with vector fields and curl operations.
  • Knowledge of line integrals and surface integrals.
  • Basic proficiency in multivariable calculus, particularly in the context of planar surfaces.
NEXT STEPS
  • Study the applications of Stokes' theorem in various coordinate systems.
  • Learn about curl and divergence in vector calculus.
  • Explore examples of line integrals and their geometric interpretations.
  • Investigate the relationship between surface integrals and line integrals in physics.
USEFUL FOR

Students and educators in mathematics and physics, particularly those focusing on vector calculus, as well as professionals applying Stokes' theorem in engineering and physical sciences.

rado5
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Homework Statement



Prove that [tex]2A=\oint \vec{r}\times d\vec{r}[/tex]

Homework Equations





The Attempt at a Solution



From stokes theorem we have [tex]\oint d\vec{r}\times \vec{r}=\int _{s}(d\vec{s}\times \nabla)\times \vec{r}= \int _{s}(2ds\frac{\partial f}{\partial x},-ds+ds\frac{\partial f}{\partial y},-2ds)[/tex]. I'm stuck here and I don't know how to continue. Please help me.

Another method is [tex]\oint \vec{r}\times d\vec{r}=\oint (ydz-zdy,zdx-xdz,xdy-ydx)[/tex]. Please help me solve it.
 
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Maybe they didn't state it very clearly, but that formula is only true for a planar surface, say a region in the xy plane. And I'm not sure what formulation you are using of Stokes theorem. If you are using the vector form, that means a unit normal points in the z direction, say k. To get the area you want to integrate 1*dx*dy. Now if you define the vector field F=(-y/2,x/2,0) what is curl(F)?
 
Dick said:
Maybe they didn't state it very clearly, but that formula is only true for a planar surface, say a region in the xy plane.

So in this case z=0 and then we have: [tex]\oint \vec{r}\times d\vec{r}=\oint (ydz-zdy,zdx-xdz,xdy-ydx)=\oint (0,0,-ydx+xdy)[/tex] so [tex]\vec{F}=(-y,x,0) and \vec{n}=(0,0,1) and \nabla \times \vec{F}=(0,0,2) and so (\nabla \times \vec{F}).\vec{n}=(0,0,2)[/tex] so we have [tex]\oint \vec{F}.d\vec{r}=\int_{s}(\nabla\times \vec{F}).\vec{n}ds=\int\int_{D}2dxdy=2A[/tex]
I think my method is now correct. Thank you very much for your kind help.
 

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