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Stokes theorem and line integral

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Dec 12, 2009 #2

    Dick

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    Homework Helper

    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)?
     
  4. Dec 12, 2009 #3
    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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