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Stokes' theorem over a tetrahedron

  1. Apr 20, 2005 #1
    Check the Stokes' theorem for the function [tex]\vec v = y\hat z[/tex]
    Here it is over a tetrahedron.

    Stokes' theorem suggests:
    [tex]\int_s {(\nabla\times \vec v).d\vec a = \oint_p\vec v.d\vec r [/tex]

    For the right hand side I computed the line integral from (a,0,0)--->(0,2a,0)--->(0,0,a)--->(a,0,0);
    which comes out to be [tex]a^2[/tex] (matches with the solution given).

    For the surface integral, one needs to obtain the expression for the area element[tex]d\vec a[/tex] of the plane given by the above points, which is where my problem lies. Can someone help me on this?
     
  2. jcsd
  3. Apr 20, 2005 #2

    Galileo

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    You don't necessarily need the expression for the area element. Remember that you may choose the surface anyway you like as long as it has the triangle as its boundary.
    Since the curl point only in the x-direction, choose your surface such that the integration will become trivial.
     
  4. Apr 20, 2005 #3
    Thank you so much for pointing it out to me. Here is how I figured it out!

    [tex]\nabla\times\vec v = \hat x[/tex]
    so,[tex]d\vec a = dydz\hat x[/tex]
    Taking x=a, y-->0 to 2a-2z & z-->0 to a
    [tex]\int(\nabla\times\vec v) .d\vec a = \int_{0}^{a}dz\int_{0}^{2a-2z}dy = a^2[/tex]
    Since intregration over other surfaces are trivial.
    Hence verifying Stokes' law.
     
  5. Apr 20, 2005 #4

    Galileo

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    Right. Geometrically, the flux through the parts in the xz-plane and the xy-plane are both zero (curl only points in the x-direction) and the flux through the remaining surface is simple the area of that surface, which is a right triangle with height a and length 2a, thus area a^2.

    Always look for ways to exploit geometry, since it will speed up your calculations. Calculating fluxes and line integrals generally involve a lot of work, so every bit helps.
     
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