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Vector Calculus question Div and Stokes Theorem

  1. Feb 26, 2008 #1
    If you start with the two dimensional green's theorem, and you want to extend this three dimensions.

    Closed line integral = Surface Integral of the partials (dP/dx + dQ/dy) da
    seems to leads the divergence theorem,
    When the space is extended to three dimensions.

    On the other hand:
    Closed line integral = Surface Integral of the partials (dQ/dx - dP/dy) da
    seems to lead to Stokes theorem when the space is extended to three dimensions.

    Would it be correct to view these two vector calculus theorems this way.
    Start with the two dimensional Green's theorem and go either way to get the Divergence
    or Stokes theorem.

    Any additional insights you could give would be appreciated.
  2. jcsd
  3. Feb 26, 2008 #2
    You can't generalize Green's theorem to get divergence theorem, since Green's theorem is not a special (2D) case of div. th.:
    vector field is parallel to the boundary of the object in the first case and perpendicular in the second case.

    Green's theorem is in fact a special (2D) case of Stokes' theorem.
  4. Feb 26, 2008 #3

    First I though this formula is wrong, but now I realized that you probably considered the integral of <vector,normal vector>*dl in this formula. This is basicly the same as divergence theorem. (the proof of both theorems is almost the same, except for the number of dimensions).
  5. Feb 27, 2008 #4
    Thanks for responding.

    I can start with a closed line integral and by multiplying by 1 either dx/dx or dy/dy and the fundamental theorem cal end up with Green's Theorem. I can also see the k component of the curl in this so I have no problem going to stokes theorem.

    I'm am though having though with showing (not my words) that The divergence theorem is an extension to R3 of the fundamental theorem of calculus and Green’s Theorem. (I do not understand the two proofs I have looked at)

    I'd appreciate it if you could list a simple proof

    thanks , bob
  6. Feb 27, 2008 #5
    Let's first calculate only the integral of <Fz,n> dS.

    Imagine an infinitezimal square in xy plane with size dx*dy and a random plane, inclined against x,y plane for angle alpha. The intersection of the plane and the square (extended in z direction) has area dx*dy/cos(alpha). A normal vector on this intersection has z component: nz=cos(alpha). So integral <Fz,n>dS over this intersection is dx*dy*Fz (angle is canceled).
    A suface of a volume can be distributed among such small dx*dy squares. The surface intersects each square twice, so the contribution of integral <Fz,n>dS over the square is dx*dy*(Fz(top)-Fz(bottom))=dx*dy*integral(dFz/dz*dz,from bottom to top). After integrating over x and y, you get integral(dFz/dz) over whole volume.

    Obviously the same calculation can be done for <Fx,n>dS and <Fy,n>dS, so
    integral <F,n>dS (over surface of V)=integral (dFx/dx+dFy/dy+dFz/dz)*dx*dy*dz (over V)
  7. Feb 28, 2008 #6


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    Staff Emeritus
    Science Advisor

    Actually, Green's theorem, the divergence theorem, Stoke's theorem as it is usually stated in calculus books, are all special cases of the "Generalized Stoke's Theorem":

    If M is a "nice" manifold with boundary [itex]\partial M[/itex], of dimension n, [itex]\omega[/itex] is an n-1 order differential form on [itex]\partial M[/itex], so that d[itex]\omega[/itex] is an n order differential form on M, then
    [tex]\int_{\partial M} \omega= \int_{M} d\omega[/tex]

    (A precise statement of "nice" here would require Algebraic Topology but roughly it means "no holes".)

    And since, once again, this has nothing to do with differential equations, I am moving it to "Calculus and Analysis".

    (People who are taking Calculus seem to think that anything about "derivatives" or "differentials" belongs in "Differential Equations"!)
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