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Using Stokes' theorem, the line integral of a vector field gives a surface integral of the curl of the vector field, and after that, if we apply Gauss' divergence theorem in that, it gives a volume integral of the divergence of the curl of that vector field. But we know the divergence of the curl of a vector field is $0$. … so how can it be possible? Where is the mistake??

Daniel Fischer
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Sahil
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  • Take a 3D object, say a solid ball. Its boundary is the surface, i.e. a sphere. Gauss relates those two integrals. Stokes relates the integral over that the sphere and its boundary curve. But, what is the boundary curve of the surface of a sphere? There isn't one! The exact same thing happens if you start with a torus. Misner, Thoren & Wheeler repeatedly state: boundary of a boundary is zero They do mean it in the context of differential forms, but it does apply geometrically as well. If it didn't we would run into problems of the type you describe! – Jyrki Lahtonen Oct 08 '15 at 15:31
  • But in case of a solid cylinder it has two circular surfaces. In those surfaces is the problem true? – Sahil Oct 08 '15 at 15:40
  • No. The surface of a solid cylinder has three parts. The bottom disk, the top disk, and the surface of that cylinder. The two circles (=the common boundaries of those three parts) are traversed in both clockwise and counterclockwise directions, so the corresponding line integrals cancel automatically from the sum of the line integrals giving all of the boundary of the boundary. No contradiction there either! – Jyrki Lahtonen Oct 08 '15 at 15:51
  • Title of the question looks very interesting....! – Empty Oct 08 '15 at 15:54
  • Related: http://physics.stackexchange.com/q/75229/2451 – Qmechanic Oct 08 '15 at 15:56

2 Answers2

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Stokes theorem relates the line integral around a curve to a surface integral through an open surface.

The divergence theorem relates the surface integral through a closed surface to a volume integral.

You cannot do what you are trying.

CStarAlgebra
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For Stokes, your curved line is closed and bounds a surface. For Gauss, your surface is closed and bounds a volume. Both basically say the same thing, relating a border to the inside. But the out of the first does not fit the in of the second, even if both are "surfaces" ;-)

Fabrice NEYRET
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  • If i take a closed volume say a cylinder. Now the circular surface of the cylinder represents a closed surface as well as a closed curve(i.e circle). So on this case or cases like this, the previous is true? – Sahil Oct 08 '15 at 15:28
  • The circle curve fit one surface that is the up part of the cylinder, and another surface that is the down part of the cylinder. You have to consider both separately. – Fabrice NEYRET Oct 08 '15 at 15:32
  • Yeah, i will consider, but my point was that is the example of cylinder a special case of the given problem or what? – Sahil Oct 08 '15 at 15:37
  • I don't see what would make the cylinder different to any other close shape able to host the circle curve. – Fabrice NEYRET Oct 08 '15 at 15:45