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CHAPTER 3 Electric Flux Density, Gauss’s Law, and Divergence    69
















                                 Figure 3.7  The divergence theorem states that the total
                                 flux crossing the closed surface is equal to the integral of
                                 the divergence of the flux density throughout the enclosed
                                 volume. The volume is shown here in cross section.


                         Again, we emphasize that the divergence theorem is true for any vector field,
                     although we have obtained it specifically for the electric flux density D, and we will
                     have occasion later to apply it to several different fields. Its benefits derive from the
                     fact that it relates a triple integration throughout some volume to a double integration
                     over the surface of that volume. For example, it is much easier to look for leaks in
                     a bottle full of some agitated liquid by inspecting the surface than by calculating the
                     velocity at every internal point.
                         The divergence theorem becomes obvious physically if we consider a volume ν,
                     shown in cross section in Figure 3.7, which is surrounded by a closed surface S.
                     Division of the volume into a number of small compartments of differential size and
                     consideration of one cell show that the flux diverging from such a cell enters,or
                     converges on, the adjacent cells unless the cell contains a portion of the outer surface.
                     In summary, the divergence of the flux density throughout a volume leads, then, to
                     the same result as determining the net flux crossing the enclosing surface.


                                                                                            EXAMPLE 3.5
                                                                                 2
                     Evaluate both sides of the divergence theorem for the field D = 2xya x + x a y C/m 2
                     and the rectangular parellelepiped formed by the planes x = 0 and 1, y = 0 and 2,
                     and z = 0 and 3.
                     Solution. Evaluating the surface integral first, we note that D is parallel to the sur-
                     faces at z = 0 and z = 3, so D · dS = 0 there. For the remaining four surfaces
                     we have



                                       3     2                  3     2
                            D · dS =      (D) x=0 · (−dy dz a x ) +  (D) x=1 · (dy dz a x )
                           S         0  0                     0  0
                                         3     1                 3     1
                                    +       (D) y=0 · (−dx dz a y ) +  (D) y=2 · (dx dz a y )
                                       0  0                     0  0
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