Page 259 - Schaum's Outline of Theory and Problems of Advanced Calculus
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250       LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS        [CHAP. 10



                          so that
                                                         ðð ð       ðð
                                                            @A 3
                                                                      A 3 k   n dS
                                                             @z
                                                     ð1Þ       dV ¼
                                                          V         S
                              Similarly, by projecting S on the other coordinate planes,
                                                         ððð        ðð
                                                            @A 1
                                                                      A 1 i   n dS
                                                             @x
                                                     ð2Þ       dV ¼
                                                          V         S
                                                         ððð        ðð
                                                            @A 2
                                                                      A 2 j   n dS
                                                     ð3Þ       dV ¼
                                                             @y
                                                          V         S
                              Adding (1), (2), and (3),
                                            ððð                   ðð
                                                 @A 1  @A 2  @A 3
                                                                    ðA 1 i þ A 2 j þ A 3 kÞ  n dS
                                                 @x  þ  @y  þ  @z  dV ¼
                                              V                    S
                                                       ðð ð       ðð
                          or                              r  A dV ¼  A   n dS
                                                        V          S
                              The theorem can be extended to surfaces which are such that lines parallel to the coordinate axes meet
                          them in more than two points.  To establish this extension, subdivide the region bounded by S into
                          subregions whose surfaces do satisfy this condition.  The procedure is analogous to that used in Green’s
                          theorem for the plane.
                                                                          2
                                                                    2
                     10.23. Verify the divergence theorem for A ¼ð2x   zÞi þ x yj   xz k taken over the region bounded by
                          x ¼ 0; x ¼ 1; y ¼ 0; y ¼ 1; z ¼ 0; z ¼ 1.
                                           ðð
                              We first evaluate  A   n dS where S is the surface of the cube in Fig. 10-19.
                                           S
                          Face DEFG:  n ¼ i; x ¼ 1.  Then
                                          1  1
                              ðð         ð ð
                                                       2
                                            fð2   zÞi þ j   z kg  i dy dz
                                 A   n dS ¼
                                          0  0
                              DEFG
                                          1  1
                                         ð ð
                                            ð2   zÞ dy dz ¼ 3=2
                                       ¼
                                          0  0
                          Face ABCO:  n ¼ i; x ¼ 0.  Then
                                              1  1
                                  ðð         ð ð
                                                ð ziÞ ð iÞ dy dz
                                     A   n dS ¼
                                              0  0
                                 ABCO
                                              1
                                             ð ð 1
                                                zdy dz ¼ 1=2
                                           ¼
                                              0  0
                                                                                   Fig. 10-19
                          Face ABEF:  n ¼ j; y ¼ 1.  Then
                                      1  1                        1  1
                           ðð         ð ð                        ð ð
                                                   2   2              2
                              A   n dS ¼  fð2x   zÞi þ x j   xz kg  j dx dz ¼  x dx dz ¼ 1=3
                                      0  0                        0  0
                          ABEF
                          Face OGDC:  n ¼ j; y ¼ 0.  Then
                                                          1  1
                                              ðð         ð ð
                                                                       2
                                                             fð2x   zÞi   xz kg ð jÞ dx dz ¼ 0
                                                 A   n dS ¼
                                                          0  0
                                              OGDC
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