Page 258 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 258

CHAP. 10]  LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS             249


                           is an example of a one-sided surface. This is sometimes called a nonorientable surface. A two-sided surface
                           is orientable.


                     THE DIVERGENCE THEOREM
                     10.22. Prove the divergence theorem. (See Fig. 10-18.)





















                                                             Fig. 10-18


                              Let S be a closed surface which is such that any line parallel to the coordinate axes cuts S in at most two
                           points. Assume the equations of the lower and upper portions, S 1 and S 2 ,tobe z ¼ f 1 ðx; yÞ and z ¼ f 2 ðx; yÞ,
                           respectively.  Denote the projection of the surface on the xy plane by r.  Consider

                                      ððð        ðð ð           ðð ð

                                                                    f 2 ðx;yÞ
                                          @A 3       @A 3                @A 3
                                                                            dz dy dx
                                          @z  dV ¼   @z  dz dy dx ¼  z¼f 1 ðx;yÞ @z
                                       V          V             r
                                                             2
                                                 ðð           f     ðð
                                                                      ½A 3 ðx; y; f 2 Þ  A 3 ðx; y; f 1 ފ dy dx
                                                ¼   A 3 ðx; y; zÞ  dy dx ¼

                                                           z¼f 1
                                                  r                 r
                              For the upper portion S 2 , dy dx ¼ cos 
 2 dS 2 ¼ k   n 2 dS 2 since the normal n 2 to S 2 makes an acute angle
                           
 2 with k.
                              For the lower portion S 1 , dy dx ¼  cos 
 1 dS 1 ¼ k   n 1 dS 1 since the normal n 1 to S 1 makes an obtuse
                           angle 
 1 with k.
                                                   ðð              ðð
                              Then                   A 3 ðx; y; f 2 Þ dy dx ¼  A 3 k   n 2 dS 2
                                                   r               S 2
                                                   ðð               ðð
                                                                       A 3 k   n 1 dS 1
                                                     A 3 ðx; y; f 1 Þ dy dx ¼
                                                   r                 S 1
                           and
                                     ðð              ðð              ðð           ðð
                                                                                    A 3 k   n 1 dS 1
                                       A 3 ðx; y; f 2 Þ dy dx    A 3 ðx; y; f 1 Þ dy dx ¼  A 3 k   n 2 dS 2 þ
                                      r               r               S 2          S 1
                                                                     ðð
                                                                        A 3 k   n dS
                                                                   ¼
                                                                      S
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