Page 263 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 263

254       LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS        [CHAP. 10



                          or
                                                      ðð              þ
                                                                        A 1 dx
                                                        ½r   ðA 1 iފ   n dS ¼
                                                                       C
                                                       S
                          Similarly, by projections on the other coordinate planes,
                                         ðð              þ         ðð              þ
                                                           A 2 dy;                   A 3 dz
                                           ½r   ðA 2 jފ   n dS ¼    ½r   ðA 3 kފ   n dS ¼
                                                          C                         C
                                         S                         S
                          Thus, by addition,
                                                       ðð            þ
                                                                       A   dr
                                                         ðr   AÞ  n dS ¼
                                                                      C
                                                       S
                              The theorem is also valid for surfaces S which may not satisfy the restrictions imposed above.  For
                          assume that S can be subdivided into surfaces S 1 ; S 2 ; ... ; S k with boundaries C 1 ; C 2 ; .. . ; C k which do satisfy
                          the restrictions. Then Stokes’ theorem holds for each such surface. Adding these surface integrals, the total
                          surface integral over S is obtained.  Adding the corresponding line integrals over C 1 ; C 2 ; ... ; C k ,the line
                          integral over C is obtained.

                                                               2
                     10.27. Verify Stoke’s theorem for A ¼ 3yi   xzj þ yz k, where S is
                                                       2   2
                          the surface of the paraboloid 2z ¼ x þ y bounded by z ¼ 2
                          and C is its boundary.  See Fig. 10-21.
                              The  boundary  C  of S  is  a  circle  with  equations
                           2
                               2
                          x þ y ¼ 4; z ¼ 2 and  parametric  equations x ¼ 2cos t; y ¼
                          2 sin t; z ¼ 2, where 0 @ t < 2 .  Then
                              þ       þ
                                                     2
                                        3ydx   xz dy þ yz dz
                                A   dr ¼
                               C       C
                                      ð 0
                                        3ð2 sin tÞð 2 sin tÞ dt  ð2cos tÞð2Þð2cos tÞ dt
                                    ¼
                                       2
                                      ð 2
                                             2
                                                    2
                                        ð12 sin t þ 8cos tÞ dt ¼ 20
                                    ¼
                                       0
                                                                                        Fig. 10-21

                                         i   j
                                                 k

                                           @  @  @
                                                       2
                          Also,  r  A ¼              ¼ðz þ xÞi  ðz þ 3Þk

                                           @x  @y  z

                                           3y  xz  yz  2
                                                         2   2       xi þ yj   k
                          and                          rðx þ y   2zÞ  ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :
                                                         2   2         2  2
                                                   n ¼
                                                      jrðx þ y   2zÞj  x þ y þ 1
                          Then
                                        ðð             ðð              ðð
                                                                 dx dy
                                                                           2
                                                                              2
                                                         ðr   AÞ  n      ðxz þ x þ z þ 3Þ dx dy
                                           ðr   AÞ  n dS ¼           ¼
                                                                jn   kj
                                         S             r               r
                                                                 ! 2
                                                         8                      9
                                                                         2
                                                             2
                                                            x þ y 2     x þ y 2  =
                                                       ðð <
                                                          x          2       þ 3 dx dy
                                                              2           2
                                                     ¼             þx þ
                                                         :                      ;
                                                       r
                          In polar coordinates this becomes
                                           ð 2   ð 2
                                                         4
                                                               2
                                                                  2
                                                                       2
                                                 fð  cos  Þð  =2Þþ   cos   þ   =2 þ 3g   d  d  ¼ 20
                                             ¼0  ¼0
   258   259   260   261   262   263   264   265   266   267   268