Page 254 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 254

CHAP. 10]  LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS             245


                           (b)Let A ¼ðx 1 ; y 1 Þ; B ¼ðx 2 ; y 2 Þ.  From part (a),
                                                              ð
                                                               ðx;yÞ
                                                                  Pdx þ Qdy
                                                         ðx; yÞ¼
                                                               ða;bÞ
                              Then omitting the integrand Pdx þ Qdy,we have
                                          B   ðx 2 ;y 2 Þ  ðx 2 ;y 2 Þ  ðx 1 ;y 1 Þ
                                         ð   ð      ð     ð
                                           ¼      ¼            ¼  ðx 2 ; y 2 Þ   ðx 1 ; y 1 Þ¼  ðBÞ   ðAÞ
                                          A   ðx 1 ;y 1 Þ  ða;bÞ  ða;bÞ
                                       ð
                                        ð3;4Þ  2  3       2      2
                     10.14. (a) Prove that  ð6xy   y Þ dx þð6x y   3xy Þ dy is independent of the path joining ð1; 2Þ and
                                       ð1;2Þ
                           ð3; 4Þ.(b) Evaluate the integral in (a).
                                              2
                                    2
                                        3
                                                                         2
                                                     2
                           (a) P ¼ 6xy   y ; Q ¼ 6x y   3xy .  Then @P=@y ¼ 12xy   3y ¼ @Q=@x and by Problem 10.12 the line
                              integral is independent of the path.
                           (b) Method 1:  Since the line integral is independent of the path, choose any path joining ð1; 2Þ and ð3; 4Þ,
                              for example that consisting of lines from ð1; 2Þ to ð3; 2Þ [along which y ¼ 2; dy ¼ 0] and then ð3; 2Þ to
                              ð3; 4Þ [along which x ¼ 3; dx ¼ 0].  Then the required integral equals
                                             ð 3           ð 4
                                                                     2
                                                              ð54y   9y Þ dy ¼ 80 þ 156 ¼ 236
                                                ð24x   8Þ dx þ
                                              x¼1           y¼2
                                            @P  @Q                   @                  @
                                                                                                    2
                                                                               3
                                                                                             2
                                                                           2
                              Method 2:  Since    ; we must have       ¼ 6xy   y ;        ¼ 6x y   3xy :
                                            @y  ¼  @x            ð1Þ  @x            ð2Þ  @y
                                                    3
                                              2 2
                                                                         2 2
                                                                               3
                                  From (1),   ¼ 3x y   xy þ f ð yÞ.  From (2),   ¼ 3x y   xy þ gðxÞ.  The only way in which
                                                                                                    3
                                                                                              2 2
                              these two expressions for   are equal is if f ð yÞ¼ gðxÞ¼ c,aconstant.  Hence   ¼ 3x y   xy þ c.
                              Then by Problem 10.13,
                                         ð                            ð
                                          ð3;4Þ                        ð3;4Þ
                                                2  3       2     2           2 2   3
                                            ð6xy   y Þ dx þð6x y   3xy Þ dy ¼  dð3x y   xy þ cÞ
                                          ð1;2Þ                        ð1;2Þ
                                                                        2 2   3   ð3;4Þ
                                                                    ¼ 3x y   xy þ cj  ¼ 236
                                                                                  ð1;2Þ
                              Note that in this evaluation the arbitrary constant c can be omitted. See also Problem 6.16, Page 131.
                                  We could also have noted by inspection that
                                         2   3       2     2       2      2      3       2
                                      ð6xy   y Þ dx þð6x y   3xy Þ dy ¼ð6xy dx þ 6x ydyÞ ð y dx þ 3xy dyÞ
                                                                   2 2     3      2 2   3
                                                              ¼ dð3x y Þ  dðxy Þ¼ dð3x y   xy Þ
                                                       2 2
                                                             3
                              from which it is clear that   ¼ 3x y   xy þ c.
                                   þ
                                                                              x
                                                                    2
                                      2
                                                          2 x
                     10.15. Evaluate  ðx y cos x þ 2xy sin x   y e Þ dx þðx sin x   2ye Þ dy around the hypocycloid
                           x 2=3  þ y 2=3  ¼ a 2=3 :
                                                   2
                                                                          2
                                                                   2 x
                                              P ¼ x y cos x þ 2xy sin x   y e ; Q ¼ x sin x   2ye x
                                                          x
                                         2
                              Then @P=@y ¼ x cos x þ 2x sin x   2ye ¼ @Q=@x,sothat by Problem 10.11 the line integral around any
                           closed path, in particular x 2=3  þ y 2=3  ¼ a 2=3  is zero.
                     SURFACE INTEGRALS
                     10.16. If 
 is the angle between the normal line to any point ðx; y; zÞ of a surface S and the
                           positive z-axis, prove that
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