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
                                             2      1     2 cos
                                                          =  2r  2  − r  4    d 
                                             0      4     0

                                             
                                             2
                                             
                                                    2
                                           =
                                                              4
                                              ( cos   − cos  d
                                                                ) 
                                                         4
                                               4
                                             0
                                               
                                               2
                                                    2
                                                            4
                                                               ) 
                                               = 4  (cos  − cos  d
                                               0
                                              
                                              2
                                                            4
                                                    2
                                                          = 4  (cos  − cos  ) d
                                              0
                                                                                                       
                                                                              3
                                               sin cos  1   2    sin cos    3  sin cos  1    2
                                           =  4         −      − 4           −             −      
                                                                    
                                                  2       2    0        4       4     2       2     0  
                                               1      3 
                                           =  . 8  .  −  . 4  .
                                               2  2    8  2
                                             5
                                          =
                                             4
                      
                      2 2 sin   4
                  2.        rdzdrd  

                      0   0  r 2

                      Penyelesaian:
                                         
                      2 2 sin  4         2 2 sin 
                           rdzdrd  =    ( ) dr    d 
                                                  4
                                               rz
                                                   2
                                                  r
                      0  0  r 2           0  0
                                       
                                       2 sin2  
                                                 =   ( r −4  r  3 )dr  d 
                                       0  0
                                      
                                      2      1    2 sin 
                                       =  2r 2  − r 4    d 
                                      0      4    0

                                      
                                      2
                                      
                                    =
                                                       4
                                                  4
                                                ( sin8  2  − sin  ) d
                                      0


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