Page 58 - Kalkulus Lanjut

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


2

=
2
4
) 
4
( cos  − cos  d
4
0

2
4
2
) 
= 4  (cos  − cos  d
0

2
2
4
) 
= 4  (cos  − cos  d
0
 
 sin cos 1  2  sin cos  3  sin cos 1   2
3
= 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  =   ( ) drrz 4 r 2 d 

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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