k) Dentro de r = 2 + Cos 2  y fuera de r = 2 + Sen  Solución: A = 2       2 1  6 2   ( ) ( ) 22222  SenCos +−+ d A =  6 2   (4 + 4 Cos 2  + Cos2 2  - 4 – 4 Sen  - Sen2 ) d A = 4  6 2   (Cos 2 ) d  6 2   (Cos2 2) d - 4  6 2   (Sen ) d  -  6 2   (Sen2 ) d A = ( ) 2 6 22    − Sen + 4 1 2 6 2 4 2     −       + Sen + ( ) 2 6 4    − Cos - 2 1 2 6 2 2     −       − Sen A = 8 3 4 3 32 16 3 16 316 +−++  A = 2 16 351 u       