6.- Hallar el área A de la región limitada por la curva: (a  0) a) 3 2 3 2 3 2 ayx =+ Sug: x = a Cos3 t , y = a Sen 3 t ; t   0, 2  Solución: x = 0 → t = AAt 4 2 =  x = a → t = 0 A = 3a2  2 0  (Cos2 t Sen 4 t) dt A = 3a2  2 0  dt tCostCos A = 8 3 2a  2 0  (1 – Cos 2t + Cos2 2t + Cos 2t – 2 Cos2 2t + Cos3 2t) dt A = 8 3 2a  2 0  (1 – Cos 2t - Cos 2 2t + Cos3 2t) dt A = 8 3 2a 2/ 0 3 6 2 2 2 2 4 2 4 1 2 2          −+      +−− tSentSentSen t tSen t A = 8 3 2a 2/ 02        t A = 32 3 2 a At = 2 2 8 3 u a        b) 5 2 5 2 5 2 ayx =+ Solución: Sug: x = a Cos3 t , y = a Sen3 t ; t   0, 2  x = 0 → t = AAt 4 2 =  x = a → t = 0 A = 5a2  2 0  (Cos4 t Sen6 t) dt A = 5a2  2 0  dt tCostCos 23 2 21 2 21       +       − A = 32 5 2a  2 0  (1 + 3 Cos 2t – 3 Cos2 2t + Cos3 2t) (1 + 2 Cos 2t + Cos2 2t)dt A = 32 5 2a  2 0  (1 + 5 cos 2t + 4 Cos2 2t – 2Cos3 2t – Cos4 2t + Cos5 2t) dt A = 32 5 2a 2/ 010 2 5 3 2 3 2 2 64 8 88 4 43 2 3 2 2 4 2 2 25                +−++−−−−−+++ tSentSentSentSenttSenttSen tSen tSen t A = 32 5 2a 2/ 0 5 10 2 8 43 22 8 21          +++ tSentSen tSen t A = 512 105 2a 2 2 128 105 u a A       =  c) 1 45 3 22 =      +      yx Sug : x = 5 Cos t, y = 4 Sen 3 t, t   0, 2  Solucion: A = -4  2 0  (4 Sen3 t) (- 5 Sen t) dt A = 80  2 0  dt tCos 2 2 21       − A = 20  2 0  (1 – 2 Cos 2t + Cos2 t) dt A = 20 2/ 08 4 2        ++ tSent t A = 20 2/ 02 3        t A = 15  u2

a) 3
b) 5
c) 1