Question-113246

Question Number 113246 by mohammad17 last updated on 11/Sep/20

Commented by mohammad17 last updated on 11/Sep/20

p l e a s e s i r h e l p m e

Answered by Aziztisffola last updated on 11/Sep/20

The area that incluses

first leaf of the curve lies in between

θ = - \frac{π}{6} and θ = \frac{π}{6}

so area A = 3 \int_{- \frac{π}{6}}^{\frac{π}{6}} \frac{1}{2} r^{2} d θ (curve has 3 simillar leaf)

= \frac{3}{2} \int_{- \frac{π}{6}}^{\frac{π}{6}} {(2 \cos (3 θ))}^{2} d θ

= \frac{3}{2} \int_{- \frac{π}{6}}^{\frac{π}{6}} {4 \cos}^{2} (3 θ) d θ

= 6 \int_{- \frac{π}{6}}^{\frac{π}{6}} \frac{\cos 6 θ + 1}{2} d θ (\cos 2 θ = {2 \cos}^{2} θ - 1

= 3 {[\frac{\sin 6 θ}{6} + θ]}_{\frac{- π}{6}}^{\frac{π}{6}} = π