Question Number 113246 by mohammad17 last updated on 11/Sep/20
Commented by mohammad17 last updated on 11/Sep/20
$${please}\:{sir}\:{help}\:{me} \\ $$
Answered by Aziztisffola last updated on 11/Sep/20
![The area that incluses first leaf of the curve lies in between θ=−(π/6) and θ=(π/6) so area A=3∫_(−(π/6)) ^( (π/6)) (1/2) r^2 dθ (curve has 3 simillar leaf) =(3/2)∫_(−(π/6)) ^( (π/6)) (2cos(3θ))^2 dθ =(3/2)∫_(−(π/6)) ^( (π/6)) 4cos^2 (3θ)dθ =6∫_(−(π/6)) ^( (π/6)) ((cos6θ+1)/2)dθ (cos2θ=2cos^2 θ−1 =3[((sin6θ)/6)+θ]_((−π)/6) ^(π/6) =π](Q113256.png)
$$\mathrm{The}\:\mathrm{area}\:\mathrm{that}\:\mathrm{incluses}\: \\ $$$$\:\mathrm{first}\:\mathrm{leaf}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{between} \\ $$$$\:\theta=−\frac{\pi}{\mathrm{6}}\:\mathrm{and}\:\theta=\frac{\pi}{\mathrm{6}} \\ $$$$\:\mathrm{so}\:\mathrm{area}\:\mathrm{A}=\mathrm{3}\int_{−\frac{\pi}{\mathrm{6}}} ^{\:\frac{\pi}{\mathrm{6}}} \frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{r}^{\mathrm{2}} \mathrm{d}\theta\:\:\left(\mathrm{curve}\:\mathrm{has}\:\mathrm{3}\:\mathrm{simillar}\:\mathrm{leaf}\right) \\ $$$$\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{−\frac{\pi}{\mathrm{6}}} ^{\:\frac{\pi}{\mathrm{6}}} \left(\mathrm{2cos}\left(\mathrm{3}\theta\right)\right)^{\mathrm{2}} \mathrm{d}\theta \\ $$$$\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{−\frac{\pi}{\mathrm{6}}} ^{\:\frac{\pi}{\mathrm{6}}} \mathrm{4cos}^{\mathrm{2}} \left(\mathrm{3}\theta\right)\mathrm{d}\theta \\ $$$$\:=\mathrm{6}\int_{−\frac{\pi}{\mathrm{6}}} ^{\:\frac{\pi}{\mathrm{6}}} \frac{\mathrm{cos6}\theta+\mathrm{1}}{\mathrm{2}}\mathrm{d}\theta\:\:\:\left(\mathrm{cos2}\theta=\mathrm{2cos}^{\mathrm{2}} \theta−\mathrm{1}\right. \\ $$$$\:=\mathrm{3}\left[\frac{\mathrm{sin6}\theta}{\mathrm{6}}+\theta\right]_{\frac{−\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} =\pi \\ $$