Question Number 79290 by 21042009 last updated on 24/Jan/20
$${What}\:{is}\:{the}\:{area}\:{of}\:{one}\:{petal}\:{of} \\ $$$${r}=\mathrm{2cos}\left(\mathrm{3}\theta\right) \\ $$
Answered by mr W last updated on 24/Jan/20
![A=∫_(−(π/6)) ^(π/6) ((r^2 dθ)/2) =∫_(−(π/6)) ^(π/6) 2 cos^2 (3θ)dθ =∫_(−(π/6)) ^(π/6) (1+cos 6θ)dθ =[θ]_(−(π/6)) ^(π/6) =(π/3)](https://www.tinkutara.com/question/Q79341.png)
$${A}=\int_{−\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} \frac{{r}^{\mathrm{2}} {d}\theta}{\mathrm{2}} \\ $$$$=\int_{−\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{3}\theta\right){d}\theta \\ $$$$=\int_{−\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{6}\theta\right){d}\theta \\ $$$$=\left[\theta\right]_{−\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} \\ $$$$=\frac{\pi}{\mathrm{3}} \\ $$