Question Number 40270 by MJS last updated on 18/Jul/18
$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{one}\:“\mathrm{leaf}''\:\mathrm{of} \\ $$$${r}=\mathrm{sin}\:{n}\theta \\ $$$${n}\in\mathbb{N} \\ $$
Answered by MrW3 last updated on 19/Jul/18
![one leaf means 0≤θ≤(π/n) A=∫_0 ^(π/n) ((r^2 dθ)/2)=(1/2)∫_0 ^(π/n) sin^2 nθ dθ =(1/4)∫_0 ^(π/n) (1−cos 2nθ)dθ =(1/(8n))∫_0 ^(π/n) (1−cos 2nθ)d(2nθ) =(1/(8n))[2nθ−sin 2nθ]_0 ^(π/n) =(π/(4n))](https://www.tinkutara.com/question/Q40317.png)
$${one}\:{leaf}\:{means}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{{n}} \\ $$$${A}=\int_{\mathrm{0}} ^{\frac{\pi}{{n}}} \frac{{r}^{\mathrm{2}} {d}\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{{n}}} \mathrm{sin}^{\mathrm{2}} \:{n}\theta\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{{n}}} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{n}\theta\right){d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{n}}\int_{\mathrm{0}} ^{\frac{\pi}{{n}}} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{n}\theta\right){d}\left(\mathrm{2}{n}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{n}}\left[\mathrm{2}{n}\theta−\mathrm{sin}\:\mathrm{2}{n}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{{n}}} \\ $$$$=\frac{\pi}{\mathrm{4}{n}} \\ $$
Commented by MJS last updated on 19/Jul/18
$$\mathrm{thank}\:\mathrm{you},\:\mathrm{also}\:\mathrm{for}\:\mathrm{the}\:\mathrm{other}\:\mathrm{answers} \\ $$