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Question Number 40270 by MJS last updated on 18/Jul/18

$$\mathrm{calculate}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{one}‘‘{\mathrm{leaf}}^{″}\mathrm{of}$$$$r=\sin n\theta$$$$n\in \mathbb{N}$$

Answered by MrW3 last updated on 19/Jul/18

$$oneleafmeans0⩽\theta ⩽\frac{\pi }{n}$$$$A={\int }_0^{\frac{\pi }{n}}\frac{r^{2}d\theta }{2}=\frac{1}{2}{\int }_0^{\frac{\pi }{n}}{\sin }^{2}n\theta d\theta$$$$=\frac{1}{4}{\int }_0^{\frac{\pi }{n}}(1-\cos 2n\theta )d\theta$$$$=\frac{1}{8n}{\int }_0^{\frac{\pi }{n}}(1-\cos 2n\theta )d\left(2n\theta \right)$$$$=\frac{1}{8n}{[2n\theta -\sin 2n\theta ]}_0^{\frac{\pi }{n}}$$$$=\frac{\pi }{4n}$$

Commented by MJS last updated on 19/Jul/18

$$\mathrm{thank}\mathrm{you},\mathrm{also}\mathrm{for}\mathrm{the}\mathrm{other}\mathrm{answers}$$

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