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

Commented by mohammad17 last updated on 11/Sep/20

$$pleasesirhelpme$$

Answered by Aziztisffola last updated on 11/Sep/20

$$\mathrm{The}\mathrm{area}\mathrm{that}\mathrm{incluses}$$$$\mathrm{first}\mathrm{leaf}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{lies}\mathrm{in}\mathrm{between}$$$$\theta =-\frac{\pi }{6}\mathrm{and}\theta =\frac{\pi }{6}$$$$\mathrm{so}\mathrm{area}\mathrm{A}=3{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{1}{2}{\mathrm{r}}^2\mathrm{d}\theta \left(\mathrm{curve}\mathrm{has}3\mathrm{simillar}\mathrm{leaf}\right)$$$$=\frac{3}{2}{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{\left(2\cos \left(3\theta \right)\right)}^2\mathrm{d}\theta$$$$=\frac{3}{2}{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}{4\cos }^2\left(3\theta \right)\mathrm{d}\theta$$$$=6{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{\cos 6\theta +1}{2}\mathrm{d}\theta (\cos 2\theta ={2\cos }^2\theta -1$$$$=3{[\frac{\sin 6\theta }{6}+\theta ]}_{\frac{-\pi }{6}}^{\frac{\pi }{6}}=\pi$$

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