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Question Number 72401 by Learner-123 last updated on 28/Oct/19

$$Findtheareaboundedbyoneleafof$$$$theroser=12\cos \left(3\theta \right).$$

Answered by ajfour last updated on 28/Oct/19

Commented by ajfour last updated on 28/Oct/19

$$A={\int }_0^{\pi /6}{a}^2\cos {}^{2}3\theta d\theta$$$$=\frac{a^2}{2}{\int }_0^{\pi /6}(1+\cos 6\theta )d\theta$$$$=\frac{a^2}{2}\left(\frac{\pi }{6}\right)$$$$fora=12$$$$A=12\pi .$$

Commented by Learner-123 last updated on 28/Oct/19

$$how,upperlimitis\theta =\frac{\pi }{6}?$$

Commented by ajfour last updated on 28/Oct/19

$$for\theta =\frac{\pi }{6},\cos 3\theta =\cos \frac{\pi }{2}=0$$$$itboundaryclosesr=0.$$$$Angularrangeofabovepetalis$$$$\theta =-\frac{\pi }{6}to\theta =\frac{\pi }{6}.Isthatallright?$$$$A={\int }_{-\pi /6}^{\pi /6}\left(\frac{r^2}{2}\right)d\theta ={\int }_0^{\pi /6}{r}^{2}d\theta .$$

Commented by Learner-123 last updated on 28/Oct/19

$$thankyousir!$$

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