Answer:
Hence the area of the region bounded by all leaves of the rose is [tex]\pi[/tex].
Step-by-step explanation:
Here,
[tex]2 cos (3\Theta )=0\\\Rightarrow cos(3\Theta)=0\\\Rightarrow cos(3\Theta )=cos\left (\frac{\pi }{2} \right )\\\Rightarrow \Theta =\left (\frac{\pi }{6},-\frac{\pi }{6} \right )[/tex]
Now consider for one leaf of rose:-
[tex]Area = 3\int_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{r^{2}}{2}d\Theta \\\\ = \frac{3}{2}\int_{-\frac{\pi }{6}}^{\frac{\pi }{6}}4 cos^{2}\left ( 3\Theta \right ) d\Theta \\\\ = \frac{12}{2}\int_{-\frac{\pi }{6}}^{\frac{\pi }{6}} \frac{cos(6\Theta +1)}{2}d\Theta\\\\ =\frac{6}{2}\left [ \frac{sin(6\Theta +\Theta )}{6} \right ]_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\\\\ =3\times \left [ \frac{\pi }{6}+\frac{\pi }{6} \right ]\\\\ =3\times \frac{2\pi }{6}\\\\ = \pi[/tex]
