Answer
Total area of the flower does not depend on $n$:
$area = \frac{\pi }{2}$
Work Step by Step
In Exercise 9 of Section 12.4, we see that the "rose" traces out a complete curve for the interval $0 \le \theta \le 2\pi $.
Using Eq. (2) of Theorem 1 of Section 12.4, the total area is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{2\pi } {\sin ^2}\left( {n\theta } \right){\rm{d}}\theta $
Since ${\sin ^2}\left( {n\theta } \right) = \frac{1}{2}\left( {1 - \cos \left( {2n\theta } \right)} \right)$, the integral becomes
$area = \frac{1}{4}\cdot\mathop \smallint \limits_0^{2\pi } \left( {1 - \cos \left( {2n\theta } \right)} \right){\rm{d}}\theta $
$area = \frac{1}{4}\left( {\theta - \frac{1}{{2n}}\sin \left( {2n\theta } \right)} \right)|_0^{2\pi }$
$area = \frac{\pi }{2}$
Hence, the total area of the flower does not depend on $n$.
