Answer
The area of one loop of the rose is equal to $\frac{\pi}{12}$
Work Step by Step
One loop of the rose $r=\cos 3\theta $, it is given by:
$$
D=\left\{(r, \theta) | 0 \leq r \leq cos 3\theta , \quad -\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6} \right\}
$$
Therefore, the area of one loop of the rose is given by
$$
\begin{aligned}
\iint_{D} d A &=\int_{-x / 6}^{\pi / 6} \int_{0}^{\cos 3 \theta} r d r d \theta \\
&=\int_{-\pi / 6}^{\pi / 6}\left[\frac{1}{2} r^{2}\right]_{r=0}^{r=\cos 3 \theta} d \theta \\
&=\int_{-\pi / 6}^{\pi / 6} \frac{1}{2} \cos ^{2} 3 \theta d \theta \\
&=2 \int_{0}^{\pi / 6} \frac{1}{2}\left(\frac{1+\cos 6 \theta}{2}\right) d \theta \\
&=\frac{1}{2}\left[\theta+\frac{1}{6} \sin 6 \theta\right]_{0}^{\pi / 6} \\
&=\frac{\pi}{12}.
\end{aligned}
$$
So, the area of one loop of the rose is equal to $\frac{\pi}{12}$
