If curve $\mathrm{r}=\mathrm{a} \sin 3 \theta$
To trace the curve, we consider the following table :
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline 3 \theta= & 0 & \frac{\pi}{2} & \pi & \frac{3 \pi}{2} & 2 \pi & \frac{5 \pi}{2} & 3 \pi \\
\hline\theta= & 0 & \frac{\pi}{6} & \frac{\pi}{3} & \frac{\pi}{2} & \frac{2 \pi}{3} & \frac{5 \pi}{6} & \pi \\
\hline \mathrm{r}= & 0 & \mathrm{a} & 0 & -\mathrm{a} & 0 & \mathrm{a} & 0 \\
\hline
\end{array}
Thus there is a loop between $\theta=0 \& \theta=\frac{\pi}{3}$ as r varies from $\mathrm{r}=0$ to $\mathrm{r}=0$.


Hence, the area of the loop lying in the
$$
\text { positive quadrant }=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} r^{2} \mathrm{~d} \theta
$$
$$
=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \sin ^{2} \phi \cdot \frac{1}{3} \mathrm{~d} \phi
$$
$\left[\right.$ On putting, $\left.3 \theta=\phi \Rightarrow \mathrm{d} \theta=\frac{1}{3} \mathrm{~d} \phi\right]$
$$
\begin{array}{c}
=\frac{a^{2}}{6} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \mathrm{d} \phi \\
=\frac{\mathrm{a}^{2}}{6} \cdot \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \phi}{2} \mathrm{~d} \phi\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]
\end{array}
$$
$=\frac{\mathrm{a}^{2}}{12} \cdot\left[\phi+\frac{\sin 2 \phi}{2}\right]_{0}^{\frac{\pi}{2}}$
$=\frac{\mathrm{a}^{2}}{12} \cdot\left[\frac{\pi}{2}+\sin \pi\right]=\frac{\mathrm{a}^{2} \pi}{24}$