91Ó°ÊÓ

Take a look at the polar function

$r=3-3\sin \mathrm{θ},r=1+\sin \mathrm{θ}$

$r=3-3\sin \mathrm{θ}$and outside the cardioid $r=1+\sin \mathrm{θ}$

The region's equivalent boundaries are $-\frac{\mathrm{Ï€}}{2}$to $\frac{\mathrm{Ï€}}{6}$

The interval is$\left[-\frac{\mathrm{Ï€}}{2},\frac{\mathrm{Ï€}}{6}\right]$
Formula to find the area is$A={∫}_a^{\mathrm{β}}\frac{1}{2}\left(f\right(\mathrm{θ}){)}^{2}d\mathrm{θ}$or$A={∫}_a^{\mathrm{β}}\frac{1}{2}{r}^{2}d\mathrm{θ}$

The space within the cardioid $r=3-3\sin \mathrm{θ}$and outside the cardioid $r=1+\sin \mathrm{θ}$

$$\begin{gathered} A=2·\frac{1}{2}{∫}_{-\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}}(3-3\sin \mathrm{θ}{)}^2-(1+\sin \mathrm{θ}{)}^{2}d\mathrm{θ} \\ ⇒A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}}\left(9+9{\sin }^2\mathrm{θ}-18\sin \mathrm{θ}\right)-\left(1+{\sin }^2\mathrm{θ}+2\sin \mathrm{θ}\right)d\mathrm{θ} \\ ⇒A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}}\left(9+9{\sin }^2\mathrm{θ}-18\sin \mathrm{θ}\right)-1-{\sin }^2\mathrm{θ}-2\sin \mathrm{θ}d\mathrm{θ}\left[\sin ce\cos 2\mathrm{θ}=2{\cos }^2\mathrm{θ}-1⇒{\cos }^2\mathrm{θ}=\frac{1+\cos 2\mathrm{θ}}{2}\right] \\ ⇒A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}}\left(8+8{\sin }^2\mathrm{θ}-20\sin \mathrm{θ}\right)d\mathrm{θ} \\ A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}}\left(8+8\frac{1-\cos 2\mathrm{θ}}{2}-20\sin \mathrm{θ}\right)d\mathrm{θ} \\ ⇒A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}}(8+4(1-\cos 2\mathrm{θ})-20\sin \mathrm{θ})d\mathrm{θ} \\ ⇒A={∫}_{\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}}(12-4\cos 2\mathrm{θ}-20\sin \mathrm{θ})d\mathrm{θ} \end{gathered}$$

On integration

$$\begin{gathered} A={\left(12\mathrm{θ}-4\frac{\sin 2\mathrm{θ}}{2}+20\cos \mathrm{θ}\right)}_{-\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}} \\ ⇒A=(12\mathrm{θ}-2\sin 2\mathrm{θ}+20\cos \mathrm{θ}{)}_{-\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}} \end{gathered}$$

By applying the limits
$$\begin{gathered} A=\left(12·\frac{\mathrm{π}}{6}-2\sin 2·\frac{\mathrm{π}}{6}+20\cos \frac{\mathrm{π}}{6}-12·-\frac{\mathrm{π}}{2}-2\sin 2·\frac{\mathrm{π}}{2}-20\cos \frac{\mathrm{π}}{2}\right) \\ ⇒A=\left(2\mathrm{π}-2·\frac{\sqrt{3}}{2}+20·\frac{\sqrt{3}}{2}+6\mathrm{π}-2\sin 2·\frac{\mathrm{π}}{2}-20\cos \frac{\mathrm{π}}{2}\right) \\ ⇒A=(8\mathrm{π}-\sqrt{3}+10\sqrt{3}) \\ ⇒A=(8\mathrm{π}+9\sqrt{3}) \end{gathered}$$

Hence the required area is

${∫}_{\frac{-\mathrm{π}}{2}}^{\frac{\mathrm{π}}{6}}(3-3\sin \mathrm{θ}{)}^2-(1+\sin \mathrm{θ}{)}^{2}d\mathrm{θ}=(8\mathrm{π}+9\sqrt{3})$