The region inside the cardioid $r=3-3\sin \theta$ and outside

the cardioid$r=1+\sin \theta$

Take a look at the polar function

$r=3-3\sin \theta ,r=1+\sin \theta$

$r=3-3\sin \theta$and outside the cardioid $r=1+\sin \theta$

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

The interval is$\left[-\frac{\pi }{2},\frac{\pi }{6}\right]$
Formula to find the area is$A={\int }_a^{\beta }\frac{1}{2}\left(f\right(\theta ){)}^{2}d\theta$or$A={\int }_a^{\beta }\frac{1}{2}{r}^{2}d\theta$

The space within the cardioid $r=3-3\sin \theta$and outside the cardioid $r=1+\sin \theta$

$$\begin{gathered} A=2·\frac{1}{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{6}}(3-3\sin \theta {)}^2-(1+\sin \theta {)}^{2}d\theta \\ \Rightarrow A={\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}\left(9+9{\sin }^2\theta -18\sin \theta \right)-\left(1+{\sin }^2\theta +2\sin \theta \right)d\theta \\ \Rightarrow A={\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}\left(9+9{\sin }^2\theta -18\sin \theta \right)-1-{\sin }^2\theta -2\sin \theta d\theta \left[\sin ce\cos 2\theta =2{\cos }^2\theta -1\Rightarrow {\cos }^2\theta =\frac{1+\cos 2\theta }{2}\right] \\ \Rightarrow A={\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}\left(8+8{\sin }^2\theta -20\sin \theta \right)d\theta \\ A={\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}\left(8+8\frac{1-\cos 2\theta }{2}-20\sin \theta \right)d\theta \\ \Rightarrow A={\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}(8+4(1-\cos 2\theta )-20\sin \theta )d\theta \\ \Rightarrow A={\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}(12-4\cos 2\theta -20\sin \theta )d\theta \end{gathered}$$

On integration

$$\begin{gathered} A={\left(12\theta -4\frac{\sin 2\theta }{2}+20\cos \theta \right)}_{-\frac{\pi }{2}}^{\frac{\pi }{6}} \\ \Rightarrow A=(12\theta -2\sin 2\theta +20\cos \theta {)}_{-\frac{\pi }{2}}^{\frac{\pi }{6}} \end{gathered}$$

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

Hence the required area is

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{6}}(3-3\sin \theta {)}^2-(1+\sin \theta {)}^{2}d\theta =(8\pi +9\sqrt{3})$