Answer
$$\frac{3\pi }{2}$$
Work Step by Step
From the graph .
The area is bounded by $r=1-\sin \theta$ for $\theta=0$ to $\theta=2\pi$.
$$
\begin{aligned}
A & =\int_0^\pi \frac{1}{2} r^2 d \theta\\
&=\frac{1}{2} \int_0^{1\pi}(1- \sin \theta)^2 d \theta\\
&=\frac{1}{2} \int_0^{2\pi} ( 1-2\sin \theta- \sin ^2 \theta )d \theta \\
&=\frac{1}{2} \int_0^{2\pi} ( \frac{3}{2}-2\sin \theta+ \cos 2 \theta )d \theta \\
&=\frac{1}{2}\left[\frac{3}{2}\theta+2\cos \theta+\frac{1}{2} \sin 2 \theta\right]_0^{2\pi}\\
&=\frac{3\pi }{2}
\end{aligned}
$$
