Answer
The area that encloses:
$$
r=1- \sin \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi
$$
is equal to:
$$
A =\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta =\frac{3 \pi}{2}
$$
Work Step by Step
The area that encloses:
$$ r=1- \sin \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta \\
&=\frac{1}{2} \int_{0}^{2\pi}(1- \sin \theta)^{2} d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left(1-2 \sin \theta+\sin ^{2} \theta\right) d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left[1-2 \sin \theta+\frac{1}{2}(1-\cos 2 \theta)\right] d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left(\frac{3}{2}-2 \sin \theta-\frac{1}{2} \cos 2 \theta\right) d \theta\\
&=\frac{1}{2}\left[\frac{3}{2} \theta+2 \cos \theta-\frac{1}{4} \sin 2 \theta\right]_{0}^{2 \pi}\\
&=\frac{1}{2}[(3 \pi+2)-(2)]\\
&=\frac{3 \pi}{2}
\end{aligned}
$$
