Answer
$\pi$
Work Step by Step
From the graph .
The area is bounded by $r=2 \sin \theta$ for $\theta=0$ to $\theta=\pi$.
$$
\begin{aligned}
A & =\int_0^\pi \frac{1}{2} r^2 d \theta\\
&=\frac{1}{2} \int_0^\pi(2 \sin \theta)^2 d \theta\\
&=\frac{1}{2} \int_0^\pi 4 \sin ^2 \theta d \theta \\
& =2 \int_0^\pi \frac{1}{2}(1-\cos 2 \theta) d \theta\\
&=\left[\theta-\frac{1}{2} \sin 2 \theta\right]_0^\pi\\
&=\pi
\end{aligned}
$$
