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