Answer
The proof is below.
Work Step by Step
In the given figure, we note that $0 \leq \theta \leq \phi$ and $0 \leq r \leq 2 a \sin \theta .$ We have the integration limits. With this:
\[
\begin{aligned}
A &\left.=\int_{0}^{\phi} \int_{0}^{2 a \sin \theta} r d r d \theta=\int_{0}^{\phi} \frac{r^{2}}{2}\right]_{0}^{2 a \sin \theta} d \theta \\
&=\int_{0}^{\phi} \frac{1}{2} 4 a^{2} \sin ^{2} \theta d \theta=2 a^{2} \int_{0}^{\phi} \frac{-\cos (2 \theta)+1}{2} d \theta \\
&\left.\left.=a^{2}[\theta]_{0}^{\phi}-\frac{1}{2} \sin (2 \theta)\right]_{0}^{\phi}\right]=a^{2} \phi-\frac{1}{2} a^{2} \sin (2 \phi)
\end{aligned}
\]
$A=a^{2} \phi-\frac{1}{2} a^{2} \sin (2 \phi) .$ QED
