Answer
$$\int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}} \frac{2}{\sqrt{4-y^{2}}} d x d y=4$$
Work Step by Step
Given$$\int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}} \frac{2}{\sqrt{4-y^{2}}} d x d y$$
So, we have
\begin{align}
\int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}} \frac{2}{\sqrt{4-y^{2}}} d x d y&=\int_{0}^{2}\left[\frac{2 x}{\sqrt{4-y^{2}}}\right]_{0}^{\sqrt{4-y^{2}}} d y\\
&=\int_{0}^{2} 2 d y\\
&=[2 y]_{0}^{2}\\
&=4
\end{align}
