Answer
$$A=\frac{8}{3}$$
Work Step by Step
Given
$$\sqrt{x}+ \sqrt y=2, \quad x=0, \quad y=0$$
First, we get the intersections points of the curves
at $x=0 \Rightarrow \sqrt y=2 \Rightarrow y=4$
at $y=0 \Rightarrow \sqrt x=2 \Rightarrow x=4$
Secondly by taking vertical slices, consider the following integration
\begin{align}
A&=\int_{x=0}^{x=4} \int_{y=0}^{y=(2-\sqrt{x})^{2}} d y d x\\
&=\int_{x=0}^{x=4}[y]_{0}^{(2-\sqrt{x})^{2}} d x\\
&=\int_{x=0}^{x=4}(4-4 \sqrt{x}+x) d x\\
&=\left[4 x-\frac{8}{3} x \sqrt{x}+\frac{x^{2}}{2}\right]_{0}^{4}\\
&=\frac{8}{3}
\end{align}
