Answer
$$
S=\{(x, y) |-2 \lt x \leq 0, \quad 0 \leq y \leq 1 / \sqrt{x+2}\}
$$
The shaded area is the region of interest.
$$
\begin{aligned} \text { Area } &=\int_{-2}^{0} \frac{1}{\sqrt{x+2}} d x \\
&=2 \sqrt{2}. \end{aligned}
$$
Work Step by Step
$$
S=\{(x, y) |-2 \lt x \leq 0, \quad 0 \leq y \leq 1 / \sqrt{x+2}\}
$$
The shaded area is the region of interest.
$$
\begin{aligned} \text { Area } &=\int_{-2}^{0} \frac{1}{\sqrt{x+2}} d x \\
&=\lim _{t \rightarrow-2^{+}} \int_{t}^{0} \frac{1}{\sqrt{x+2}} d x \\
&=\lim _{t \rightarrow-2^{+}}[2 \sqrt{x+2}]_{t}^{0}
\\
&= \lim _{t \rightarrow-2^{+}}(2 \sqrt{2}-2 \sqrt{t+2})\\
&=2 \sqrt{2}. \end{aligned}
$$
