Answer
$$
\begin{aligned}
\iint_{D} e^{-x^{2}-y^{2}} d A =\frac{\pi}{2}\left(1-e^{-4}\right).
\end{aligned}
$$
Work Step by Step
$$
\iint_{D} e^{-x^{2}-y^{2}} d A
$$
The region $D$ can be described as
$$
D=\left\{(x, y) | \quad x \geq 0, \quad a \leq x^{2}+y^{2} \leq 4 \right\}
$$
and in polar coordinates it is given by
$$
D=\left\{(r, \theta) | 0 \leq r \leq 2 , \quad -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \right\}
$$
Therefore,
$$
\begin{aligned}
\iint_{D} e^{-x^{2}-y^{2}} d A &=\int_{-\pi / 2}^{\pi / 2} \int_{0}^{2} e^{-r^{2}} r d r d \theta \\
&=\int_{-\pi / 2}^{\pi / 2} d \theta \int_{0}^{2} r e^{-r^{2}} d r \\
& =[\theta]_{-\pi / 2}^{\pi / 2}\left[-\frac{1}{2} e^{-r^{2}}\right]_{0}^{2} \\
&=\pi\left(-\frac{1}{2}\right)\left(e^{-4}-e^{0}\right) \\
&=\frac{\pi}{2}\left(1-e^{-4}\right).
\end{aligned}
$$
