Answer
$$\frac{e^9-1}{2e^9}$$
Work Step by Step
Given $$\iint_{D} e^{-y^2} d A, \quad D=\{(x, y) | 0 \leqslant x \leqslant y,0 \leqslant y \leqslant 3\}$$
From $D$, we have
\begin{align*}
\iint_{D} e^{-y^2} d A&=\int_{0}^{3}\int_{0}^{y}e^{-y^2} dxdy\\
&=\int_{0}^{3}[x]_{0}^{y}e^{-y^2} dxdy\\
&= \int_{0}^{3}ye^{-y^2} dy \\
&=\frac{-1}{2}e^{-y^2}\bigg|_{0}^{3}\\
&=\frac{1}{2}(1-e^{-9})\\
&=\frac{e^9-1}{2e^9}
\end{align*}
