Answer
$$\frac{4}{3}$$
Work Step by Step
Given $$\iint_{D}y\sqrt{x^2-y^2} d A, \quad D=\{(x, y) | 0 \leqslant x \leqslant 2,0 \leqslant y \leqslant x\}$$
From $D$, we have
\begin{align*}
\iint_{D} e^{-y^2} d A&=\int_{0}^{2}\int_{0}^{x}y\sqrt{x^2-y^2} dydx\\
&=\frac{-1}{3}\int_{0}^{2} (x^2-y^2)^{3/2}\bigg|_{0}^{x} dx\\
&=\frac{1}{3}\int_{0}^{2} x^{3} dx\\
&=\frac{1}{12}x^4\bigg|_{0}^{2}\\
&= \frac{4}{3}
\end{align*}
