1. **Problem statement:** Evaluate the double integral $$\iint_D y^2 \, dA$$ where $$D = \{(x,y) \mid -1 \leq y \leq 1, -y - 2 \leq x \leq y\}$$. 2. **Set up the integral:** The region $$D$$ is described by $$y$$ between $$-1$$ and $$1$$, and for each fixed $$y$$, $$x$$ ranges from $$-y - 2$$ to $$y$$. 3. **Integral expression:** $$ \iint_D y^2 \, dA = \int_{y=-1}^1 \int_{x=-y-2}^y y^2 \, dx \, dy $$ 4. **Integrate with respect to $$x$$ first:** Since $$y^2$$ is constant with respect to $$x$$, $$ \int_{x=-y-2}^y y^2 \, dx = y^2 \cdot \left(y - (-y - 2)\right) = y^2 (y + y + 2) = y^2 (2y + 2) = 2y^3 + 2y^2 $$ 5. **Rewrite the integral:** $$ \int_{-1}^1 (2y^3 + 2y^2) \, dy = 2 \int_{-1}^1 y^3 \, dy + 2 \int_{-1}^1 y^2 \, dy $$ 6. **Evaluate each integral:** - $$\int_{-1}^1 y^3 \, dy = 0$$ because $$y^3$$ is an odd function over symmetric limits. - $$\int_{-1}^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-1}^1 = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}$$ 7. **Calculate the total:** $$ 2 \cdot 0 + 2 \cdot \frac{2}{3} = \frac{4}{3} $$ **Final answer:** $$\boxed{\frac{4}{3}}$$