\( \left. \begin{array} { l } { \int \int _ { D } x ^ { 2 } y d A = ? } \\ { D = \{ ( x , y ) | - 1 \leq x \leq 1,0 \leq y \leq 2 \} } \end{array} \right. \)

To solve the double integral \( \int \int_{D} x^2 y \, dA \) over the region \( D = \{ (x, y) | -1 \leq x \leq 1, 0 \leq y \leq 2 \} \), we first set up the bounds of integration. The area can be expressed as: \[ \int_{-1}^{1} \int_{0}^{2} x^2 y \, dy \, dx. \] Now, let's first integrate with respect to \( y \): \[ \int_{0}^{2} x^2 y \, dy = x^2 \left[ \frac{y^2}{2} \right]_{0}^{2} = x^2 \cdot \frac{2^2}{2} = 2x^2. \] Now, we substitute this back into the integral with respect to \( x \): \[ \int_{-1}^{1} 2x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_{-1}^{1} = 2 \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = 2 \left( \frac{1}{3} - \frac{-1}{3} \right) = 2 \left( \frac{2}{3} \right) = \frac{4}{3}. \] So, the value of the double integral is \( \frac{4}{3} \).