The problem asks us to evaluate the double integral $\int_0^2 \int_1^3 x^2 y \, dy \, dx$.

1. Problem Description

The problem asks us to evaluate the double integral $\int_0^2 \int_1^3 x^2 y \, dy \, dx$.

2. Solution Steps

First, we evaluate the inner integral with respect to $y$:

$$\int_1^3 x^2 y \, dy = x^2 \int_1^3 y \, dy$$

Using the power rule for integration:

$$\int y \, dy = \frac{y^2}{2} + C$$

Thus, we have:

$$x^2 \int_1^3 y \, dy = x^2 \left[ \frac{y^2}{2} \right]_1^3 = x^2 \left( \frac{3^2}{2} - \frac{1^2}{2} \right) = x^2 \left( \frac{9}{2} - \frac{1}{2} \right) = x^2 \left( \frac{8}{2} \right) = 4x^2$$

Now, we evaluate the outer integral with respect to $x$:

$$\int_0^2 4x^2 \, dx = 4 \int_0^2 x^2 \, dx$$

Using the power rule for integration:

$$\int x^2 \, dx = \frac{x^3}{3} + C$$

Thus, we have:

$$4 \int_0^2 x^2 \, dx = 4 \left[ \frac{x^3}{3} \right]_0^2 = 4 \left( \frac{2^3}{3} - \frac{0^3}{3} \right) = 4 \left( \frac{8}{3} - 0 \right) = 4 \left( \frac{8}{3} \right) = \frac{32}{3}$$

3. Final Answer

The final answer is $\frac{32}{3}$.