1. **Problem:** Find the volume under the surface $z = 3x^3 + 3x^2 y$ over the rectangle $R = \{(x,y): 1 \leq x \leq 3, 0 \leq y \leq 2\}$. 2. **Formula:** The volume under a surface $z = f(x,y)$ over a region $R$ is given by the double integral $$V = \iint_R f(x,y) \, dA.$$ For a rectangular region, this becomes $$V = \int_{x=a}^{b} \int_{y=c}^{d} f(x,y) \, dy \, dx.$$ 3. **Set up the integral:** Here, $$V = \int_1^3 \int_0^2 (3x^3 + 3x^2 y) \, dy \, dx.$$ 4. **Integrate with respect to $y$: $$\int_0^2 (3x^3 + 3x^2 y) \, dy = \int_0^2 3x^3 \, dy + \int_0^2 3x^2 y \, dy = 3x^3 y \Big|_0^2 + 3x^2 \frac{y^2}{2} \Big|_0^2 = 3x^3 (2) + 3x^2 \frac{4}{2} = 6x^3 + 6x^2.$$ 5. **Now integrate with respect to $x$: $$\int_1^3 (6x^3 + 6x^2) \, dx = 6 \int_1^3 x^3 \, dx + 6 \int_1^3 x^2 \, dx = 6 \left( \frac{x^4}{4} \Big|_1^3 \right) + 6 \left( \frac{x^3}{3} \Big|_1^3 \right) = 6 \left( \frac{81 - 1}{4} \right) + 6 \left( \frac{27 - 1}{3} \right) = 6 \times 20 + 6 \times 8.6667 = 120 + 52 = 172.$$ 6. **Answer:** The volume under the surface over the given rectangle is $$\boxed{172}.$$