1. **Problem statement:** Calculate the double integral of the function $f(x,y) = 12xy^2 - 8x^3$ over the region $R = \{(x,y) : 1 \leq x \leq 2, -1 \leq y \leq 2\}$. 2. **Formula and explanation:** The double integral over a rectangular region is given by $$\iint_R f(x,y) \, dA = \int_{x=a}^{b} \int_{y=c}^{d} f(x,y) \, dy \, dx$$ where $a=1$, $b=2$, $c=-1$, and $d=2$. 3. **Set up the integral:** $$\int_{x=1}^{2} \int_{y=-1}^{2} (12xy^2 - 8x^3) \, dy \, dx$$ 4. **Integrate with respect to $y$ first:** $$\int_{y=-1}^{2} (12xy^2 - 8x^3) \, dy = \int_{y=-1}^{2} 12xy^2 \, dy - \int_{y=-1}^{2} 8x^3 \, dy$$ 5. **Calculate each integral:** - For $12xy^2$: $$12x \int_{-1}^{2} y^2 \, dy = 12x \left[ \frac{y^3}{3} \right]_{-1}^{2} = 12x \left( \frac{2^3}{3} - \frac{(-1)^3}{3} \right) = 12x \left( \frac{8}{3} + \frac{1}{3} \right) = 12x \cdot 3 = 36x$$ - For $8x^3$: $$8x^3 \int_{-1}^{2} dy = 8x^3 [y]_{-1}^{2} = 8x^3 (2 - (-1)) = 8x^3 \cdot 3 = 24x^3$$ 6. **Combine results:** $$\int_{y=-1}^{2} (12xy^2 - 8x^3) \, dy = 36x - 24x^3$$ 7. **Now integrate with respect to $x$:** $$\int_{1}^{2} (36x - 24x^3) \, dx = \int_{1}^{2} 36x \, dx - \int_{1}^{2} 24x^3 \, dx$$ 8. **Calculate each integral:** - For $36x$: $$36 \int_{1}^{2} x \, dx = 36 \left[ \frac{x^2}{2} \right]_{1}^{2} = 36 \left( \frac{4}{2} - \frac{1}{2} \right) = 36 \cdot \frac{3}{2} = 54$$ - For $24x^3$: $$24 \int_{1}^{2} x^3 \, dx = 24 \left[ \frac{x^4}{4} \right]_{1}^{2} = 24 \left( \frac{16}{4} - \frac{1}{4} \right) = 24 \cdot \frac{15}{4} = 90$$ 9. **Combine results:** $$54 - 90 = -36$$ **Final answer for part (a):** $$\boxed{-36}$$