1. **State the problem:** Find the area of the surface $S$ defined by $f(x,y) = 1 + x - 2y$ over the region $R$, which is a square with vertices $(0,0)$, $(3,0)$, $(0,3)$, and $(3,3)$. 2. **Formula for surface area:** The surface area $A$ of $z = f(x,y)$ over a region $R$ is given by: $$A = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dA$$ 3. **Calculate partial derivatives:** $$\frac{\partial f}{\partial x} = 1$$ $$\frac{\partial f}{\partial y} = -2$$ 4. **Substitute into formula:** $$\sqrt{1 + (1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$ 5. **Determine the area of region $R$:** $R$ is a square with side length 3, so $$\text{Area}(R) = 3 \times 3 = 9$$ 6. **Calculate surface area:** $$A = \sqrt{6} \times 9 = 9\sqrt{6}$$ **Final answer:** $$\boxed{9\sqrt{6}}$$