1. **Problem statement:** Find the volume under the surface $$z = 6 - x - y$$ above the triangular region bounded by $$x=0$$, $$y=0$$, and $$x + y = 2$$. 2. **Understanding the region:** The triangular region $$D$$ is defined by $$x \geq 0$$, $$y \geq 0$$, and $$x + y \leq 2$$. This means $$y$$ ranges from $$0$$ to $$2 - x$$ for each $$x$$ in $$[0,2]$$. 3. **Volume formula:** The volume under a surface $$z = f(x,y)$$ over a region $$D$$ is given by the double integral: $$ V = \iint_D f(x,y) \, dA $$ where $$dA$$ is the area element. 4. **Set up the integral:** Here, $$ V = \int_0^2 \int_0^{2-x} (6 - x - y) \, dy \, dx $$ 5. **Integrate with respect to $$y$$:** $$ \int_0^{2-x} (6 - x - y) \, dy = \left[(6 - x)y - \frac{y^2}{2}\right]_0^{2-x} = (6 - x)(2 - x) - \frac{(2 - x)^2}{2} $$ 6. **Simplify the expression:** $$ (6 - x)(2 - x) = 12 - 6x - 2x + x^2 = 12 - 8x + x^2 $$ $$ \frac{(2 - x)^2}{2} = \frac{4 - 4x + x^2}{2} = 2 - 2x + \frac{x^2}{2} $$ So, $$ (6 - x)(2 - x) - \frac{(2 - x)^2}{2} = (12 - 8x + x^2) - (2 - 2x + \frac{x^2}{2}) = 10 - 6x + \frac{x^2}{2} $$ 7. **Integrate with respect to $$x$$:** $$ V = \int_0^2 \left(10 - 6x + \frac{x^2}{2}\right) dx = \left[10x - 3x^2 + \frac{x^3}{6}\right]_0^2 $$ 8. **Evaluate the definite integral:** $$ 10(2) - 3(2)^2 + \frac{(2)^3}{6} = 20 - 12 + \frac{8}{6} = 8 + \frac{4}{3} = \frac{24}{3} + \frac{4}{3} = \frac{28}{3} $$ 9. **Final answer:** The volume under the surface above the triangular region is $$ \boxed{\frac{28}{3}} $$