1. **State the problem:** We need to find the area of the region bounded by the x-axis, y-axis, and the curve $$\sqrt{x} + \sqrt{y} = 3$$. 2. **Rewrite the curve equation:** Solve for $$y$$ in terms of $$x$$: $$\sqrt{y} = 3 - \sqrt{x} \implies y = (3 - \sqrt{x})^2 = 9 - 6\sqrt{x} + x$$. 3. **Set the limits of integration:** The curve intersects the axes at $$x=0$$ and $$y=0$$. - When $$x=0$$, $$\sqrt{y} = 3 \Rightarrow y=9$$. - When $$y=0$$, $$\sqrt{x} = 3 \Rightarrow x=9$$. 4. **Area calculation:** The area under the curve from $$x=0$$ to $$x=9$$ above the x-axis is: $$\text{Area} = \int_0^9 y \, dx = \int_0^9 (9 - 6\sqrt{x} + x) \, dx$$. 5. **Evaluate the integral:** $$\int_0^9 9 \, dx = 9x \Big|_0^9 = 81$$ $$\int_0^9 6\sqrt{x} \, dx = 6 \int_0^9 x^{1/2} \, dx = 6 \cdot \frac{2}{3} x^{3/2} \Big|_0^9 = 4 \cdot 9^{3/2} = 4 \cdot 27 = 108$$ $$\int_0^9 x \, dx = \frac{x^2}{2} \Big|_0^9 = \frac{81}{2} = 40.5$$ 6. **Combine the results:** $$\text{Area} = 81 - 108 + 40.5 = 13.5$$ 7. **Express area as $$\frac{k}{2}$$:** $$13.5 = \frac{27}{2}$$ so $$k = 27$$. **Final answer:** $$k = 27$$.