1. **State the problem:** We want to find the double integral $$\iint_D y \, dA$$ where $D$ is the region enclosed by the $x$-axis, the line $y=\frac{1}{3}x$, and the parabola $y=\sqrt{18-x}$.\n\n2. **Understand the region $D$:** The $x$-axis is $y=0$. The line is $y=\frac{1}{3}x$. The parabola is $y=\sqrt{18-x}$. We need to find the intersection points to determine the limits.\n\n3. **Find intersections:**\n- Between $y=0$ and $y=\frac{1}{3}x$: at $y=0$, $\frac{1}{3}x=0 \Rightarrow x=0$.\n- Between $y=0$ and $y=\sqrt{18-x}$: at $y=0$, $\sqrt{18-x}=0 \Rightarrow 18-x=0 \Rightarrow x=18$.\n- Between $y=\frac{1}{3}x$ and $y=\sqrt{18-x}$: set equal:\n$$\frac{1}{3}x = \sqrt{18-x}$$\nSquare both sides:\n$$\left(\frac{1}{3}x\right)^2 = 18 - x$$\n$$\frac{x^2}{9} = 18 - x$$\nMultiply both sides by 9:\n$$x^2 = 162 - 9x$$\nBring all terms to one side:\n$$x^2 + 9x - 162 = 0$$\nSolve quadratic:\n$$x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-162)}}{2} = \frac{-9 \pm \sqrt{81 + 648}}{2} = \frac{-9 \pm \sqrt{729}}{2} = \frac{-9 \pm 27}{2}$$\nTwo solutions:\n- $x = \frac{-9 + 27}{2} = 9$\n- $x = \frac{-9 - 27}{2} = -18$ (discard since $x \geq 0$ in region)\nSo intersection at $x=9$.\n\n4. **Set up the integral:** The region $D$ is bounded between $x=0$ and $x=9$ by the line and parabola, and between $x=9$ and $x=18$ by the parabola and $x$-axis.\n\nWe split the integral into two parts:\n$$\iint_D y \, dA = \int_0^9 \int_{y=\frac{1}{3}x}^{y=\sqrt{18-x}} y \, dy \, dx + \int_9^{18} \int_0^{\sqrt{18-x}} y \, dy \, dx$$\n\n5. **Evaluate inner integrals:**\n- For $0 \leq x \leq 9$:\n$$\int_{\frac{1}{3}x}^{\sqrt{18-x}} y \, dy = \left[ \frac{y^2}{2} \right]_{\frac{1}{3}x}^{\sqrt{18-x}} = \frac{(\sqrt{18-x})^2}{2} - \frac{(\frac{1}{3}x)^2}{2} = \frac{18 - x}{2} - \frac{x^2}{18}$$\n- For $9 \leq x \leq 18$:\n$$\int_0^{\sqrt{18-x}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\sqrt{18-x}} = \frac{18 - x}{2}$$\n\n6. **Write the full integral:**\n$$\int_0^9 \left( \frac{18 - x}{2} - \frac{x^2}{18} \right) dx + \int_9^{18} \frac{18 - x}{2} dx$$\n\n7. **Calculate each integral:**\n- First integral:\n$$\int_0^9 \left( \frac{18 - x}{2} - \frac{x^2}{18} \right) dx = \int_0^9 \frac{18 - x}{2} dx - \int_0^9 \frac{x^2}{18} dx$$\nCalculate separately:\n$$\int_0^9 \frac{18 - x}{2} dx = \frac{1}{2} \int_0^9 (18 - x) dx = \frac{1}{2} \left[18x - \frac{x^2}{2} \right]_0^9 = \frac{1}{2} (18 \cdot 9 - \frac{81}{2}) = \frac{1}{2} (162 - 40.5) = \frac{121.5}{2} = 60.75$$\n$$\int_0^9 \frac{x^2}{18} dx = \frac{1}{18} \int_0^9 x^2 dx = \frac{1}{18} \left[ \frac{x^3}{3} \right]_0^9 = \frac{1}{18} \cdot \frac{729}{3} = \frac{1}{18} \cdot 243 = 13.5$$\nSo first integral is:\n$$60.75 - 13.5 = 47.25$$\n- Second integral:\n$$\int_9^{18} \frac{18 - x}{2} dx = \frac{1}{2} \int_9^{18} (18 - x) dx = \frac{1}{2} \left[18x - \frac{x^2}{2} \right]_9^{18}$$\nCalculate:\nAt $x=18$:\n$$18 \cdot 18 - \frac{18^2}{2} = 324 - \frac{324}{2} = 324 - 162 = 162$$\nAt $x=9$:\n$$18 \cdot 9 - \frac{9^2}{2} = 162 - \frac{81}{2} = 162 - 40.5 = 121.5$$\nDifference:\n$$162 - 121.5 = 40.5$$\nMultiply by $\frac{1}{2}$:\n$$\frac{40.5}{2} = 20.25$$\n\n8. **Add both parts:**\n$$47.25 + 20.25 = 67.5$$\n\n**Final answer:**\n$$\boxed{67.5}$$