1. **State the problem:** Find the points of intersection of the line $y=\frac{1}{2}x$ and the parabola $y=2x-\frac{1}{6}x^2$. 2. **Set the equations equal to find intersection points:** $$\frac{1}{2}x = 2x - \frac{1}{6}x^2$$ 3. **Rearrange the equation:** $$0 = 2x - \frac{1}{6}x^2 - \frac{1}{2}x = \left(2x - \frac{1}{2}x\right) - \frac{1}{6}x^2 = \frac{3}{2}x - \frac{1}{6}x^2$$ 4. **Multiply through by 6 to clear denominators:** $$0 = 6 \times \left(\frac{3}{2}x - \frac{1}{6}x^2\right) = 9x - x^2$$ 5. **Rewrite as:** $$x^2 - 9x = 0$$ 6. **Factor:** $$x(x - 9) = 0$$ 7. **Solve for $x$:** $$x=0 \quad \text{or} \quad x=9$$ 8. **Find corresponding $y$ values using $y=\frac{1}{2}x$:** - For $x=0$, $y=\frac{1}{2} \times 0 = 0$ - For $x=9$, $y=\frac{1}{2} \times 9 = \frac{9}{2} = 4.5$ 9. **Points of intersection:** $$(0,0) \quad \text{and} \quad (9,4.5)$$ 10. **Representative strip for area calculation:** The vertical strip at position $x$ has height equal to the difference between the upper curve and lower curve: $$h(x) = \left(2x - \frac{1}{6}x^2\right) - \frac{1}{2}x = \frac{3}{2}x - \frac{1}{6}x^2$$ 11. **Calculate the area bounded by the curves:** $$A = \int_0^9 \left(2x - \frac{1}{6}x^2 - \frac{1}{2}x\right) dx = \int_0^9 \left(\frac{3}{2}x - \frac{1}{6}x^2\right) dx$$ 12. **Integrate term-by-term:** $$\int_0^9 \frac{3}{2}x \, dx = \frac{3}{2} \times \frac{x^2}{2} \Big|_0^9 = \frac{3}{2} \times \frac{81}{2} = \frac{3}{2} \times 40.5 = 60.75$$ $$\int_0^9 \frac{1}{6}x^2 \, dx = \frac{1}{6} \times \frac{x^3}{3} \Big|_0^9 = \frac{1}{6} \times \frac{729}{3} = \frac{1}{6} \times 243 = 40.5$$ 13. **Calculate area:** $$A = 60.75 - 40.5 = 20.25$$ 14. **Calculate the second moment of area about the y-axis:** The second moment of area $I_y$ is given by: $$I_y = \int_0^9 x^2 \times h(x) \, dx = \int_0^9 x^2 \left(\frac{3}{2}x - \frac{1}{6}x^2\right) dx = \int_0^9 \left(\frac{3}{2}x^3 - \frac{1}{6}x^4\right) dx$$ 15. **Integrate term-by-term:** $$\int_0^9 \frac{3}{2}x^3 \, dx = \frac{3}{2} \times \frac{x^4}{4} \Big|_0^9 = \frac{3}{2} \times \frac{6561}{4} = \frac{3}{2} \times 1640.25 = 2460.375$$ $$\int_0^9 \frac{1}{6}x^4 \, dx = \frac{1}{6} \times \frac{x^5}{5} \Big|_0^9 = \frac{1}{6} \times \frac{59049}{5} = \frac{1}{6} \times 11809.8 = 1968.3$$ 16. **Calculate $I_y$:** $$I_y = 2460.375 - 1968.3 = 492.075$$ **Final answers:** - Points of intersection: $(0,0)$ and $(9,4.5)$ - Area bounded by the curves: $20.25$ - Second moment of area about the y-axis: $492.075$