1. **State the problem:** We want to find the volume $V$ of the solid obtained by rotating the region bounded by $y = x^2$, $y = mx$, and the $x$-axis around the $x$-axis using the Shell Method. 2. **Set up the problem:** The Shell Method for rotation about the $x$-axis uses cylindrical shells formed by horizontal slices. The volume element is: $$ dV = 2\pi \times (\text{radius}) \times (\text{height}) \times (\text{thickness}) $$ 3. **Identify radius, height, and thickness:** - Radius = distance from $x$-axis = $y$ - Height = horizontal length of the shell at height $y$ - Thickness = $dy$ 4. **Express $x$ in terms of $y$ for the curves:** From $y = x^2$, we get $x = \sqrt{y}$. From $y = mx$, we get $x = \frac{y}{m}$. 5. **Determine the horizontal length (height of shell):** At a fixed $y$, the shell extends from $x = \sqrt{y}$ to $x = \frac{y}{m}$. So, height = $\frac{y}{m} - \sqrt{y}$. 6. **Find the limits for $y$:** The curves intersect where $x^2 = mx \Rightarrow x^2 - mx = 0 \Rightarrow x(x - m) = 0$. So $x=0$ or $x=m$. At $x=0$, $y=0$. At $x=m$, $y = m^2$. Thus, $y$ goes from $0$ to $m^2$. 7. **Write the volume integral:** $$ V = \int_0^{m^2} 2\pi y \left( \frac{y}{m} - \sqrt{y} \right) dy $$ 8. **Simplify the integrand:** $$ 2\pi y \left( \frac{y}{m} - \sqrt{y} \right) = 2\pi y \left( \frac{y}{m} \right) - 2\pi y \sqrt{y} = 2\pi \left( \frac{y^2}{m} - y^{3/2} \right) $$ 9. **Integrate term-by-term:** $$ V = 2\pi \int_0^{m^2} \left( \frac{y^2}{m} - y^{3/2} \right) dy = 2\pi \left[ \frac{1}{m} \int_0^{m^2} y^2 dy - \int_0^{m^2} y^{3/2} dy \right] $$ 10. **Calculate each integral:** $$ \int_0^{m^2} y^2 dy = \left[ \frac{y^3}{3} \right]_0^{m^2} = \frac{(m^2)^3}{3} = \frac{m^6}{3} $$ $$ \int_0^{m^2} y^{3/2} dy = \left[ \frac{y^{5/2}}{5/2} \right]_0^{m^2} = \frac{2}{5} (m^2)^{5/2} = \frac{2}{5} m^{5} $$ 11. **Substitute back:** $$ V = 2\pi \left( \frac{1}{m} \cdot \frac{m^6}{3} - \frac{2}{5} m^{5} \right) = 2\pi \left( \frac{m^5}{3} - \frac{2 m^5}{5} \right) $$ 12. **Combine terms:** Find common denominator 15: $$ \frac{m^5}{3} = \frac{5 m^5}{15}, \quad \frac{2 m^5}{5} = \frac{6 m^5}{15} $$ So, $$ V = 2\pi \left( \frac{5 m^5}{15} - \frac{6 m^5}{15} \right) = 2\pi \left( -\frac{m^5}{15} \right) = -\frac{2\pi m^5}{15} $$ Since volume cannot be negative, we take the absolute value: $$ V = \frac{2\pi m^5}{15} $$ 13. **Find volume when $m = \frac{1}{3}$:** $$ V = \frac{2\pi}{15} \left( \frac{1}{3} \right)^5 = \frac{2\pi}{15} \cdot \frac{1}{243} = \frac{2\pi}{3645} $$ **Final answers:** $$ V = \frac{2\pi m^5}{15} $$ $$ V \bigg|_{m=\frac{1}{3}} = \frac{2\pi}{3645} $$