1. **State the problem:** Find the volume of the solid generated by revolving the region bounded by the curves $y=2x-1$, $y=\sqrt{x}$, and $x=0$ about the y-axis using the shell method. 2. **Recall the shell method formula:** The volume $V$ when revolving around the y-axis is given by: $$V = 2\pi \int_a^b (\text{radius})(\text{height}) \, dx$$ where: - radius = distance from the y-axis = $x$ - height = difference between the top and bottom functions in terms of $x$ 3. **Identify the bounds and functions:** - The region is bounded by $y=2x-1$ and $y=\sqrt{x}$. - Find intersection points to determine limits $a$ and $b$: Set $2x-1 = \sqrt{x}$. 4. **Solve for intersection:** $$2x - 1 = \sqrt{x}$$ Square both sides: $$ (2x - 1)^2 = x $$ $$4x^2 - 4x + 1 = x$$ $$4x^2 - 5x + 1 = 0$$ Use quadratic formula: $$x = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8}$$ So, $$x=1$$ or $$x=\frac{1}{4}$$ 5. **Determine which function is on top between $x=\frac{1}{4}$ and $x=1$:** At $x=0.5$, $2(0.5)-1=0$, $\sqrt{0.5} \approx 0.707$, so $y=\sqrt{x}$ is on top. At $x=0.9$, $2(0.9)-1=0.8$, $\sqrt{0.9} \approx 0.95$, so $\sqrt{x}$ is still on top. 6. **Set up the integral:** Height = top - bottom = $\sqrt{x} - (2x -1) = \sqrt{x} - 2x + 1$ Bounds: $a=\frac{1}{4}$, $b=1$ 7. **Write the volume integral:** $$V = 2\pi \int_{1/4}^1 x \left(\sqrt{x} - 2x + 1\right) dx$$ 8. **Simplify the integrand:** $$x \sqrt{x} = x^{3/2}$$ So integrand: $$x^{3/2} - 2x^2 + x$$ 9. **Integrate term-by-term:** $$\int x^{3/2} dx = \frac{2}{5} x^{5/2}$$ $$\int x^2 dx = \frac{x^3}{3}$$ $$\int x dx = \frac{x^2}{2}$$ 10. **Evaluate the definite integral:** $$\int_{1/4}^1 \left(x^{3/2} - 2x^2 + x\right) dx = \left[ \frac{2}{5} x^{5/2} - 2 \cdot \frac{x^3}{3} + \frac{x^2}{2} \right]_{1/4}^1$$ Calculate at $x=1$: $$\frac{2}{5} (1)^{5/2} - \frac{2}{3} (1)^3 + \frac{1}{2} = \frac{2}{5} - \frac{2}{3} + \frac{1}{2}$$ Common denominator 30: $$\frac{12}{30} - \frac{20}{30} + \frac{15}{30} = \frac{7}{30}$$ Calculate at $x=\frac{1}{4}$: $$\frac{2}{5} \left(\frac{1}{4}\right)^{5/2} - \frac{2}{3} \left(\frac{1}{4}\right)^3 + \frac{1}{2} \left(\frac{1}{4}\right)^2$$ Calculate powers: $$\left(\frac{1}{4}\right)^{5/2} = \left(\frac{1}{4}\right)^2 \cdot \left(\frac{1}{4}\right)^{1/2} = \frac{1}{16} \cdot \frac{1}{2} = \frac{1}{32}$$ $$\left(\frac{1}{4}\right)^3 = \frac{1}{64}$$ $$\left(\frac{1}{4}\right)^2 = \frac{1}{16}$$ Substitute: $$\frac{2}{5} \cdot \frac{1}{32} - \frac{2}{3} \cdot \frac{1}{64} + \frac{1}{2} \cdot \frac{1}{16} = \frac{2}{160} - \frac{2}{192} + \frac{1}{32} = \frac{1}{80} - \frac{1}{96} + \frac{1}{32}$$ Common denominator 480: $$\frac{6}{480} - \frac{5}{480} + \frac{15}{480} = \frac{16}{480} = \frac{1}{30}$$ 11. **Subtract:** $$\frac{7}{30} - \frac{1}{30} = \frac{6}{30} = \frac{1}{5}$$ 12. **Multiply by $2\pi$ to get volume:** $$V = 2\pi \cdot \frac{1}{5} = \frac{2\pi}{5}$$ **Final answer:** $$\boxed{\frac{2\pi}{5}}$$