Volume Rotation 340D98

1. **State the problem:** We need to find the volume of the solid formed by rotating the region enclosed by the curves $y = \sqrt{x}$ and $y = 2x - 1$ around the y-axis. 2. **Find the points of intersection:** Set $\sqrt{x} = 2x - 1$ to find the limits of integration. $$\sqrt{x} = 2x - 1$$ Square both sides: $$x = (2x - 1)^2 = 4x^2 - 4x + 1$$ Rearranged: $$4x^2 - 5x + 1 = 0$$ Use quadratic formula: $$x = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8}$$ So, $$x_1 = \frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{4}$$ $$x_2 = \frac{5 + 3}{8} = \frac{8}{8} = 1$$ 3. **Express $x$ as a function of $y$ for each curve:** - For $y = \sqrt{x}$, $x = y^2$ - For $y = 2x - 1$, solve for $x$: $$y = 2x - 1 \implies x = \frac{y + 1}{2}$$ 4. **Find the $y$-values corresponding to the intersection points:** - At $x = \frac{1}{4}$, $y = \sqrt{\frac{1}{4}} = \frac{1}{2}$ - At $x = 1$, $y = \sqrt{1} = 1$ So the region is bounded between $y = \frac{1}{2}$ and $y = 1$. 5. **Set up the volume integral using the shell method (around y-axis):** The volume $V$ is given by: $$V = 2\pi \int_{y=\frac{1}{2}}^{1} (\text{radius})(\text{height}) \, dy$$ - Radius = distance from y-axis = $x$ - Height = difference between $x$ values of the two curves at given $y$: $$\text{height} = x_{right} - x_{left} = \frac{y + 1}{2} - y^2$$ 6. **Write the integral:** $$V = 2\pi \int_{\frac{1}{2}}^{1} y \left( \frac{y + 1}{2} - y^2 \right) dy$$ Note: Radius is $y$ because shells are vertical slices at height $y$ with radius $y$ (distance from y-axis). Actually, since we rotate around y-axis, the shell radius is $x$, but here we integrate with respect to $y$, so the shell radius is $x$. Correction: Using shell method with respect to $x$ is simpler here. 7. **Use shell method with respect to $x$:** - Height = difference in $y$ values: $$h = \sqrt{x} - (2x - 1)$$ - Radius = $x$ - Limits: $x$ from $\frac{1}{4}$ to $1$ Volume: $$V = 2\pi \int_{\frac{1}{4}}^{1} x \left( \sqrt{x} - (2x - 1) \right) dx$$ 8. **Simplify the integrand:** $$x \sqrt{x} = x^{3/2}$$ $$x (2x - 1) = 2x^2 - x$$ So integrand: $$x^{3/2} - (2x^2 - x) = x^{3/2} - 2x^2 + x$$ 9. **Write the integral:** $$V = 2\pi \int_{\frac{1}{4}}^{1} \left( x^{3/2} - 2x^2 + x \right) dx$$ 10. **Integrate term by term:** $$\int x^{3/2} dx = \frac{2}{5} x^{5/2}$$ $$\int x^2 dx = \frac{x^3}{3}$$ $$\int x dx = \frac{x^2}{2}$$ 11. **Evaluate the integral:** $$V = 2\pi \left[ \frac{2}{5} x^{5/2} - 2 \cdot \frac{x^3}{3} + \frac{x^2}{2} \right]_{\frac{1}{4}}^{1}$$ 12. **Calculate at $x=1$:** $$\frac{2}{5} (1)^{5/2} - \frac{2}{3} (1)^3 + \frac{1}{2} = \frac{2}{5} - \frac{2}{3} + \frac{1}{2}$$ 13. **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: $$\left( \frac{1}{4} \right)^{5/2} = \left( \frac{1}{4} \right)^2 \cdot \left( \frac{1}{4} \right)^{1/2} = \frac{1}{16} \cdot \frac{1}{2} = \frac{1}{32}$$ $$\left( \frac{1}{4} \right)^3 = \frac{1}{64}$$ $$\left( \frac{1}{4} \right)^2 = \frac{1}{16}$$ So, $$\frac{2}{5} \cdot \frac{1}{32} - \frac{2}{3} \cdot \frac{1}{64} + \frac{1}{2} \cdot \frac{1}{16} = \frac{2}{160} - \frac{2}{192} + \frac{1}{32} = \frac{1}{80} - \frac{1}{96} + \frac{1}{32}$$ Find common denominator 480: $$\frac{1}{80} = \frac{6}{480}, \quad \frac{1}{96} = \frac{5}{480}, \quad \frac{1}{32} = \frac{15}{480}$$ Sum: $$6/480 - 5/480 + 15/480 = (6 - 5 + 15)/480 = 16/480 = \frac{1}{30}$$ 14. **Subtract:** $$\left( \frac{2}{5} - \frac{2}{3} + \frac{1}{2} \right) - \frac{1}{30}$$ Find common denominator 30: $$\frac{2}{5} = \frac{12}{30}, \quad \frac{2}{3} = \frac{20}{30}, \quad \frac{1}{2} = \frac{15}{30}$$ Sum inside parentheses: $$12/30 - 20/30 + 15/30 = (12 - 20 + 15)/30 = 7/30$$ Subtract $1/30$: $$7/30 - 1/30 = 6/30 = 1/5$$ 15. **Multiply by $2\pi$:** $$V = 2\pi \times \frac{1}{5} = \frac{2\pi}{5}$$ **Final answer:** $$\boxed{\frac{2\pi}{5}}$$