1. **Problem Statement:** Find the volume of the solid generated by rotating the region bounded by the curves $y = x^2$ and $y = x + 2$ about the $y$-axis through 360°. 2. **Identify the curves and limits:** Given curves: $$y = x^2$$ $$y = x + 2$$ Find intersection points by setting $x^2 = x + 2$: $$x^2 - x - 2 = 0$$ Factor: $$(x - 2)(x + 1) = 0$$ So, $x = 2$ and $x = -1$. Corresponding $y$ values: $$y_1 = (2)^2 = 4$$ $$y_2 = (-1)^2 = 1$$ So the limits for $y$ are from 1 to 4. 3. **Express $x$ as functions of $y$ for rotation about $y$-axis:** From $y = x^2$, we get: $$x = \sqrt{y}$$ From $y = x + 2$, solve for $x$: $$x = y - 2$$ 4. **Volume formula for rotation about $y$-axis:** $$V = \pi \int_{c}^{d} \left[ R(y)^2 - r(y)^2 \right] dy$$ where $R(y)$ is the outer radius and $r(y)$ is the inner radius. 5. **Determine outer and inner radius:** Since $x = \sqrt{y}$ is to the right of $x = y - 2$ in the interval $y \in [1,4]$, outer radius is $\sqrt{y}$ and inner radius is $y - 2$. 6. **Set up the integral:** $$V = \pi \int_1^4 \left( (\sqrt{y})^2 - (y - 2)^2 \right) dy = \pi \int_1^4 \left( y - (y^2 - 4y + 4) \right) dy$$ 7. **Simplify the integrand:** $$y - y^2 + 4y - 4 = -y^2 + 5y - 4$$ 8. **Integrate:** $$V = \pi \int_1^4 (-y^2 + 5y - 4) dy = \pi \left[ -\frac{y^3}{3} + \frac{5y^2}{2} - 4y \right]_1^4$$ 9. **Evaluate the definite integral:** Calculate at upper limit $y=4$: $$-\frac{4^3}{3} + \frac{5 \cdot 4^2}{2} - 4 \cdot 4 = -\frac{64}{3} + \frac{80}{2} - 16 = -\frac{64}{3} + 40 - 16 = -\frac{64}{3} + 24$$ Calculate at lower limit $y=1$: $$-\frac{1^3}{3} + \frac{5 \cdot 1^2}{2} - 4 \cdot 1 = -\frac{1}{3} + \frac{5}{2} - 4 = -\frac{1}{3} + 2.5 - 4 = -\frac{1}{3} - 1.5 = -\frac{1}{3} - \frac{3}{2} = -\frac{11}{6}$$ Subtract lower from upper: $$\left(-\frac{64}{3} + 24\right) - \left(-\frac{11}{6}\right) = -\frac{64}{3} + 24 + \frac{11}{6}$$ Find common denominator 6: $$-\frac{128}{6} + \frac{144}{6} + \frac{11}{6} = \frac{27}{6} = \frac{9}{2}$$ 10. **Final volume:** $$V = \pi \times \frac{9}{2} = \frac{9\pi}{2}$$ **Answer:** The volume of the solid is $\boxed{\frac{9\pi}{2}}$ cubic units.