1. **State the problem:** We need to find the volume of the solid formed when the region enclosed by the curves $y = x^2 + 1$ and $y = 2x + 1$ is rotated about the y-axis. 2. **Find the points of intersection:** Set $x^2 + 1 = 2x + 1$ to find the limits of integration. $$x^2 + 1 = 2x + 1 \implies x^2 - 2x = 0 \implies x(x - 2) = 0$$ So, $x = 0$ and $x = 2$. 3. **Set up the volume integral using the shell method:** When rotating around the y-axis, the shell radius is $r = x$ and the height is the difference between the functions: $$h = (2x + 1) - (x^2 + 1) = 2x + 1 - x^2 - 1 = 2x - x^2$$ The volume is: $$V = 2\pi \int_0^2 (\text{radius})(\text{height}) \, dx = 2\pi \int_0^2 x(2x - x^2) \, dx$$ 4. **Simplify the integrand:** $$x(2x - x^2) = 2x^2 - x^3$$ 5. **Integrate:** $$V = 2\pi \int_0^2 (2x^2 - x^3) \, dx = 2\pi \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^2$$ 6. **Evaluate the definite integral:** $$\left[ \frac{2(2)^3}{3} - \frac{(2)^4}{4} \right] - \left[0\right] = \left[ \frac{2 \times 8}{3} - \frac{16}{4} \right] = \left[ \frac{16}{3} - 4 \right] = \frac{16}{3} - \frac{12}{3} = \frac{4}{3}$$ 7. **Calculate the volume:** $$V = 2\pi \times \frac{4}{3} = \frac{8\pi}{3}$$ **Final answer:** The volume of the solid is $\boxed{\frac{8\pi}{3}}$.