1. **State the problem:** Find the volume of the region bounded by the cylinder $x^2 + y^2 = 2y$ and the planes $z + y = 8$ and $z = 2y$. 2. **Rewrite the cylinder equation:** The cylinder equation is $x^2 + y^2 = 2y$. Complete the square for $y$: $$x^2 + y^2 - 2y = 0 \implies x^2 + (y^2 - 2y + 1) = 1 \implies x^2 + (y - 1)^2 = 1$$ This represents a cylinder with circular cross-section of radius 1 centered at $(0,1)$ in the $xy$-plane. 3. **Identify the planes:** - Plane 1: $z + y = 8 \implies z = 8 - y$ - Plane 2: $z = 2y$ 4. **Find the height of the region:** The height between the planes at a given $y$ is: $$h = (8 - y) - (2y) = 8 - 3y$$ 5. **Set up the volume integral:** The volume is the integral over the circular region in the $xy$-plane of the height: $$V = \iint_{D} (8 - 3y) \, dA$$ where $D$ is the disk $x^2 + (y - 1)^2 \leq 1$. 6. **Use polar coordinates centered at $(0,1)$:** Let $x = r \cos \theta$, $y = 1 + r \sin \theta$, with $r \in [0,1]$, $\theta \in [0, 2\pi]$. 7. **Express the height in polar coordinates:** $$h = 8 - 3y = 8 - 3(1 + r \sin \theta) = 8 - 3 - 3r \sin \theta = 5 - 3r \sin \theta$$ 8. **The area element:** $$dA = r \, dr \, d\theta$$ 9. **Set up the integral:** $$V = \int_0^{2\pi} \int_0^1 (5 - 3r \sin \theta) r \, dr \, d\theta = \int_0^{2\pi} \int_0^1 (5r - 3r^2 \sin \theta) \, dr \, d\theta$$ 10. **Integrate with respect to $r$:** $$\int_0^1 (5r - 3r^2 \sin \theta) dr = \left[ \frac{5r^2}{2} - r^3 \sin \theta \right]_0^1 = \frac{5}{2} - \sin \theta$$ 11. **Integrate with respect to $\theta$:** $$V = \int_0^{2\pi} \left( \frac{5}{2} - \sin \theta \right) d\theta = \int_0^{2\pi} \frac{5}{2} d\theta - \int_0^{2\pi} \sin \theta d\theta = \frac{5}{2} \times 2\pi - 0 = 5\pi$$ 12. **Final answer:** $$\boxed{5\pi}$$ The volume of the region bounded by the given cylinder and planes is $5\pi$.