1. **Problem Statement:** Determine the volume of the solid obtained by rotating the region bounded by the curves $y = x^2 - 2x$ and $y = x$ about the line $y = 4$. 2. **Identify the region and axis of rotation:** The region is bounded by $y = x^2 - 2x$ and $y = x$. We rotate this region about the horizontal line $y = 4$. 3. **Find points of intersection:** Set $x^2 - 2x = x$ to find the limits of integration. $$x^2 - 2x = x \implies x^2 - 3x = 0 \implies x(x - 3) = 0$$ So, $x = 0$ and $x = 3$. 4. **Set up the volume integral using the washer method:** The volume $V$ is given by $$V = \pi \int_0^3 \left(R(x)^2 - r(x)^2\right) dx$$ where $R(x)$ is the outer radius and $r(x)$ is the inner radius from the axis of rotation $y=4$. 5. **Determine radii:** - Outer radius $R(x)$ is the distance from $y=4$ to the lower curve $y = x^2 - 2x$: $$R(x) = 4 - (x^2 - 2x) = 4 - x^2 + 2x$$ - Inner radius $r(x)$ is the distance from $y=4$ to the upper curve $y = x$: $$r(x) = 4 - x$$ 6. **Write the integral explicitly:** $$V = \pi \int_0^3 \left[(4 - x^2 + 2x)^2 - (4 - x)^2\right] dx$$ 7. **Expand the squares:** $$(4 - x^2 + 2x)^2 = ( - x^2 + 2x + 4)^2 = x^4 - 4x^3 - 4x^2 + 16x + 16$$ $$(4 - x)^2 = 16 - 8x + x^2$$ 8. **Substitute and simplify the integrand:** $$\left[(4 - x^2 + 2x)^2 - (4 - x)^2\right] = (x^4 - 4x^3 - 4x^2 + 16x + 16) - (16 - 8x + x^2)$$ $$= x^4 - 4x^3 - 4x^2 + 16x + 16 - 16 + 8x - x^2 = x^4 - 4x^3 - 5x^2 + 24x$$ 9. **Integral becomes:** $$V = \pi \int_0^3 (x^4 - 4x^3 - 5x^2 + 24x) dx$$ 10. **Integrate term-by-term:** $$\int_0^3 x^4 dx = \left[\frac{x^5}{5}\right]_0^3 = \frac{3^5}{5} = \frac{243}{5}$$ $$\int_0^3 -4x^3 dx = -4 \left[\frac{x^4}{4}\right]_0^3 = -4 \cdot \frac{81}{4} = -81$$ $$\int_0^3 -5x^2 dx = -5 \left[\frac{x^3}{3}\right]_0^3 = -5 \cdot 9 = -45$$ $$\int_0^3 24x dx = 24 \left[\frac{x^2}{2}\right]_0^3 = 24 \cdot \frac{9}{2} = 108$$ 11. **Sum the results:** $$V = \pi \left( \frac{243}{5} - 81 - 45 + 108 \right) = \pi \left( \frac{243}{5} - 18 \right)$$ Convert $-18$ to fifths: $$-18 = -\frac{90}{5}$$ So, $$V = \pi \left( \frac{243}{5} - \frac{90}{5} \right) = \pi \cdot \frac{153}{5} = \frac{153\pi}{5}$$ **Final answer:** $$\boxed{\frac{153\pi}{5}}$$