1. **Problem statement:** We have the region bounded by the curves $y=4x - x^2$ and $y=x$, rotated about the line $x=4$. We need to find: (a) The area of the basin's horizontal cross-section at its widest point. (b) The volume of water needed to fill the basin. (c) The lateral surface area generated by rotating the upper curve $y=4x - x^2$ about $x=4$ between the intersection points. 2. **Find the intersection points:** Set $4x - x^2 = x$ to find the bounds: $$4x - x^2 = x \implies 4x - x^2 - x = 0 \implies -x^2 + 3x = 0 \implies x(-x + 3) = 0$$ So, $x=0$ or $x=3$. 3. **(a) Area of the horizontal cross-section at widest point:** The widest horizontal cross-section is at the maximum $y$ value in the region between $x=0$ and $x=3$. Find the maximum $y$ of the region bounded by $y=x$ and $y=4x - x^2$. The upper curve is $y=4x - x^2$ and the lower is $y=x$. Find the maximum $y$ on the upper curve between $0$ and $3$: $$\frac{dy}{dx} = 4 - 2x = 0 \implies x=2$$ At $x=2$, $y=4(2) - 2^2 = 8 - 4 = 4$. At $x=2$, the lower curve $y=x=2$. The horizontal cross-section is the horizontal distance between the two curves at $y=4$. We solve for $x$ in each curve at $y=4$: - For $y=4x - x^2$, set $4 = 4x - x^2 \implies x^2 - 4x + 4=0 \implies (x-2)^2=0 \implies x=2$. - For $y=x$, set $4=x \implies x=4$. The horizontal cross-section length at $y=4$ is $4 - 2 = 2$ meters. 4. **(b) Volume of water to fill the basin:** Use the shell method for volume about $x=4$. Shell radius: $r = 4 - x$. Shell height: difference between curves $h = (4x - x^2) - x = 3x - x^2$. Volume: $$V = 2\pi \int_0^3 r h \, dx = 2\pi \int_0^3 (4 - x)(3x - x^2) \, dx$$ Expand integrand: $$(4 - x)(3x - x^2) = 12x - 4x^2 - 3x^2 + x^3 = 12x - 7x^2 + x^3$$ Integrate: $$\int_0^3 (12x - 7x^2 + x^3) dx = \left[6x^2 - \frac{7}{3}x^3 + \frac{1}{4}x^4\right]_0^3$$ Calculate at $x=3$: $$6(9) - \frac{7}{3}(27) + \frac{1}{4}(81) = 54 - 63 + 20.25 = 11.25$$ So, $$V = 2\pi \times 11.25 = 22.5\pi$$ 5. **(c) Lateral surface area generated by rotating $y=4x - x^2$ about $x=4$ between $x=0$ and $x=3$:** Formula for lateral surface area when rotating about vertical line $x=4$: $$S = 2\pi \int_0^3 r \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ Where radius $r = 4 - x$ and $$\frac{dy}{dx} = 4 - 2x$$ Calculate the integrand: $$\sqrt{1 + (4 - 2x)^2} = \sqrt{1 + (4 - 2x)^2} = \sqrt{1 + 16 - 16x + 4x^2} = \sqrt{17 - 16x + 4x^2}$$ So, $$S = 2\pi \int_0^3 (4 - x) \sqrt{17 - 16x + 4x^2} \, dx$$ This integral does not simplify easily and is best evaluated numerically. **Final answers:** (a) The widest horizontal cross-section length is $2$ meters. (b) The volume of water needed is $$22.5\pi \approx 70.69$$ cubic meters. (c) The lateral surface area is $$S = 2\pi \int_0^3 (4 - x) \sqrt{17 - 16x + 4x^2} \, dx$$ which can be approximated numerically if needed.