1. **Problem statement:** We have a region bounded by the curves $y=4x - x^2$ and $y=x$, rotated about the vertical 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 limits: $$4x - x^2 = x \implies 4x - x^2 - x = 0 \implies -x^2 + 3x = 0 \implies x(-x + 3) = 0$$ So, $x=0$ or $x=3$. 3. **(a) Area of the horizontal cross-section at widest point:** The widest horizontal cross-section corresponds to the maximum vertical distance between the curves $y=4x - x^2$ and $y=x$. The vertical distance is: $$D(x) = (4x - x^2) - x = 3x - x^2$$ Find maximum of $D(x)$ on $[0,3]$: $$\frac{dD}{dx} = 3 - 2x = 0 \implies x = \frac{3}{2} = 1.5$$ Evaluate $D(1.5)$: $$D(1.5) = 3(1.5) - (1.5)^2 = 4.5 - 2.25 = 2.25$$ So the widest vertical cross-section height is $2.25$ meters. 4. **(b) Volume of water (solid of revolution about $x=4$):** Use the shell method since rotation is about vertical line $x=4$: Shell radius: $r = 4 - x$ Shell height: $h = (4x - x^2) - x = 3x - x^2$ 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 volume: $$V = 2\pi \times 11.25 = 22.5\pi \approx 70.69$$ 5. **(c) Lateral surface area of the upper curve rotated about $x=4$ between $x=0$ and $x=3$:** Formula for surface area when rotating curve $y=f(x)$ about vertical line $x=a$: $$S = 2\pi \int_a^b r(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ Here, $r(x) = |4 - x| = 4 - x$ (since $x$ in $[0,3]$), and $f(x) = 4x - x^2$. Calculate derivative: $$\frac{dy}{dx} = 4 - 2x$$ Calculate integrand: $$\sqrt{1 + (4 - 2x)^2} = \sqrt{1 + (4 - 2x)^2}$$ So surface area: $$S = 2\pi \int_0^3 (4 - x) \sqrt{1 + (4 - 2x)^2} \, dx$$ This integral does not simplify easily; it can be evaluated numerically if needed. **Final answers:** (a) Widest horizontal cross-section area height: $2.25$ meters. (b) Volume of basin: $22.5\pi \approx 70.69$ cubic meters. (c) Lateral surface area: $$S = 2\pi \int_0^3 (4 - x) \sqrt{1 + (4 - 2x)^2} \, dx$$