1. **Problem statement:** Find the area of the basin's horizontal cross-section at its widest point, the volume of water needed to fill the basin, and the lateral surface area generated by rotating the curve $y=4x - x^2$ about the line $x=4$ between the intersection points of $y=4x - x^2$ and $y=x$. 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 widest horizontal cross-section:** 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: $$D(x) = (4x - x^2) - x = 3x - x^2$$ To find the maximum, differentiate: $$D'(x) = 3 - 2x$$ Set $D'(x)=0$: $$3 - 2x = 0 \implies x = \frac{3}{2} = 1.5$$ Calculate $D(1.5)$: $$D(1.5) = 3(1.5) - (1.5)^2 = 4.5 - 2.25 = 2.25$$ The horizontal cross-section at the widest point is the region between the curves rotated about $x=4$, so the radius of the cross-section is the horizontal distance from $x=4$ to the curves. At $x=1.5$, the horizontal distances from $x=4$ to the curves are: - For $y=4x - x^2$: radius $R_1 = 4 - 1.5 = 2.5$ - For $y=x$: radius $R_2 = 4 - 1.5 = 2.5$ But the cross-section is horizontal, so the width is the vertical distance $D(1.5) = 2.25$ meters. 4. **(b) Volume of water to fill the basin:** Use the washer method rotating the region bounded by $y=x$ and $y=4x - x^2$ about $x=4$. The outer radius is the distance from $x=4$ to the left curve (the smaller $x$ value at each $y$), and the inner radius is the distance from $x=4$ to the right curve. Express $x$ in terms of $y$ for both curves: - From $y=x$, $x=y$ - From $y=4x - x^2$, rewrite as $x^2 - 4x + y = 0$ Solve quadratic for $x$: $$x = \frac{4 \pm \sqrt{16 - 4y}}{2} = 2 \pm \sqrt{4 - y}$$ The two branches correspond to the parabola's left and right sides. The region is between $x=y$ and $x=2 - \sqrt{4 - y}$ or $x=2 + \sqrt{4 - y}$. Since $y=x$ intersects parabola at $x=0$ and $x=3$, $y$ ranges from 0 to 3. The outer radius $R(y) = 4 - (2 - \sqrt{4 - y}) = 2 + \sqrt{4 - y}$ The inner radius $r(y) = 4 - y$ Volume: $$V = \pi \int_0^3 \left[R(y)^2 - r(y)^2\right] dy = \pi \int_0^3 \left[(2 + \sqrt{4 - y})^2 - (4 - y)^2\right] dy$$ Expand: $$(2 + \sqrt{4 - y})^2 = 4 + 4\sqrt{4 - y} + (4 - y) = 8 - y + 4\sqrt{4 - y}$$ $$(4 - y)^2 = 16 - 8y + y^2$$ So integrand: $$8 - y + 4\sqrt{4 - y} - (16 - 8y + y^2) = -8 + 7y - y^2 + 4\sqrt{4 - y}$$ Integral: $$V = \pi \int_0^3 (-8 + 7y - y^2 + 4\sqrt{4 - y}) dy$$ Integrate term by term: - $\int_0^3 (-8) dy = -8 \times 3 = -24$ - $\int_0^3 7y dy = 7 \times \frac{3^2}{2} = 7 \times \frac{9}{2} = 31.5$ - $\int_0^3 -y^2 dy = - \frac{3^3}{3} = -9$ - $\int_0^3 4\sqrt{4 - y} dy$ Use substitution $u = 4 - y$, $du = -dy$, when $y=0$, $u=4$, when $y=3$, $u=1$: $$\int_0^3 4\sqrt{4 - y} dy = 4 \int_4^1 \sqrt{u} (-du) = 4 \int_1^4 u^{1/2} du = 4 \left[ \frac{2}{3} u^{3/2} \right]_1^4 = \frac{8}{3} (4^{3/2} - 1)$$ Calculate $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$: $$\frac{8}{3} (8 - 1) = \frac{8}{3} \times 7 = \frac{56}{3} \approx 18.6667$$ Sum integrals: $$-24 + 31.5 - 9 + 18.6667 = 17.1667$$ Volume: $$V = \pi \times 17.1667 \approx 53.93$$ 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 $x=a$: $$S = 2\pi \int_a^b r(x) \sqrt{1 + (f'(x))^2} dx$$ Here, $r(x) = |4 - x| = 4 - x$ (since $x$ in $[0,3]$), and $f(x) = 4x - x^2$. Calculate derivative: $$f'(x) = 4 - 2x$$ Surface area: $$S = 2\pi \int_0^3 (4 - x) \sqrt{1 + (4 - 2x)^2} dx = 2\pi \int_0^3 (4 - x) \sqrt{1 + (4 - 2x)^2} dx$$ Simplify inside the root: $$(4 - 2x)^2 = 16 - 16x + 4x^2$$ So: $$\sqrt{1 + 16 - 16x + 4x^2} = \sqrt{17 - 16x + 4x^2}$$ This integral does not simplify easily and is best evaluated numerically. **Final answers:** - (a) Widest horizontal cross-section width: $2.25$ meters - (b) Volume of water: approximately $53.93$ cubic meters - (c) Lateral surface area: $S = 2\pi \int_0^3 (4 - x) \sqrt{17 - 16x + 4x^2} dx$ (numerical evaluation needed)