Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 6: work (variable force, spring, pumping, cable), hydrostatic force and pressure, moments and center of mass, the centroid, average value, probability density functions第 1 至 6 节：功（变力、弹簧、抽水、缆绳），静水压力，矩与质心，形心，平均值，概率密度函数CALC II
A force of $30\,\text{N}$ holds a spring stretched $0.15\,\text{m}$ beyond its natural length. (a) Find $k$. (b) Work to stretch from $0.1\,\text{m}$ to $0.4\,\text{m}$. (c) Find how far a second spring ($k_2 = 500\,\text{N/m}$) must be stretched from rest to do the same work.一个 $30\,\text{N}$ 的力将弹簧从自然长度拉伸了 $0.15\,\text{m}$。(a) 求 $k$。(b) 从 $0.1\,\text{m}$ 拉伸到 $0.4\,\text{m}$ 所做的功。(c) 求第二根弹簧（$k_2 = 500\,\text{N/m}$）从自然长度拉伸，做相同功时需拉伸多远。
Hooke's law gives $F = kx$, so $30 = k(0.15)$. (M1) Solving, $k = 30/0.15 = 200\,\text{N/m}$. (A1)胡克定律给出 $F = kx$，故 $30 = k(0.15)$。(M1) 解得 $k = 30/0.15 = 200\,\text{N/m}$。(A1)
The work element is $dW = F\,dx = 200x\,dx$. (M1) Integrate from $x = 0.1$ to $x = 0.4$: (M1)功的元素为 $dW = F\,dx = 200x\,dx$。(M1) 从 $x = 0.1$ 积分到 $x = 0.4$：(M1)
$$ W = \int_{0.1}^{0.4} 200x\,dx = \Big[100x^2\Big]_{0.1}^{0.4} = 100(0.16) - 100(0.01) = 16 - 1 = 15\,\text{J}. $$(A1 for antiderivative, A1 for answer)（A1 得原函数，A1 得最终答案）
For the second spring, $W_2 = \int_0^d 500x\,dx = \tfrac{1}{2}(500)d^2 = 250d^2$. (M1) Setting $250d^2 = 15$ gives $d^2 = \tfrac{15}{250} = 0.06$, so $d = \sqrt{0.06} = \tfrac{\sqrt{6}}{10} \approx 0.245\,\text{m}$. (A1)对第二根弹簧，$W_2 = \int_0^d 500x\,dx = \tfrac{1}{2}(500)d^2 = 250d^2$。(M1) 令 $250d^2 = 15$ 得 $d^2 = \tfrac{15}{250} = 0.06$，故 $d = \sqrt{0.06} = \tfrac{\sqrt{6}}{10} \approx 0.245\,\text{m}$。(A1)
Vertical cylinder: radius $3\,\text{m}$, height $8\,\text{m}$, water filled to $6\,\text{m}$, $w = 9800\,\text{N/m}^3$. Pump all water to the top rim. (a) Write $dW$. (b) Evaluate $W$. (c) Explain why pumping to the current water level saves zero work and what the correct comparison is.竖直圆柱形水箱：半径 $3\,\text{m}$，高 $8\,\text{m}$，水深 $6\,\text{m}$，$w = 9800\,\text{N/m}^3$。将全部水抽至顶部边缘。(a) 写出 $dW$。(b) 计算 $W$。(c) 解释为何将水抽至当前水面节省零功，并说明正确的比较方式。
A slab at height $y$ (with $0 \le y \le 6$) has cross-sectional area $A = \pi(3)^2 = 9\pi\,\text{m}^2$ (M1) and must rise a distance $8 - y\,\text{m}$ to clear the rim. (A1) The weight of the slab is $w\cdot A\,dy = 9800\cdot 9\pi\,dy$, so高度 $y$（$0 \le y \le 6$）处的薄层横截面面积为 $A = \pi(3)^2 = 9\pi\,\text{m}^2$ (M1)，需提升 $8 - y\,\text{m}$ 才能越过顶部边缘。(A1) 该薄层的重力为 $w\cdot A\,dy = 9800\cdot 9\pi\,dy$，故
$$ dW = 9800\cdot 9\pi\,(8 - y)\,dy = 88\,200\pi\,(8-y)\,dy. $$(A1)
Water occupies $0 \le y \le 6$, so (M1)水占 $0 \le y \le 6$，故 (M1)
$$ W = 88\,200\pi\int_0^6 (8 - y)\,dy = 88\,200\pi\Big[8y - \tfrac{y^2}{2}\Big]_0^6. $$ $$ = 88\,200\pi\Big[(48 - 18) - 0\Big] = 88\,200\pi\cdot 30 = 2\,646\,000\pi\,\text{J}. $$(A1 for antiderivative, A1 for final answer) Numerically, $W \approx 8.31\times 10^6\,\text{J}$.（A1 得原函数，A1 得最终答案）数值上，$W \approx 8.31\times 10^6\,\text{J}$。
The water surface is already at $y = 6$. Pumping to $y = 6$ is not pumping at all: every slab is already at its destination, so the integral has zero length and the work is zero. (R1) The meaningful comparison is: how much work is saved by pumping to a lower exit, say through a pipe outlet at $y = 4$ rather than the rim at $y = 8$? In that case the lift distance changes from $8 - y$ to $4 - y$ (for $0 \le y \le 4$), giving a different integral. (A1)水面已在 $y = 6$ 处。将水抽至 $y = 6$ 等于没有抽：每片水层已在目标位置，积分区间长度为零，做功为零。(R1) 有意义的比较是：若不从顶部边缘（$y = 8$）抽出，而从较低出口（如 $y = 4$ 的管道）抽出，能节省多少功？此时提升距离从 $8 - y$ 变为 $4 - y$（$0 \le y \le 4$），得到不同的积分。(A1)
Cable: $15\,\text{m}$, $\rho = 4\,\text{kg/m}$, $g = 9.8\,\text{m/s}^2$; $60\,\text{kg}$ load at free end. (a) Work to wind the cable to the top. (b) Work for the load, total, and interpretation via $mg\bar{x}$.缆绳：$15\,\text{m}$，$\rho = 4\,\text{kg/m}$，$g = 9.8\,\text{m/s}^2$；自由端挂 $60\,\text{kg}$ 重物。(a) 将缆绳收至顶部所做的功。(b) 提升重物的功、总功及通过 $mg\bar{x}$ 的解释。
Let $x$ be the distance below the top to a cable element of length $dx$. That element has mass $\rho\,dx = 4\,dx$ and must be lifted a distance $x$ to the top. (M1) The work element is设 $x$ 为大楼顶部到长度为 $dx$ 的缆绳元素的向下距离，该元素质量为 $\rho\,dx = 4\,dx$，需提升距离 $x$ 至顶部。(M1) 功的元素为
$$ dW = (\rho g\,dx)\cdot x = 4(9.8)\,x\,dx = 39.2\,x\,dx. $$(A1) Integrating from $x = 0$ (top) to $x = 15$ (free end):(A1) 从 $x = 0$（顶部）积分到 $x = 15$（自由端）：
$$ W_{\text{cable}} = \int_0^{15} 39.2\,x\,dx = 39.2\Big[\tfrac{x^2}{2}\Big]_0^{15} = 39.2\cdot \tfrac{225}{2} = 39.2\cdot 112.5 = 4410\,\text{J}. $$(A1)
The $60\,\text{kg}$ load hangs at $x = 15$ and is lifted the full $15\,\text{m}$ at constant weight: (M1)$60\,\text{kg}$ 重物挂在 $x = 15$ 处，以恒定重力提升整个 $15\,\text{m}$：(M1)
$$ W_{\text{load}} = 60(9.8)(15) = 8820\,\text{J}. $$Total work $W_{\text{total}} = 4410 + 8820 = 13\,230\,\text{J}$. (A1)总功 $W_{\text{total}} = 4410 + 8820 = 13\,230\,\text{J}$。(A1)
Interpretation via center of mass: the cable (mass $m_c = 4\times 15 = 60\,\text{kg}$) has its center of mass at $x = 15/2 = 7.5\,\text{m}$ from the top; the load (mass $m_L = 60\,\text{kg}$) is at $x = 15\,\text{m}$. The combined center of mass from the top is通过质心解释：缆绳（质量 $m_c = 4\times 15 = 60\,\text{kg}$）的质心在距顶部 $x = 15/2 = 7.5\,\text{m}$ 处；重物（质量 $m_L = 60\,\text{kg}$）在 $x = 15\,\text{m}$ 处。合质心距顶部为
$$ \bar{x} = \frac{60(7.5) + 60(15)}{60 + 60} = \frac{450 + 900}{120} = \frac{1350}{120} = 11.25\,\text{m}. $$Check: $mg\bar{x} = 120(9.8)(11.25) = 13\,230\,\text{J}$. (R1) The total work equals the work that would lift the entire mass from its combined center of mass.验算：$mg\bar{x} = 120(9.8)(11.25) = 13\,230\,\text{J}$。(R1) 总功等于将全部质量从其合质心处提升相同高度所做的功。
$T(x) = 20 + 80e^{-x/5}$ on $[0,10]$. (a) Find $T_{\text{avg}}$. (b) Find $c$ with $T(c) = T_{\text{avg}}$.$T(x) = 20 + 80e^{-x/5}$，$x \in [0,10]$。(a) 求 $T_{\text{avg}}$。(b) 求满足 $T(c) = T_{\text{avg}}$ 的 $c$。
Apply $f_{\text{avg}} = \tfrac{1}{b-a}\int_a^b f\,dx$ with $a=0$, $b=10$: (M1)用平均值公式 $f_{\text{avg}} = \tfrac{1}{b-a}\int_a^b f\,dx$，取 $a=0$，$b=10$：(M1)
$$ T_{\text{avg}} = \frac{1}{10}\int_0^{10}\!\big(20 + 80e^{-x/5}\big)\,dx = \frac{1}{10}\Big[20x - 400e^{-x/5}\Big]_0^{10}. $$(M1 for antiderivative $\int e^{-x/5}dx = -5e^{-x/5}$)（M1 得原函数 $\int e^{-x/5}dx = -5e^{-x/5}$）
$$ = \frac{1}{10}\Big[\big(200 - 400e^{-2}\big) - \big(0 - 400\big)\Big] = \frac{1}{10}(600 - 400e^{-2}) = 60 - 40e^{-2}. $$(A1) Numerically, $T_{\text{avg}} \approx 60 - 40(0.1353) \approx 54.6\,^\circ\text{C}$.(A1) 数值上，$T_{\text{avg}} \approx 60 - 40(0.1353) \approx 54.6\,^\circ\text{C}$。
Set $20 + 80e^{-c/5} = 60 - 40e^{-2}$. (M1) Then $80e^{-c/5} = 40 - 40e^{-2} = 40(1 - e^{-2})$, so令 $20 + 80e^{-c/5} = 60 - 40e^{-2}$。(M1) 则 $80e^{-c/5} = 40 - 40e^{-2} = 40(1 - e^{-2})$，故
$$ e^{-c/5} = \frac{1 - e^{-2}}{2}. $$Taking natural logarithm: $-c/5 = \ln\!\left(\tfrac{1-e^{-2}}{2}\right)$. (M1) Therefore取自然对数：$-c/5 = \ln\!\left(\tfrac{1-e^{-2}}{2}\right)$。(M1) 故
$$ c = -5\ln\!\left(\frac{1 - e^{-2}}{2}\right) = 5\ln\!\left(\frac{2}{1 - e^{-2}}\right). $$(A1) Since $\tfrac{1-e^{-2}}{2} \approx 0.432 < 1$, its logarithm is negative, confirming $c > 0$; numerically $c \approx 5(0.839) \approx 4.2\,\text{cm}$, which lies in $[0,10]$ as the theorem guarantees.(A1) 由于 $\tfrac{1-e^{-2}}{2} \approx 0.432 < 1$，其对数为负，确认 $c > 0$；数值上 $c \approx 5(0.839) \approx 4.2\,\text{cm}$，在 $[0,10]$ 内，符合定理的保证。
(a) Starting from $dm = \delta f(x)\,dx$ and strip center $(x, \tfrac{1}{2}f(x))$, derive the centroid formulas for $\bar{x}$ and $\bar{y}$. Explain why $\delta$ cancels. (b) State and justify the $\bar{y}$ formula for a region between two curves $f \ge g$.(a) 从 $dm = \delta f(x)\,dx$ 及条形中心 $(x, \tfrac{1}{2}f(x))$ 出发，推导 $\bar{x}$ 和 $\bar{y}$ 的形心公式，并解释为何 $\delta$ 可消去。(b) 对 $f \ge g$ 的两曲线间区域，写出 $\bar{y}$ 的公式并给出理由。
The mass element of a vertical strip at position $x$ of width $dx$ is $dm = \delta f(x)\,dx$. (M1) The total mass is位置 $x$ 处宽度为 $dx$ 的竖直条的质量元素为 $dm = \delta f(x)\,dx$。(M1) 总质量为
$$ m = \int_a^b \delta f(x)\,dx = \delta\int_a^b f(x)\,dx = \delta\,A, $$where $A = \int_a^b f(x)\,dx$ is the area.其中 $A = \int_a^b f(x)\,dx$ 为面积。
The moment about the $y$-axis uses $x$ as the distance to that axis: $dM_y = x\,dm = x\,\delta\,f(x)\,dx$. (M1) Therefore对 $y$ 轴的矩以 $x$ 为到轴的距离：$dM_y = x\,dm = x\,\delta\,f(x)\,dx$。(M1) 故
$$ M_y = \delta\int_a^b x\,f(x)\,dx, \qquad \bar{x} = \frac{M_y}{m} = \frac{\delta\int_a^b x\,f(x)\,dx}{\delta\int_a^b f(x)\,dx} = \frac{\int_a^b x\,f(x)\,dx}{\int_a^b f(x)\,dx}. $$(A1) The constant density $\delta$ cancels from numerator and denominator, leaving a formula that depends only on geometry.(A1) 常数密度 $\delta$ 在分子和分母中消去，留下仅依赖几何形状的公式。
The moment about the $x$-axis uses the vertical distance to that axis, which for the strip is its center height $\tfrac{1}{2}f(x)$: $dM_x = \tfrac{1}{2}f(x)\,dm = \tfrac{1}{2}f(x)\cdot\delta\,f(x)\,dx = \tfrac{\delta}{2}[f(x)]^2\,dx$. (M1) Therefore对 $x$ 轴的矩以竖直距离为臂，对于该条形即其中心高度 $\tfrac{1}{2}f(x)$：$dM_x = \tfrac{1}{2}f(x)\,dm = \tfrac{1}{2}f(x)\cdot\delta\,f(x)\,dx = \tfrac{\delta}{2}[f(x)]^2\,dx$。(M1) 故
$$ M_x = \frac{\delta}{2}\int_a^b [f(x)]^2\,dx, \qquad \bar{y} = \frac{M_x}{m} = \frac{\tfrac{\delta}{2}\int_a^b [f(x)]^2\,dx}{\delta\int_a^b f(x)\,dx} = \frac{\int_a^b \tfrac{1}{2}[f(x)]^2\,dx}{\int_a^b f(x)\,dx}. $$(A1) Again $\delta$ cancels; the centroid is a purely geometric quantity.(A1) $\delta$ 再次消去；形心是纯几何量。
For a strip between $y = g(x)$ (bottom) and $y = f(x)$ (top), the strip has width $dx$, height $f(x) - g(x)$, and its center lies at the midpoint height $\tfrac{1}{2}(f(x) + g(x))$. (M1) So对于 $y = g(x)$（下方）与 $y = f(x)$（上方）之间的条形，宽度为 $dx$，高度为 $f(x) - g(x)$，中心在中点高度 $\tfrac{1}{2}(f(x) + g(x))$ 处。(M1) 故
$$ dM_x = \tfrac{1}{2}\big[f(x) + g(x)\big]\cdot\delta\big[f(x) - g(x)\big]\,dx = \frac{\delta}{2}\big[f(x)^2 - g(x)^2\big]\,dx. $$Therefore $\bar{y} = \dfrac{\int_a^b \tfrac{1}{2}[f(x)^2 - g(x)^2]\,dx}{\int_a^b [f(x) - g(x)]\,dx}$. (A1)故 $\bar{y} = \dfrac{\int_a^b \tfrac{1}{2}[f(x)^2 - g(x)^2]\,dx}{\int_a^b [f(x) - g(x)]\,dx}$。(A1)
The expression $\tfrac{1}{2}[f^2 - g^2]$ is NOT $\tfrac{1}{2}[f-g]^2$: the first is $\tfrac{1}{2}(f+g)(f-g)$, the product of midpoint times height; the second squares only the height and discards the midpoint information. (R1)表达式 $\tfrac{1}{2}[f^2 - g^2]$ 不是 $\tfrac{1}{2}[f-g]^2$：前者为 $\tfrac{1}{2}(f+g)(f-g)$，即中点高度乘以条带高度；后者仅对高度取平方，丢失了中点信息。(R1)
$f(x) = c\,x^2(3-x)$ on $[0,3]$, zero elsewhere. (a) Find $c$. (b) Find $\mu = E[X]$. (c) Find $P(1 \le X \le 2)$. (d) Show $P(X \le 2) = 16/27 > 1/2$, and hence locate the median.$f(x) = c\,x^2(3-x)$，$x \in [0,3]$，其余为零。(a) 求 $c$。(b) 求 $\mu = E[X]$。(c) 求 $P(1 \le X \le 2)$。(d) 证明 $P(X \le 2) = 16/27 > 1/2$，并据此确定中位数的范围。
Require $\int_0^3 c\,x^2(3-x)\,dx = 1$. (M1) Expand: $\int_0^3 (3x^2 - x^3)\,dx = \Big[x^3 - \tfrac{x^4}{4}\Big]_0^3 = 27 - \tfrac{81}{4} = \tfrac{108 - 81}{4} = \tfrac{27}{4}$. So $c\cdot\tfrac{27}{4} = 1$ and $c = \tfrac{4}{27}$. (A1)要求 $\int_0^3 c\,x^2(3-x)\,dx = 1$。(M1) 展开：$\int_0^3 (3x^2 - x^3)\,dx = \Big[x^3 - \tfrac{x^4}{4}\Big]_0^3 = 27 - \tfrac{81}{4} = \tfrac{108 - 81}{4} = \tfrac{27}{4}$。故 $c\cdot\tfrac{27}{4} = 1$，$c = \tfrac{4}{27}$。(A1)
$\mu = \int_0^3 x\cdot f(x)\,dx = \tfrac{4}{27}\int_0^3 x^3(3-x)\,dx$. (M1) Expand: $\int_0^3 (3x^3 - x^4)\,dx = \Big[\tfrac{3x^4}{4} - \tfrac{x^5}{5}\Big]_0^3$. (M1)$\mu = \int_0^3 x\cdot f(x)\,dx = \tfrac{4}{27}\int_0^3 x^3(3-x)\,dx$。(M1) 展开：$\int_0^3 (3x^3 - x^4)\,dx = \Big[\tfrac{3x^4}{4} - \tfrac{x^5}{5}\Big]_0^3$。(M1)
$$ = \tfrac{3(81)}{4} - \tfrac{243}{5} = \tfrac{243}{4} - \tfrac{243}{5} = 243\!\left(\tfrac{1}{4} - \tfrac{1}{5}\right) = 243\cdot\tfrac{1}{20} = \tfrac{243}{20}. $$ $$ \mu = \tfrac{4}{27}\cdot\tfrac{243}{20} = \tfrac{4\cdot 243}{27\cdot 20} = \tfrac{4\cdot 9}{20} = \tfrac{36}{20} = \tfrac{9}{5}. $$(A1)
$P(1 \le X \le 2) = \tfrac{4}{27}\int_1^2 (3x^2 - x^3)\,dx = \tfrac{4}{27}\Big[x^3 - \tfrac{x^4}{4}\Big]_1^2$. (M1)$P(1 \le X \le 2) = \tfrac{4}{27}\int_1^2 (3x^2 - x^3)\,dx = \tfrac{4}{27}\Big[x^3 - \tfrac{x^4}{4}\Big]_1^2$。(M1)
$$ = \tfrac{4}{27}\Big[\big(8 - 4\big) - \big(1 - \tfrac{1}{4}\big)\Big] = \tfrac{4}{27}\Big[4 - \tfrac{3}{4}\Big] = \tfrac{4}{27}\cdot\tfrac{13}{4} = \tfrac{13}{27}. $$(M1 for antiderivative, A1 for answer)（M1 得原函数，A1 得答案）
The median $m$ satisfies $\int_0^m f(x)\,dx = \tfrac{1}{2}$. (M1) Compute:中位数 $m$ 满足 $\int_0^m f(x)\,dx = \tfrac{1}{2}$。(M1) 计算：
$$ P(X \le 2) = \tfrac{4}{27}\int_0^2 (3x^2 - x^3)\,dx = \tfrac{4}{27}\Big[x^3 - \tfrac{x^4}{4}\Big]_0^2 = \tfrac{4}{27}\Big[8 - 4\Big] = \tfrac{4}{27}\cdot 4 = \tfrac{16}{27}. $$(A1) Since $\tfrac{16}{27} > \tfrac{1}{2}$, exactly half the probability mass lies to the left of some point less than $2$, so $m < 2$. Since $f(x) > 0$ for $x \in (0,3)$, we also have $P(X \le 0) = 0 < \tfrac{1}{2}$, so $m > 0$. Therefore the median satisfies $m \in (0, 2)$.(A1) 由于 $\tfrac{16}{27} > \tfrac{1}{2}$，恰好一半的概率质量在某个小于 $2$ 的点左侧，故 $m < 2$。由于 $f(x) > 0$（$x \in (0,3)$），$P(X \le 0) = 0 < \tfrac{1}{2}$，故 $m > 0$。因此中位数满足 $m \in (0, 2)$。
Vertical triangular gate: base $6\,\text{m}$ at surface, apex at depth $4\,\text{m}$, $w = 9800\,\text{N/m}^3$. (a) Write $L(h)$ and $dF$. (b) Evaluate $F$. (c) Find $\bar{h}$ using $F = w\bar{h}A$ and compare to the geometric centroid of the triangle.竖直三角形闸门：底边宽 $6\,\text{m}$ 位于水面，顶点在深度 $4\,\text{m}$ 处，$w = 9800\,\text{N/m}^3$。(a) 写出 $L(h)$ 和 $dF$。(b) 计算 $F$。(c) 用 $F = w\bar{h}A$ 求 $\bar{h}$，并与三角形的几何形心比较。
The triangle has its base (width $6$) at $h = 0$ and its apex (width $0$) at $h = 4$. It narrows linearly, so $L(h) = 6\cdot\tfrac{4-h}{4} = \tfrac{3}{2}(4-h)$. (M1·A1)三角形底边（宽 $6$）在 $h = 0$ 处，顶点（宽 $0$）在 $h = 4$ 处，线性收窄，故 $L(h) = 6\cdot\tfrac{4-h}{4} = \tfrac{3}{2}(4-h)$。(M1·A1)
A horizontal strip at depth $h$ of thickness $dh$ has area $L(h)\,dh$ and lies at depth $h$, so the pressure on it is $w\,h$. The force element is (M1)深度 $h$ 处厚度为 $dh$ 的水平条带，面积为 $L(h)\,dh$，所受压强为 $w\,h$。压力元素为 (M1)
$$ dF = w\,h\,L(h)\,dh = 9800\,h\cdot\tfrac{3}{2}(4-h)\,dh. $$(A1)
(M1 for setting up integral, A1 for antiderivative, A1 for final answer)（M1 建立积分，A1 得原函数，A1 得最终答案）
The area of the triangle is $A = \tfrac{1}{2}\cdot 6\cdot 4 = 12\,\text{m}^2$. (M1) The centroid shortcut gives三角形面积为 $A = \tfrac{1}{2}\cdot 6\cdot 4 = 12\,\text{m}^2$。(M1) 形心捷径给出
$$ \bar{h} = \frac{F}{w\,A} = \frac{156\,800}{9800\cdot 12} = \frac{156\,800}{117\,600} = \frac{4}{3}\,\text{m}. $$(A1) The base of the triangle is at the surface ($h = 0$) and the apex is at $h = 4$. The centroid of a triangle lies one-third of the height measured from the base, i.e., at $h = \tfrac{1}{3}\cdot 4 = \tfrac{4}{3}$ from the surface. This is consistent. The result confirms both the integral calculation and the shortcut $F = w\bar{h}A$.(A1) 三角形底边在水面（$h = 0$），顶点在 $h = 4$ 处。三角形的形心在距底边三分之一高处，即从水面起 $h = \tfrac{1}{3}\cdot 4 = \tfrac{4}{3}$ 处。结果一致。这既验证了积分计算，也验证了捷径公式 $F = w\bar{h}A$。
Inverted cone: height $12\,\text{m}$, top radius $3\,\text{m}$, liquid ($w = 8000\,\text{N/m}^3$) to depth $9\,\text{m}$ from vertex; pump to top rim. (a) $r(y)$. (b) $A(y)$, $d(y)$, $dW$. (c) Evaluate $W$. (d) Factor increase if full to $12\,\text{m}$.倒置正圆锥：高 $12\,\text{m}$，顶部半径 $3\,\text{m}$，液体（$w = 8000\,\text{N/m}^3$）从顶点量起深度 $9\,\text{m}$；将液体抽至顶部边缘。(a) $r(y)$。(b) $A(y)$，$d(y)$，$dW$。(c) 计算 $W$。(d) 若注满至 $12\,\text{m}$，功增大多少倍。
The cone widens linearly from radius $0$ at $y = 0$ (vertex) to radius $3$ at $y = 12$ (rim). By similar triangles, $r/y = 3/12$, so $r(y) = y/4$. (M1·A1)锥形从 $y = 0$（顶点，半径 $0$）到 $y = 12$（顶部，半径 $3$）线性扩大。由相似三角形，$r/y = 3/12$，故 $r(y) = y/4$。(M1·A1)
The cross-sectional area at height $y$ is $A(y) = \pi r(y)^2 = \pi(y/4)^2 = \pi y^2/16$. (M1) The lift distance to the top rim is $d(y) = 12 - y$. (A1) The work element is高度 $y$ 处的横截面面积为 $A(y) = \pi r(y)^2 = \pi(y/4)^2 = \pi y^2/16$。(M1) 到顶部边缘的提升距离为 $d(y) = 12 - y$。(A1) 功元素为
$$ dW = w\cdot A(y)\cdot d(y)\,dy = 8000\cdot\frac{\pi y^2}{16}\cdot(12-y)\,dy = 500\pi\,y^2(12-y)\,dy. $$(A1)
(M1 for setting up the expanded integral, M1 for antiderivative, A1 for final answer)（M1 建立展开积分，M1 得原函数，A1 得最终答案）
If filled to $y = 12$, the integral becomes $500\pi\int_0^{12}(12y^2 - y^3)\,dy$. (M1)若注满至 $y = 12$，积分变为 $500\pi\int_0^{12}(12y^2 - y^3)\,dy$。(M1)
$$ \int_0^{12}(12y^2 - y^3)\,dy = \Big[4y^3 - \tfrac{y^4}{4}\Big]_0^{12} = 4(1728) - \tfrac{20736}{4} = 6912 - 5184 = 1728. $$The factor increase is $\dfrac{W_{12}}{W_9} = \dfrac{500\pi\cdot 1728}{500\pi\cdot 5103/4} = \dfrac{1728\cdot 4}{5103} = \dfrac{6912}{5103} = \dfrac{256}{189} \approx 1.354$. (A1)功的增大倍数为 $\dfrac{W_{12}}{W_9} = \dfrac{500\pi\cdot 1728}{500\pi\cdot 5103/4} = \dfrac{1728\cdot 4}{5103} = \dfrac{6912}{5103} = \dfrac{256}{189} \approx 1.354$。(A1)
Dam: $8\,\text{m}$ wide, $5\,\text{m}$ deep water, $w = 9800\,\text{N/m}^3$; circular window radius $1\,\text{m}$, center at depth $3\,\text{m}$. (a) Force on the full rectangular dam face. (b) Set up the force integral for the window. (c) Evaluate using the given integrals; verify with the centroid shortcut.坝墙：宽 $8\,\text{m}$，水深 $5\,\text{m}$，$w = 9800\,\text{N/m}^3$；圆形窗半径 $1\,\text{m}$，圆心在水面下 $3\,\text{m}$。(a) 整个矩形坝面所受压力。(b) 建立圆形窗的压力积分。(c) 用所给积分求值，并用形心捷径验算。
Each horizontal strip at depth $h$ (with $0 \le h \le 5$) has constant width $L = 8$ and area $8\,dh$. (M1) The force element is $dF = w\,h\,(8)\,dh$:深度 $h$（$0 \le h \le 5$）处每条水平条带宽度恒为 $L = 8$，面积为 $8\,dh$。(M1) 压力元素为 $dF = w\,h\,(8)\,dh$：
$$ F_{\text{rect}} = 9800\cdot 8\int_0^5 h\,dh = 78\,400\cdot\Big[\tfrac{h^2}{2}\Big]_0^5 = 78\,400\cdot 12.5 = 980\,000\,\text{N}. $$(A1 for antiderivative, A1 for answer)（A1 得原函数，A1 得答案）
Place the origin at the center of the circle. A strip at height $y$ (with $-1 \le y \le 1$) is at depth $h = 3 - y$ below the surface (M1) and has half-width $\sqrt{1-y^2}$, giving strip width $L(y) = 2\sqrt{1-y^2}$. The force element is以圆心为原点。高度 $y$（$-1 \le y \le 1$）处的条带距水面深度 $h = 3 - y$ (M1)，半宽为 $\sqrt{1-y^2}$，条带宽度 $L(y) = 2\sqrt{1-y^2}$。压力元素为
$$ dF = w(3-y)\cdot 2\sqrt{1-y^2}\,dy. $$(M1) Splitting by linearity:(M1) 按线性性质拆分：
$$ F_{\text{win}} = 2w\int_{-1}^{1}(3-y)\sqrt{1-y^2}\,dy = 2w\left[3\int_{-1}^{1}\sqrt{1-y^2}\,dy - \int_{-1}^{1}y\sqrt{1-y^2}\,dy\right]. $$(A1 for correct split)（A1 得正确拆分）
The first integral is the area of the upper half of the unit circle: $\int_{-1}^{1}\sqrt{1-y^2}\,dy = \tfrac{\pi}{2}$. (M1) The second integrand $y\sqrt{1-y^2}$ is an odd function on the symmetric interval $[-1,1]$, so it integrates to $0$. Therefore第一个积分是单位圆上半部面积：$\int_{-1}^{1}\sqrt{1-y^2}\,dy = \tfrac{\pi}{2}$。(M1) 第二个被积函数 $y\sqrt{1-y^2}$ 在对称区间 $[-1,1]$ 上为奇函数，积分为 $0$。故
$$ F_{\text{win}} = 2w\left[3\cdot\tfrac{\pi}{2} - 0\right] = 3\pi w = 3\pi(9800) = 29\,400\pi \approx 92\,360\,\text{N}. $$(A1)
Centroid shortcut: the circle has area $A = \pi(1)^2 = \pi\,\text{m}^2$ and its center (which is also its centroid) is at depth $\bar{h} = 3\,\text{m}$. So $w\bar{h}A = 9800\cdot 3\cdot\pi = 29\,400\pi$. (A1) The two methods agree exactly.形心捷径：圆的面积 $A = \pi(1)^2 = \pi\,\text{m}^2$，其圆心（也是形心）在深度 $\bar{h} = 3\,\text{m}$ 处。故 $w\bar{h}A = 9800\cdot 3\cdot\pi = 29\,400\pi$。(A1) 两种方法结果完全一致。
Region $R$: above $y = x^2 - 2x$, below $y = 4 - x^2$, on $[0,2]$. (a) Area $A$. (b) $\bar{x}$. (c) $\bar{y}$ using the two-curve formula.区域 $R$：在 $[0,2]$ 上，上方为 $y = 4 - x^2$，下方为 $y = x^2 - 2x$。(a) 面积 $A$。(b) $\bar{x}$。(c) 用两曲线公式求 $\bar{y}$。
The height of a strip at $x$ is $f(x) - g(x) = (4-x^2)-(x^2-2x) = 4 + 2x - 2x^2$. (M1)$x$ 处条带的高度为 $f(x) - g(x) = (4-x^2)-(x^2-2x) = 4 + 2x - 2x^2$。(M1)
$$ A = \int_0^2 (4 + 2x - 2x^2)\,dx = \Big[4x + x^2 - \tfrac{2x^3}{3}\Big]_0^2 = 8 + 4 - \tfrac{16}{3} = 12 - \tfrac{16}{3} = \tfrac{36 - 16}{3} = \tfrac{20}{3}. $$(A1)
(M1 for correct integrand $x(f-g)$, M1 for antiderivative, A1 for answer)（M1 得正确被积函数 $x(f-g)$，M1 得原函数，A1 得答案）
With $f = 4 - x^2$ and $g = x^2 - 2x$, compute $f^2 - g^2$: (M1)取 $f = 4 - x^2$，$g = x^2 - 2x$，计算 $f^2 - g^2$：(M1)
$$ f^2 = (4-x^2)^2 = 16 - 8x^2 + x^4, $$ $$ g^2 = (x^2-2x)^2 = x^4 - 4x^3 + 4x^2, $$ $$ f^2 - g^2 = (16 - 8x^2 + x^4) - (x^4 - 4x^3 + 4x^2) = 16 + 4x^3 - 12x^2. $$(M1 for correct expansion)（M1 得正确展开）
$$ \bar{y} = \frac{1}{A}\int_0^2 \tfrac{1}{2}(f^2 - g^2)\,dx = \frac{3}{20}\int_0^2 \tfrac{1}{2}(16 + 4x^3 - 12x^2)\,dx = \frac{3}{20}\int_0^2 (8 + 2x^3 - 6x^2)\,dx. $$ $$ \int_0^2 (8 + 2x^3 - 6x^2)\,dx = \Big[8x + \tfrac{x^4}{2} - 2x^3\Big]_0^2 = 16 + 8 - 16 = 8. $$ $$ \bar{y} = \frac{3}{20}\cdot 8 = \frac{24}{20} = \frac{6}{5}. $$(A1)