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
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PART I · CORE TECHNIQUES第一部分 · 核心技术Computational fluency · 28 marks计算能力 · 28分
Work, Center of Mass, and Average Value功、质心与平均值
Show all working. For every pumping or work problem, name the variable and write the element $dW$ before integrating. Carry SI units throughout.展示全部解题过程。对每道抽水或做功题，先命名变量，写出元素 $dW$，再积分。全程使用国际单位制。
Q1MEDIUMCOREHooke's law and spring work胡克定律与弹簧做功[8 marks]
A force of $30\,\text{N}$ holds a spring stretched $0.15\,\text{m}$ beyond its natural length. Use Hooke's law $F = kx$ throughout.一个 $30\,\text{N}$ 的力将弹簧从自然长度拉伸了 $0.15\,\text{m}$。全题使用胡克定律 $F = kx$。
(a)Find the spring constant $k$.求弹簧的劲度系数 $k$。[2]
(b)Find the work done in stretching the spring from $0.1\,\text{m}$ to $0.4\,\text{m}$ beyond its natural length.求将弹簧从超出自然长度 $0.1\,\text{m}$ 拉伸到 $0.4\,\text{m}$ 所做的功。[4]
(c)A second spring has constant $k_2 = 500\,\text{N/m}$. Find how far beyond its natural length this spring must be stretched, starting from rest, so that the work done equals the answer in part (b).第二根弹簧的劲度系数为 $k_2 = 500\,\text{N/m}$。求从自然长度开始拉伸，使所做的功等于 (b) 中答案时，弹簧需要拉伸多远。[2]
Q2MEDIUMCOREpumping water from a cylindrical tank从圆柱形水箱抽水[8 marks]
A vertical cylindrical tank has radius $3\,\text{m}$ and height $8\,\text{m}$. It contains water (weight density $w = 9800\,\text{N/m}^3$) filled to a depth of $6\,\text{m}$. Water is to be pumped out over the top rim. Let $y$ measure height above the base of the tank.一个竖直圆柱形水箱，半径为 $3\,\text{m}$，高为 $8\,\text{m}$，其中注有深度 $6\,\text{m}$ 的水（重量密度 $w = 9800\,\text{N/m}^3$）。水将从顶部边缘抽出。设 $y$ 为水箱底部以上的高度。
(a)Write the element of work $dW$ done in lifting a thin horizontal slab of water at height $y$ of thickness $dy$ to the top rim. State the cross-sectional area and the lift distance explicitly.写出将高度 $y$ 处厚度为 $dy$ 的薄水层提升至顶部边缘所做功的元素 $dW$。明确说明横截面面积和提升距离。[3]
(b)Set up and evaluate the integral for the total work to pump all the water out. Give an exact answer and a decimal approximation.建立并计算将全部水抽出所需总功的积分。给出精确答案和十进制近似值。[3]
(c)What fraction of the work in part (b) would be saved if, instead of pumping to the rim, the water were pumped only to the level of the water surface (height $6\,\text{m}$) inside the tank? Explain why this fraction is zero and what the correct comparison should be.如果不把水抽到顶部边缘，而只抽到水箱内水面高度（$6\,\text{m}$），可以节省 (b) 中所做功的多大比例？解释为何该比例为零，并说明正确的比较方式。[2]
Q3HARDCOREcable work and center of mass of a rod缆绳做功与杆的质心[6 marks]
A uniform cable $15\,\text{m}$ long with linear density $\rho = 4\,\text{kg/m}$ hangs vertically from the top of a building. Take $g = 9.8\,\text{m/s}^2$.一根均匀缆绳长 $15\,\text{m}$，线密度 $\rho = 4\,\text{kg/m}$，竖直悬挂于大楼顶部。取 $g = 9.8\,\text{m/s}^2$。
(a)Let $x$ be the distance below the top of the building to a small segment of cable. Write the work element $dW$ to lift that segment to the top, and integrate to find the total work to wind the entire cable to the top.设 $x$ 为大楼顶部到缆绳小段的向下距离。写出将该段提升至顶部的功元素 $dW$，并积分求将整根缆绳收到顶部所需的总功。[3]
(b)A $60\,\text{kg}$ load is attached to the free end of the cable. Find the additional work needed to lift this load to the top of the building, and hence the total work (cable plus load). Interpret the total by noting it equals $mg\bar{x}$, where $m$ is the total mass and $\bar{x}$ is the combined center of mass height.一个 $60\,\text{kg}$ 的重物附在缆绳自由端。求将该重物提升至大楼顶部所需的额外功，进而求总功（缆绳加重物）。注意总功等于 $mg\bar{x}$（$m$ 为总质量，$\bar{x}$ 为合质心高度），据此解释结果。[3]
Q4MEDIUMCOREaverage value and the Mean Value Theorem for Integrals平均值与积分中值定理[6 marks]
The temperature (in degrees Celsius) in a metal rod at position $x\,\text{cm}$ from one end is given by $T(x) = 20 + 80e^{-x/5}$ for $0 \le x \le 10$.一根金属杆从一端起 $x\,\text{cm}$ 处的温度（摄氏度）为 $T(x) = 20 + 80e^{-x/5}$，$0 \le x \le 10$。
(a)Find the average temperature $T_{\text{avg}}$ along the rod. Give an exact answer.求金属杆上的平均温度 $T_{\text{avg}}$。给出精确答案。[3]
(b)The Mean Value Theorem for Integrals guarantees a point $c \in [0,10]$ where $T(c) = T_{\text{avg}}$. Find $c$ exactly, expressing it in terms of natural logarithms.积分中值定理保证存在点 $c \in [0,10]$ 使得 $T(c) = T_{\text{avg}}$。用自然对数精确表示 $c$。[3]
PART II · DEFINITIONS AND PROOF第二部分 · 定义与证明Rigorous arguments · 26 marks严谨论证 · 26分
Deriving the Centroid Formulas and Working with Densities推导形心公式与密度的应用
These items are graded on the logic of the argument. For centroid derivations, start from the moment definition and carry each step. For probability, verify normalization before computing probabilities.本部分按论证逻辑评分。推导形心公式时，从矩的定义出发，逐步推进。概率题中，先验证归一化条件，再计算概率。
Q5HARDPROOFderiving centroid formulas via moments通过矩推导形心公式[8 marks]
Let $R$ be the region under the curve $y = f(x) \ge 0$ on $[a,b]$, with uniform density $\delta$ (mass per unit area). The centroid $(\bar{x},\bar{y})$ is defined by $\bar{x} = M_y / m$ and $\bar{y} = M_x / m$, where $m$ is total mass and $M_y$, $M_x$ are moments about the $y$- and $x$-axes respectively.设 $R$ 为曲线 $y = f(x) \ge 0$ 在 $[a,b]$ 上方的区域，均匀密度为 $\delta$（单位面积的质量）。形心 $(\bar{x},\bar{y})$ 定义为 $\bar{x} = M_y / m$，$\bar{y} = M_x / m$，其中 $m$ 为总质量，$M_y$、$M_x$ 分别为对 $y$ 轴和 $x$ 轴的矩。
(a)A vertical strip at position $x$ of width $dx$ has area $f(x)\,dx$ and mass $dm = \delta f(x)\,dx$. Its center is at the point $(x,\, \tfrac{1}{2}f(x))$. Write down the contributions $dM_y$ and $dM_x$ from this strip, integrate, and hence derive the formulas
$$ \bar{x} = \frac{\displaystyle\int_a^b x\,f(x)\,dx}{\displaystyle\int_a^b f(x)\,dx}, \qquad \bar{y} = \frac{\displaystyle\int_a^b \tfrac{1}{2}[f(x)]^2\,dx}{\displaystyle\int_a^b f(x)\,dx}. $$
Explain why $\delta$ cancels.位置 $x$ 处宽度为 $dx$ 的竖直条，面积为 $f(x)\,dx$，质量为 $dm = \delta f(x)\,dx$，其中心在 $(x,\, \tfrac{1}{2}f(x))$。写出该条对 $dM_y$ 和 $dM_x$ 的贡献，积分后推导出公式
$$ \bar{x} = \frac{\displaystyle\int_a^b x\,f(x)\,dx}{\displaystyle\int_a^b f(x)\,dx}, \qquad \bar{y} = \frac{\displaystyle\int_a^b \tfrac{1}{2}[f(x)]^2\,dx}{\displaystyle\int_a^b f(x)\,dx}. $$
解释为何 $\delta$ 可以消去。[5]
(b)For the region between $f(x)$ (top) and $g(x)$ (bottom) with $f(x) \ge g(x)$ on $[a,b]$, state without full re-derivation the correct formula for $\bar{y}$, and explain in one sentence why it uses $\tfrac{1}{2}[f(x)^2 - g(x)^2]$ rather than $\tfrac{1}{2}[f(x)-g(x)]^2$.对于 $[a,b]$ 上 $f(x)$（上方）与 $g(x)$（下方）之间的区域（$f(x) \ge g(x)$），无需完整重新推导，直接写出 $\bar{y}$ 的正确公式，并用一句话解释为何使用 $\tfrac{1}{2}[f(x)^2 - g(x)^2]$ 而非 $\tfrac{1}{2}[f(x)-g(x)]^2$。[3]
A continuous random variable $X$ has probability density function $f(x) = c\,x^2(3 - x)$ for $0 \le x \le 3$ (and $f(x) = 0$ otherwise), where $c$ is a positive constant.连续随机变量 $X$ 的概率密度函数为 $f(x) = c\,x^2(3 - x)$（$0 \le x \le 3$，其余为零），其中 $c$ 为正常数。
(a)Find the constant $c$ that makes $f$ a valid probability density function.求使 $f$ 成为有效概率密度函数的常数 $c$。[2]
(b)Find the mean $\mu = E[X]$.求均值 $\mu = E[X]$。[3]
(c)Find the probability $P(1 \le X \le 2)$. Give an exact answer.求概率 $P(1 \le X \le 2)$，给出精确答案。[3]
(d)State the defining equation for the median $m$ of $X$. Compute $P(X \le 2)$ exactly and use it to show that the median satisfies $m \in (0, 2)$.写出 $X$ 的中位数 $m$ 的定义方程。精确计算 $P(X \le 2)$，并据此说明中位数满足 $m \in (0, 2)$。[2]
Q7HARDPROOFhydrostatic force: slicing argument and the centroid shortcut静水压力：切片论证与形心捷径[8 marks]
A vertical triangular gate is submerged in water (weight density $w = 9800\,\text{N/m}^3$). The gate is an isosceles triangle with its horizontal base of width $6\,\text{m}$ at the water surface and its apex pointing downward at depth $4\,\text{m}$. Let $h$ denote depth below the water surface, with $0 \le h \le 4$.一块竖直三角形闸门浸没在水中（重量密度 $w = 9800\,\text{N/m}^3$）。闸门为等腰三角形，宽 $6\,\text{m}$ 的水平底边位于水面，顶点向下，深度 $4\,\text{m}$。设 $h$ 为水面以下深度，$0 \le h \le 4$。
(a)Write a formula for the strip width $L(h)$ at depth $h$, and hence write the hydrostatic force element $dF$ on a horizontal strip of thickness $dh$ at depth $h$.写出深度 $h$ 处条带宽度 $L(h)$ 的公式，进而写出深度 $h$ 处厚度为 $dh$ 的水平条带所受静水压力元素 $dF$。[3]
(b)Set up and evaluate $\displaystyle F = \int_0^4 dF$ to find the total hydrostatic force on the gate.建立并计算 $\displaystyle F = \int_0^4 dF$，求闸门所受总静水压力。[3]
(c)The centroid shortcut states $F = w\,\bar{h}\,A$, where $\bar{h}$ is the depth of the centroid and $A$ is the area of the gate. Compute $A$ and hence find $\bar{h}$ from your answer to (b). State the depth of the centroid of this triangle and comment on whether it is consistent with the formula for the centroid of a triangle (which lies one-third of the height from the base).形心捷径公式为 $F = w\,\bar{h}\,A$，其中 $\bar{h}$ 为形心深度，$A$ 为闸门面积。计算 $A$，并由 (b) 的答案求 $\bar{h}$。说明该三角形形心的深度，并评论是否与三角形形心公式（距底边三分之一高处）一致。[2]
PART III · APPLICATIONS AND SYNTHESIS第三部分 · 应用与综合Extended problems · 28 marks综合题 · 28分
Pumping, Hydrostatic Force, and Centroids抽水、静水压力与形心
Set up each problem from first principles: name the variable, write the element, integrate. Carry exact values. A diagram showing the variable and limits earns method credit.从基本原理出发建立每道题：命名变量，写出元素，积分。保持精确值。画出变量与积分限的示意图可获得方法分。
Q8HARDAPPLIEDpumping from an inverted conical tank从倒置锥形容器抽液体[10 marks]
An inverted right-circular cone (vertex pointing downward) has height $12\,\text{m}$ and top radius $3\,\text{m}$. It is filled with a liquid of weight density $w = 8000\,\text{N/m}^3$ to a depth of $9\,\text{m}$ measured from the vertex. Water is to be pumped out over the top rim of the cone. Let $y$ measure height above the vertex.一个倒置的直圆锥（顶点向下），高 $12\,\text{m}$，顶部半径 $3\,\text{m}$，注有重量密度 $w = 8000\,\text{N/m}^3$ 的液体，从顶点量起深度为 $9\,\text{m}$。液体将从锥形容器顶部边缘抽出。设 $y$ 为顶点以上的高度。
(a)Use similar triangles to find the radius $r(y)$ of the liquid's cross-section at height $y$ above the vertex.用相似三角形求顶点以上高度 $y$ 处液体横截面的半径 $r(y)$。[2]
(b)Write the cross-sectional area $A(y)$ and the lift distance $d(y)$ of a thin horizontal slab at height $y$. Hence write $dW$.写出高度 $y$ 处薄水平层的横截面面积 $A(y)$ 和提升距离 $d(y)$，进而写出 $dW$。[3]
(c)Set up and evaluate $W = \displaystyle\int_0^9 dW$. Give an exact answer in terms of $\pi$ and a decimal approximation.建立并计算 $W = \displaystyle\int_0^9 dW$。给出以 $\pi$ 表示的精确答案和十进制近似值。[3]
(d)If the cone were instead full to the brim ($9\,\text{m}$ becomes $12\,\text{m}$, all dimensions the same), by what factor would the total work increase? You do not need to evaluate the new integral, only to identify how the upper limit changes and what the ratio of the two integrals is.若锥形容器注满至顶（$9\,\text{m}$ 变为 $12\,\text{m}$，其他尺寸不变），总功将增大多少倍？无需计算新积分，只需说明上限如何变化以及两个积分之比。[2]
Q9HARDAPPLIEDhydrostatic force on a vertical submerged plate竖直浸没平板上的静水压力[10 marks]
A vertical rectangular dam wall is $8\,\text{m}$ wide and retains water to a depth of $5\,\text{m}$. A circular observation window of radius $1\,\text{m}$ is set into the dam wall so that its center is $3\,\text{m}$ below the water surface. The weight density of water is $w = 9800\,\text{N/m}^3$.一面竖直矩形坝墙宽 $8\,\text{m}$，拦截水深 $5\,\text{m}$。坝墙上有一个半径 $1\,\text{m}$ 的圆形观察窗，圆心位于水面以下 $3\,\text{m}$ 处。水的重量密度 $w = 9800\,\text{N/m}^3$。
(a)Find the total hydrostatic force on the rectangular dam wall (the full $8\,\text{m}$ by $5\,\text{m}$ face).求矩形坝墙（整个 $8\,\text{m} \times 5\,\text{m}$ 面）所受的总静水压力。[3]
(b)Place the origin at the center of the circular window, with $y$ increasing upward. A horizontal strip of the window at height $y$ (where $-1 \le y \le 1$) is at depth $h = 3 - y$ below the surface and has width $L(y) = 2\sqrt{1-y^2}$. Set up the integral for the hydrostatic force on the circular window and split it into two parts by linearity.以圆形窗圆心为原点，$y$ 轴向上。窗口高度 $y$（$-1 \le y \le 1$）处的水平条带距水面深度 $h = 3 - y$，宽度 $L(y) = 2\sqrt{1-y^2}$。建立圆形窗所受静水压力的积分，并按线性性质拆分为两部分。[4]
(c)Evaluate each part using the results $\displaystyle\int_{-1}^{1}\sqrt{1-y^2}\,dy = \frac{\pi}{2}$ (area of a unit semicircle) and $\displaystyle\int_{-1}^{1}y\sqrt{1-y^2}\,dy = 0$ (odd integrand). Hence state the force on the window, and verify it equals $w \cdot \bar{h} \cdot A$ where $A = \pi$ and $\bar{h} = 3$.利用 $\displaystyle\int_{-1}^{1}\sqrt{1-y^2}\,dy = \frac{\pi}{2}$（单位半圆面积）和 $\displaystyle\int_{-1}^{1}y\sqrt{1-y^2}\,dy = 0$（奇函数）计算各部分。据此写出窗口所受压力，并验证其等于 $w \cdot \bar{h} \cdot A$（$A = \pi$，$\bar{h} = 3$）。[3]
Q10HARDAPPLIEDcentroid of a region between two curves两曲线间区域的形心[8 marks]
Let $R$ be the region bounded above by $y = 4 - x^2$ and below by $y = x^2 - 2x$ on the interval $[0, 2]$. (You may check that the top curve is always at or above the bottom curve on this interval.)设 $R$ 为区间 $[0, 2]$ 上，上方由 $y = 4 - x^2$、下方由 $y = x^2 - 2x$ 围成的区域。（可验证在此区间上上方曲线始终不低于下方曲线。）
(a)Find the area $A$ of the region $R$.求区域 $R$ 的面积 $A$。[2]
(b)Find $\bar{x}$, the $x$-coordinate of the centroid of $R$.求 $R$ 的形心 $x$ 坐标 $\bar{x}$。[3]
(c)Find $\bar{y}$, the $y$-coordinate of the centroid of $R$, using
$$ \bar{y} = \frac{1}{A}\int_0^2 \tfrac{1}{2}\!\left[(4-x^2)^2 - (x^2-2x)^2\right]dx. $$
Show the algebraic expansion clearly.用下式求 $R$ 的形心 $y$ 坐标 $\bar{y}$：
$$ \bar{y} = \frac{1}{A}\int_0^2 \tfrac{1}{2}\!\left[(4-x^2)^2 - (x^2-2x)^2\right]dx. $$
清晰展示代数展开过程。[3]