Companion to the University-Style Practice Set大学风格练习题配套解析
Sections 1 to 7: area between curves, disk and washer volumes, cylindrical shells, cross-sections, arc length, surface area第1至7节：曲线间面积、圆盘与垫圈体积、柱壳法、截面法、弧长、曲面面积CALC II
Region $R$ enclosed by $y=4-x^{2}$ and $y=x+2$: (a) area integral in $x$; (b) rewrite as integral in $y$.$y=4-x^{2}$与$y=x+2$围成的区域$R$：(a) 关于$x$的面积积分；(b) 改写为关于$y$的积分。
Set the curves equal to locate the intersection points: $4-x^{2}=x+2$, so $x^{2}+x-2=(x+2)(x-1)=0$, giving $x=-2$ and $x=1$. (M1) On the interval $(-2,1)$ test $x=0$: the parabola gives $4$ and the line gives $2$, so $y=4-x^{2}$ lies above $y=x+2$.令两曲线相等以求交点：$4-x^{2}=x+2$，故$x^{2}+x-2=(x+2)(x-1)=0$，得$x=-2$与$x=1$。(M1) 在区间$(-2,1)$上取$x=0$验证：抛物线给出$4$，直线给出$2$，故$y=4-x^{2}$在$y=x+2$上方。
$$A = \int_{-2}^{1}\!\bigl[(4-x^{2})-(x+2)\bigr]\,dx = \int_{-2}^{1}(2-x-x^{2})\,dx.$$Evaluate: $\left[2x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{-2}^{1} = \!\left(2-\tfrac{1}{2}-\tfrac{1}{3}\right)-\!\left(-4-2+\tfrac{8}{3}\right) = \tfrac{7}{6}-\!\left(-6+\tfrac{8}{3}\right) = \tfrac{7}{6}+\tfrac{10}{3} = \tfrac{7}{6}+\tfrac{20}{6} = \dfrac{9}{2}.$ (A1·A1)计算：$\left[2x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{-2}^{1} = \!\left(2-\tfrac{1}{2}-\tfrac{1}{3}\right)-\!\left(-4-2+\tfrac{8}{3}\right) = \tfrac{7}{6}-\!\left(-6+\tfrac{8}{3}\right) = \tfrac{7}{6}+\tfrac{10}{3} = \tfrac{7}{6}+\tfrac{20}{6} = \dfrac{9}{2}.$ (A1·A1)
The region $R$ spans $y\in[0,4]$: the lowest point of the enclosed boundary is the intersection at $(-2,0)$ ($y=0$), and the highest point is the top of the parabola at $(0,4)$ ($y=4$). (M1) The parabola $y=4-x^{2}$ gives two branches: $x=\sqrt{4-y}$ (right) and $x=-\sqrt{4-y}$ (left). The line gives $x=y-2$.区域$R$的$y$范围为$[0,4]$：围合边界的最低点为交点$(-2,0)$（$y=0$），最高点为抛物线顶点$(0,4)$（$y=4$）。(M1) 抛物线$y=4-x^{2}$给出两个分支：$x=\sqrt{4-y}$（右侧）和$x=-\sqrt{4-y}$（左侧）。直线给出$x=y-2$。
A horizontal slice at height $y$ must satisfy $y\ge x+2$ (above the line, i.e. $x\le y-2$) and $y\le 4-x^{2}$ (below the parabola, i.e. $-\sqrt{4-y}\le x\le\sqrt{4-y}$). Combining:在高度$y$处的水平切片需满足$y\ge x+2$（直线上方，即$x\le y-2$）与$y\le 4-x^{2}$（抛物线下方，即$-\sqrt{4-y}\le x\le\sqrt{4-y}$）。综合得：
Hence the area integral in $y$ requires a split: (A1 for identifying the split at $y=3$, A1 for correct integrands and limits)因此关于$y$的面积积分需要分段：（A1 识别在$y=3$处分段，A1 被积函数与积分上下限正确）
$$A = \int_{0}^{3}\!\Bigl[(y-2)-(-\sqrt{4-y})\Bigr]\,dy + \int_{3}^{4}\!\Bigl[\sqrt{4-y}-(-\sqrt{4-y})\Bigr]\,dy$$ $$= \int_{0}^{3}(y-2+\sqrt{4-y})\,dy + \int_{3}^{4}2\sqrt{4-y}\,dy.$$Verification: $\displaystyle\int_{0}^{3}(y-2)\,dy+\int_{0}^{3}\sqrt{4-y}\,dy = -\tfrac{3}{2}+\tfrac{14}{3}=\tfrac{19}{6}$; $\displaystyle\int_{3}^{4}2\sqrt{4-y}\,dy=\tfrac{4}{3}$; total $=\tfrac{19}{6}+\tfrac{4}{3}=\tfrac{27}{6}=\tfrac{9}{2}$. ✓验证：$\displaystyle\int_{0}^{3}(y-2)\,dy+\int_{0}^{3}\sqrt{4-y}\,dy = -\tfrac{3}{2}+\tfrac{14}{3}=\tfrac{19}{6}$；$\displaystyle\int_{3}^{4}2\sqrt{4-y}\,dy=\tfrac{4}{3}$；合计$=\tfrac{19}{6}+\tfrac{4}{3}=\tfrac{27}{6}=\tfrac{9}{2}$。✓
$y=\sin x$ and $y=\cos x$ on $\left[0,\tfrac{3\pi}{2}\right]$: (a) intersections and sign of gap; (b) total area as a sum of integrals.$\left[0,\tfrac{3\pi}{2}\right]$上的$y=\sin x$与$y=\cos x$：(a) 交点与差值符号；(b) 总面积表示为积分之和。
$\sin x = \cos x \Rightarrow \tan x=1$, so $x=\tfrac{\pi}{4}+k\pi$. On $\left[0,\tfrac{3\pi}{2}\right]$ these are $x=\tfrac{\pi}{4}$ and $x=\tfrac{5\pi}{4}$. (M1) Test values: at $x=0$, $\cos 0=1>\sin 0=0$, so $\cos x>\sin x$ on $\left[0,\tfrac{\pi}{4}\right)$; at $x=\tfrac{\pi}{2}$, $\sin>0=\cos$, so $\sin x>\cos x$ on $\left(\tfrac{\pi}{4},\tfrac{5\pi}{4}\right)$; at $x=\tfrac{4\pi}{3}$, $\cos(-60^{\circ})=\tfrac{1}{2}>-\tfrac{\sqrt{3}}{2}=\sin$, so $\cos x>\sin x$ on $\left(\tfrac{5\pi}{4},\tfrac{3\pi}{2}\right]$. (A1)$\sin x = \cos x \Rightarrow \tan x=1$，故$x=\tfrac{\pi}{4}+k\pi$。在$\left[0,\tfrac{3\pi}{2}\right]$上这些点为$x=\tfrac{\pi}{4}$和$x=\tfrac{5\pi}{4}$。(M1) 取验证值：在$x=0$处，$\cos 0=1>\sin 0=0$，故$\left[0,\tfrac{\pi}{4}\right)$上$\cos x>\sin x$；在$x=\tfrac{\pi}{2}$处，$\sin>\cos=0$，故$\left(\tfrac{\pi}{4},\tfrac{5\pi}{4}\right)$上$\sin x>\cos x$；在$x=\tfrac{4\pi}{3}$处，$\cos(-60^{\circ})=\tfrac{1}{2}>-\tfrac{\sqrt{3}}{2}=\sin$，故$\left(\tfrac{5\pi}{4},\tfrac{3\pi}{2}\right]$上$\cos x>\sin x$。(A1)
Total area $= \displaystyle\int_{0}^{\pi/4}(\cos x-\sin x)\,dx + \int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx + \int_{5\pi/4}^{3\pi/2}(\cos x-\sin x)\,dx.$ (M1)总面积 $= \displaystyle\int_{0}^{\pi/4}(\cos x-\sin x)\,dx + \int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx + \int_{5\pi/4}^{3\pi/2}(\cos x-\sin x)\,dx.$ (M1)
Antiderivative of $\cos x-\sin x$ is $\sin x+\cos x$; antiderivative of $\sin x-\cos x$ is $-\cos x-\sin x$.$\cos x-\sin x$的原函数为$\sin x+\cos x$；$\sin x-\cos x$的原函数为$-\cos x-\sin x$。
First integral: $\bigl[\sin x+\cos x\bigr]_{0}^{\pi/4} = \bigl(\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}\bigr)-(0+1) = \sqrt{2}-1.$ (A1)第一个积分：$\bigl[\sin x+\cos x\bigr]_{0}^{\pi/4} = \bigl(\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}\bigr)-(0+1) = \sqrt{2}-1.$ (A1)
Second integral: $\bigl[-\cos x-\sin x\bigr]_{\pi/4}^{5\pi/4} = \bigl(\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}\bigr)-\bigl(-\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\bigr) = \sqrt{2}+\sqrt{2} = 2\sqrt{2}.$第二个积分：$\bigl[-\cos x-\sin x\bigr]_{\pi/4}^{5\pi/4} = \bigl(\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}\bigr)-\bigl(-\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\bigr) = \sqrt{2}+\sqrt{2} = 2\sqrt{2}.$
Third integral: $\bigl[\sin x+\cos x\bigr]_{5\pi/4}^{3\pi/2} = (-1+0)-\bigl(-\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\bigr) = -1+\sqrt{2} = \sqrt{2}-1.$第三个积分：$\bigl[\sin x+\cos x\bigr]_{5\pi/4}^{3\pi/2} = (-1+0)-\bigl(-\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\bigr) = -1+\sqrt{2} = \sqrt{2}-1.$
Total $= (\sqrt{2}-1)+2\sqrt{2}+(\sqrt{2}-1) = 4\sqrt{2}-2.$ (A1)合计 $= (\sqrt{2}-1)+2\sqrt{2}+(\sqrt{2}-1) = 4\sqrt{2}-2.$ (A1)
Region $R$ bounded by $y=\sqrt{x}$, $y=0$, $x=4$: (a) disk about $x$-axis; (b) washer about $y$-axis integrating in $y$.由$y=\sqrt{x}$、$y=0$、$x=4$围成的区域$R$：(a) 绕$x$轴的圆盘法；(b) 绕$y$轴的垫圈法（关于$y$积分）。
Each cross-section perpendicular to the $x$-axis is a disk with radius $f(x)=\sqrt{x}$ and area $\pi(\sqrt{x})^{2}=\pi x$. (M1)垂直于$x$轴的每个截面是半径为$f(x)=\sqrt{x}$、面积为$\pi(\sqrt{x})^{2}=\pi x$的圆盘。(M1)
$$V = \pi\int_{0}^{4} x\,dx = \pi\!\left[\frac{x^{2}}{2}\right]_{0}^{4} = \pi\cdot 8 = 8\pi.$$(A1·A1) The volume $8\pi$ is positive and dimensionally consistent with a solid of diameter $2\sqrt{4}=4$, confirming the order of magnitude. ✓（A1·A1）体积$8\pi$为正，与直径$2\sqrt{4}=4$的旋转体在数量级上一致，结果合理。✓
Rewrite the boundary in terms of $y$. The curve $y=\sqrt{x}$ becomes $x=y^{2}$; the line $x=4$ is constant. (M1) When revolving about the $y$-axis, integrate with respect to $y$ from $y=0$ to $y=\sqrt{4}=2$. At height $y$ the washer has outer radius $R=4$ (the flat wall at $x=4$) and inner radius $r=y^{2}$ (the curve $x=y^{2}$). (M1)用$y$改写边界。曲线$y=\sqrt{x}$变为$x=y^{2}$；直线$x=4$为常数。(M1) 绕$y$轴旋转时，关于$y$从$y=0$到$y=\sqrt{4}=2$积分。在高度$y$处，垫圈的外半径为$R=4$（$x=4$处的平面壁），内半径为$r=y^{2}$（曲线$x=y^{2}$）。(M1)
$$V = \pi\int_{0}^{2}\!\bigl[4^{2}-(y^{2})^{2}\bigr]\,dy = \pi\int_{0}^{2}(16-y^{4})\,dy.$$ $$= \pi\!\left[16y-\frac{y^{5}}{5}\right]_{0}^{2} = \pi\!\left(32-\frac{32}{5}\right) = \pi\cdot\frac{128}{5} = \frac{128\pi}{5}.$$(A1 integrand, A1 limits, A1 answer) Sanity check: the enclosing cylinder has volume $\pi(4)^{2}(2)=32\pi$; the actual solid is $\tfrac{128\pi}{5}\approx 80.4$, less than $32\pi\approx 100.5$. ✓（A1 被积函数，A1 积分限，A1 答案）合理性检验：包围圆柱的体积为$\pi(4)^{2}(2)=32\pi$；实际体积$\tfrac{128\pi}{5}\approx 80.4$，小于$32\pi\approx 100.5$。✓
Region $R$ enclosed by $y=x^{2}$ and $y=2x$: (a) intersection points; (b) volume revolving $R$ about $y=-1$ by the washer method.$y=x^{2}$与$y=2x$围成的区域$R$：(a) 交点；(b) 用垫圈法求$R$绕$y=-1$旋转的体积。
Set $x^{2}=2x$: $x^{2}-2x=x(x-2)=0$, so $x=0$ and $x=2$. (M1·A1) On $(0,2)$ the line $y=2x$ lies above the parabola $y=x^{2}$ (test $x=1$: $2>1$).令$x^{2}=2x$：$x^{2}-2x=x(x-2)=0$，故$x=0$与$x=2$。(M1·A1) 在$(0,2)$上，直线$y=2x$位于抛物线$y=x^{2}$上方（取$x=1$验证：$2>1$）。
The axis of revolution is $y=-1$, one unit below the $x$-axis. Both curves lie at or above $y=0$, so they are at or above the axis. The outer radius at position $x$ is the distance from $y=-1$ to the farther curve, which is the upper curve $y=2x$: (M1)旋转轴为$y=-1$，位于$x$轴下方一个单位。两条曲线均在$y=0$处或其上方，故均在旋转轴上方。位置$x$处的外半径是从$y=-1$到较远曲线（即上曲线$y=2x$）的距离：(M1)
$$R(x) = 2x-(-1) = 2x+1.$$The inner radius is the distance from $y=-1$ to the nearer curve, which is the lower curve $y=x^{2}$: (M1)内半径是从$y=-1$到较近曲线（即下曲线$y=x^{2}$）的距离：(M1)
$$r(x) = x^{2}-(-1) = x^{2}+1.$$Both radii are positive on $[0,2]$. (A1 for each radius expression) The washer formula gives:两个半径在$[0,2]$上均为正。（每个半径表达式各得A1）垫圈公式给出：
$$V = \pi\int_{0}^{2}\!\bigl[(2x+1)^{2}-(x^{2}+1)^{2}\bigr]\,dx.$$Expand: $(2x+1)^{2}=4x^{2}+4x+1$ and $(x^{2}+1)^{2}=x^{4}+2x^{2}+1$. Their difference is $-x^{4}+2x^{2}+4x$. (A1)展开：$(2x+1)^{2}=4x^{2}+4x+1$，$(x^{2}+1)^{2}=x^{4}+2x^{2}+1$。二者之差为$-x^{4}+2x^{2}+4x$。(A1)
$$V = \pi\int_{0}^{2}(-x^{4}+2x^{2}+4x)\,dx = \pi\!\left[-\frac{x^{5}}{5}+\frac{2x^{3}}{3}+2x^{2}\right]_{0}^{2}$$ $$= \pi\!\left(-\frac{32}{5}+\frac{16}{3}+8\right) = \pi\!\left(\frac{-96+80+120}{15}\right) = \frac{104\pi}{15}.$$(A1) Sanity check: the enclosing washer cylinder (outer radius $2(2)+1=5$, inner radius $0^{2}+1=1$, length $2$) has volume $\pi(25-1)(2)=48\pi$; the actual volume $\tfrac{104\pi}{15}\approx 21.8$ is smaller. ✓（A1）合理性检验：包围垫圈圆柱（外半径$2(2)+1=5$，内半径$0^{2}+1=1$，长度$2$）的体积为$\pi(25-1)(2)=48\pi$；实际体积$\tfrac{104\pi}{15}\approx 21.8$较小。✓
(a) Derive the shell volume element $2\pi r h\,dr$ from exact shell volume. (b) Apply the shell formula to revolve the region under $y=x(2-x)$, $0\le x\le 2$, about the $y$-axis.(a) 从精确壳体积推导体积元$2\pi r h\,dr$。(b) 将柱壳公式应用于$y=x(2-x)$（$0\le x\le 2$）下方区域绕$y$轴旋转的问题。
A cylindrical shell is the region between two coaxial cylinders of the same height $h$, inner radius $r$, and outer radius $r+\Delta r$. Its volume equals the volume of the outer cylinder minus the volume of the inner cylinder: (M1)柱壳是两个同轴圆柱之间的区域，其高度均为$h$，内径为$r$，外径为$r+\Delta r$。其体积等于外圆柱体积减去内圆柱体积：(M1)
$$V_{\text{shell}} = \pi(r+\Delta r)^{2}h - \pi r^{2}h = \pi h\bigl[(r+\Delta r)^{2}-r^{2}\bigr] = \pi h(2r\,\Delta r+(\Delta r)^{2}).$$Factor: $V_{\text{shell}} = \pi(2r+\Delta r)h\,\Delta r$. (A1)提取公因子：$V_{\text{shell}} = \pi(2r+\Delta r)h\,\Delta r$。(A1)
As $\Delta r\to 0$ the term $\Delta r$ in $(2r+\Delta r)$ vanishes, and summing infinitely many such shells via the Riemann integral gives the volume element $dV = 2\pi r h\,dr$. (R1) The resulting integral formula is $V = \displaystyle\int_{a}^{b} 2\pi r(x)\,h(x)\,dx$, where $r(x)$ is the shell radius and $h(x)$ is the shell height at position $x$.当$\Delta r\to 0$时，$(2r+\Delta r)$中的$\Delta r$项趋于零，通过黎曼积分对无限多个这样的壳求和，得到体积元$dV = 2\pi r h\,dr$。(R1) 由此得到积分公式$V = \displaystyle\int_{a}^{b} 2\pi r(x)\,h(x)\,dx$，其中$r(x)$为壳半径，$h(x)$为位置$x$处的壳高。
For a vertical shell at position $x\in[0,2]$: the shell radius is $r(x)=x$ (distance to the $y$-axis) and the shell height is $h(x)=x(2-x)=2x-x^{2}$. (M1)对于位置$x\in[0,2]$处的竖直壳：壳半径为$r(x)=x$（到$y$轴的距离），壳高为$h(x)=x(2-x)=2x-x^{2}$。(M1)
$$V = 2\pi\int_{0}^{2}x\cdot(2x-x^{2})\,dx = 2\pi\int_{0}^{2}(2x^{2}-x^{3})\,dx$$ $$= 2\pi\!\left[\frac{2x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{2} = 2\pi\!\left(\frac{16}{3}-4\right) = 2\pi\cdot\frac{4}{3} = \frac{8\pi}{3}.$$(A1 integrand, A1 antiderivative, A1 answer) Sanity check: the enclosing cylinder has radius $2$ and height $1$ (the maximum of $x(2-x)$ at $x=1$), giving volume $4\pi$; the actual volume $\tfrac{8\pi}{3}\approx 8.38$ is slightly above the inscribed cylinder $2\pi(1)(1)=2\pi\approx 6.28$ and far below the enclosing value, which is reasonable. ✓（A1 被积函数，A1 原函数，A1 答案）合理性检验：包围圆柱半径为$2$，高为$1$（$x(2-x)$在$x=1$处的最大值），体积为$4\pi$；实际体积$\tfrac{8\pi}{3}\approx 8.38$略高于内切圆柱$2\pi(1)(1)=2\pi\approx 6.28$，远低于包围值，结果合理。✓
(a) Derive the arc-length integral from the chord-length approximation. (b) Find the exact arc length of $y=\ln(\cos x)$ from $x=0$ to $x=\tfrac{\pi}{4}$.(a) 从弦长近似推导弧长积分。(b) 求$y=\ln(\cos x)$从$x=0$到$x=\tfrac{\pi}{4}$的精确弧长。
Partition $[a,b]$ into subintervals of width $\Delta x$. The chord joining $(x,f(x))$ to $(x+\Delta x, f(x+\Delta x))$ has horizontal run $\Delta x$ and vertical rise $\Delta y = f(x+\Delta x)-f(x)$. By the Pythagorean theorem, the chord length is (M1)将$[a,b]$分成宽度为$\Delta x$的子区间。连接$(x,f(x))$与$(x+\Delta x, f(x+\Delta x))$的弦，水平跨度为$\Delta x$，竖直升差为$\Delta y = f(x+\Delta x)-f(x)$。由勾股定理，弦长为(M1)
$$\sqrt{(\Delta x)^{2}+(\Delta y)^{2}} = \sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^{2}}\,\Delta x.$$(M1) As $\Delta x\to 0$, the difference quotient $\tfrac{\Delta y}{\Delta x}\to f'(x)$ by the definition of the derivative, so each chord element tends to $\sqrt{1+[f'(x)]^{2}}\,dx$. (A1) Summing and passing to the limit gives the arc-length integral (R1):(M1) 当$\Delta x\to 0$时，由导数的定义，差商$\tfrac{\Delta y}{\Delta x}\to f'(x)$，故每个弦元趋于$\sqrt{1+[f'(x)]^{2}}\,dx$。(A1) 求和并取极限得弧长积分(R1)：
$$L = \int_{a}^{b}\sqrt{1+[f'(x)]^{2}}\,dx.$$Differentiate: $f'(x)=\dfrac{-\sin x}{\cos x}=-\tan x$, so $[f'(x)]^{2}=\tan^{2}x$. (M1)求导：$f'(x)=\dfrac{-\sin x}{\cos x}=-\tan x$，故$[f'(x)]^{2}=\tan^{2}x$。(M1)
Simplify under the radical using the Pythagorean identity $1+\tan^{2}x=\sec^{2}x$: (A1)利用勾股恒等式$1+\tan^{2}x=\sec^{2}x$化简根号内表达式：(A1)
$$\sqrt{1+\tan^{2}x} = \sqrt{\sec^{2}x} = |\sec x| = \sec x \quad \text{for } x\in\left[0,\tfrac{\pi}{4}\right].$$Hence: (A1)故：(A1)
$$L = \int_{0}^{\pi/4}\sec x\,dx = \Bigl[\ln|\sec x+\tan x|\Bigr]_{0}^{\pi/4}.$$At $x=\tfrac{\pi}{4}$: $\sec\tfrac{\pi}{4}=\sqrt{2}$ and $\tan\tfrac{\pi}{4}=1$, so the upper limit gives $\ln(\sqrt{2}+1)$. At $x=0$: $\sec 0+\tan 0=1$, so the lower limit gives $\ln 1=0$. (A1)在$x=\tfrac{\pi}{4}$处：$\sec\tfrac{\pi}{4}=\sqrt{2}$，$\tan\tfrac{\pi}{4}=1$，故上限给出$\ln(\sqrt{2}+1)$。在$x=0$处：$\sec 0+\tan 0=1$，故下限给出$\ln 1=0$。(A1)
$$L = \ln(\sqrt{2}+1).$$(a) Square cross-sections on base bounded by $y=\sqrt{x}$ and $y=\tfrac{x}{2}$. (b) Equilateral-triangle cross-sections on base disk $x^{2}+y^{2}\le 4$. (c) Why $V=\int A(x)\,dx$ encompasses disks and washers.(a) 以$y=\sqrt{x}$与$y=\tfrac{x}{2}$围成区域为底面的正方形截面体积。(b) 以圆盘$x^{2}+y^{2}\le 4$为底面的等边三角形截面体积。(c) 为何$V=\int A(x)\,dx$将圆盘法与垫圈法作为特例包含其中。
Intersection of $y=\sqrt{x}$ and $y=\tfrac{x}{2}$: $\sqrt{x}=\tfrac{x}{2}\Rightarrow x=\tfrac{x^{2}}{4}\Rightarrow x^{2}-4x=0\Rightarrow x=0$ or $x=4$. On $(0,4)$ test $x=1$: $\sqrt{1}=1>\tfrac{1}{2}$, so $y=\sqrt{x}$ is the upper boundary. (M1) The side length of the square at position $x$ equals the vertical gap:$y=\sqrt{x}$与$y=\tfrac{x}{2}$的交点：$\sqrt{x}=\tfrac{x}{2}\Rightarrow x=\tfrac{x^{2}}{4}\Rightarrow x^{2}-4x=0\Rightarrow x=0$或$x=4$。在$(0,4)$上取$x=1$验证：$\sqrt{1}=1>\tfrac{1}{2}$，故$y=\sqrt{x}$为上边界。(M1) 位置$x$处正方形的边长等于竖向间距：
$$s(x) = \sqrt{x}-\frac{x}{2},\qquad A(x) = s(x)^{2} = \left(\sqrt{x}-\frac{x}{2}\right)^{2} = x - x^{3/2}+\frac{x^{2}}{4}.$$ $$V = \int_{0}^{4}\!\left(x-x^{3/2}+\frac{x^{2}}{4}\right)dx = \left[\frac{x^{2}}{2}-\frac{2}{5}x^{5/2}+\frac{x^{3}}{12}\right]_{0}^{4}. \quad \text{(A1)}$$ $$= \left(8-\frac{2}{5}\cdot 32+\frac{64}{12}\right) = 8-\frac{64}{5}+\frac{16}{3} = \frac{120-192+80}{15} = \frac{8}{15}. \quad \text{(A1·A1)}$$At position $x\in[-2,2]$, the chord of the disk $x^{2}+y^{2}\le 4$ perpendicular to the $x$-axis has half-length $\sqrt{4-x^{2}}$, so the full base side is $s(x)=2\sqrt{4-x^{2}}$. (M1) The area of an equilateral triangle with side $s$ is $\tfrac{\sqrt{3}}{4}s^{2}$:在位置$x\in[-2,2]$处，圆盘$x^{2}+y^{2}\le 4$垂直于$x$轴的弦半长为$\sqrt{4-x^{2}}$，故完整底边长为$s(x)=2\sqrt{4-x^{2}}$。(M1) 边长为$s$的等边三角形面积为$\tfrac{\sqrt{3}}{4}s^{2}$：
$$A(x) = \frac{\sqrt{3}}{4}\cdot\bigl(2\sqrt{4-x^{2}}\bigr)^{2} = \frac{\sqrt{3}}{4}\cdot 4(4-x^{2}) = \sqrt{3}(4-x^{2}). \quad \text{(A1)}$$ $$V = \int_{-2}^{2}\sqrt{3}(4-x^{2})\,dx = \sqrt{3}\!\left[4x-\frac{x^{3}}{3}\right]_{-2}^{2} = \sqrt{3}\!\left[\!\left(8-\frac{8}{3}\right)-\!\left(-8+\frac{8}{3}\right)\!\right] \quad \text{(A1)}$$ $$= \sqrt{3}\!\left(16-\frac{16}{3}\right) = \sqrt{3}\cdot\frac{32}{3} = \frac{32\sqrt{3}}{3}. \quad \text{(A1)}$$When a region is revolved about the $x$-axis, every cross-section perpendicular to the $x$-axis is a disk (area $\pi R(x)^{2}$) or a washer (area $\pi[R(x)^{2}-r(x)^{2}]$); substituting either formula into $V=\int A(x)\,dx$ recovers the disk and washer formulas respectively. (R1·A1) The general formula covers any shape of cross-section, not only circles.当区域绕$x$轴旋转时，垂直于$x$轴的每个截面是圆盘（面积$\pi R(x)^{2}$）或垫圈（面积$\pi[R(x)^{2}-r(x)^{2}]$）；将任一公式代入$V=\int A(x)\,dx$，分别还原为圆盘公式和垫圈公式。(R1·A1) 一般公式适用于任意形状的截面，不仅限于圆形。
Region $R$ bounded by $y=x^{3}$, $x=0$, $y=8$, revolved about the $y$-axis: (a) sketch; (b) washer method; (c) shell method.由$y=x^{3}$、$x=0$、$y=8$围成的区域$R$绕$y$轴旋转：(a) 草图；(b) 垫圈法；(c) 柱壳法。
The boundary consists of three pieces: the curve $y=x^{3}$ from $(0,0)$ to $(2,8)$; the horizontal line $y=8$ from $(0,8)$ to $(2,8)$; and the $y$-axis ($x=0$) from $(0,0)$ to $(0,8)$. The region is bounded on the left by the $y$-axis, on the right by the cubic, and on top by the horizontal line. (A1 for each of two distinct features correctly identified.)边界由三部分组成：从$(0,0)$到$(2,8)$的曲线$y=x^{3}$；从$(0,8)$到$(2,8)$的水平线$y=8$；以及从$(0,0)$到$(0,8)$的$y$轴（$x=0$）。区域左侧以$y$轴为界，右侧以三次曲线为界，上方以水平线为界。（正确识别两个不同特征各得A1。）
Rewrite the right boundary as $x=y^{1/3}$ (the cube root). Revolving about the $y$-axis: at height $y\in[0,8]$ the cross-section is a disk (no hole, since the left boundary is the axis itself) with radius $R(y)=y^{1/3}$. (M1)将右边界改写为$x=y^{1/3}$（立方根）。绕$y$轴旋转：在高度$y\in[0,8]$处，截面是圆盘（无孔，因为左边界即为旋转轴），半径$R(y)=y^{1/3}$。(M1)
$$V = \pi\int_{0}^{8}\!\left(y^{1/3}\right)^{2}\,dy = \pi\int_{0}^{8}y^{2/3}\,dy = \pi\!\left[\frac{y^{5/3}}{5/3}\right]_{0}^{8} = \frac{3\pi}{5}\!\left[y^{5/3}\right]_{0}^{8}. \quad \text{(A1)}$$ $$= \frac{3\pi}{5}\cdot 8^{5/3} = \frac{3\pi}{5}\cdot (2^{3})^{5/3} = \frac{3\pi}{5}\cdot 2^{5} = \frac{3\pi}{5}\cdot 32 = \frac{96\pi}{5}. \quad \text{(A1·A1)}$$A vertical shell at position $x\in[0,2]$ has radius $r(x)=x$ and height $h(x)=8-x^{3}$ (distance from the top line $y=8$ down to the curve $y=x^{3}$). (M1)位置$x\in[0,2]$处的竖直壳，半径$r(x)=x$，高度$h(x)=8-x^{3}$（从上方直线$y=8$到曲线$y=x^{3}$的距离）。(M1)
$$V = 2\pi\int_{0}^{2}x(8-x^{3})\,dx = 2\pi\int_{0}^{2}(8x-x^{4})\,dx$$ $$= 2\pi\!\left[4x^{2}-\frac{x^{5}}{5}\right]_{0}^{2} = 2\pi\!\left(16-\frac{32}{5}\right) = 2\pi\cdot\frac{48}{5} = \frac{96\pi}{5}. \quad \text{(A1·A1·A1)}$$Both methods agree: $V=\dfrac{96\pi}{5}$. ✓两种方法结果一致：$V=\dfrac{96\pi}{5}$。✓
Curve $y=\dfrac{x^{3}}{6}+\dfrac{1}{2x}$ on $[1,3]$: (a) show $1+(y')^{2}$ is a perfect square; (b) exact arc length; (c) structural reason for the simplification.$[1,3]$上的曲线$y=\dfrac{x^{3}}{6}+\dfrac{1}{2x}$：(a) 证明$1+(y')^{2}$为完全平方式；(b) 精确弧长；(c) 化简成立的结构性原因。
Differentiate term by term: (M1)逐项求导：(M1)
$$\frac{dy}{dx} = \frac{x^{2}}{2}-\frac{1}{2x^{2}}.$$Square the derivative:对导数平方：
$$\left(\frac{dy}{dx}\right)^{2} = \left(\frac{x^{2}}{2}-\frac{1}{2x^{2}}\right)^{2} = \frac{x^{4}}{4}-2\cdot\frac{x^{2}}{2}\cdot\frac{1}{2x^{2}}+\frac{1}{4x^{4}} = \frac{x^{4}}{4}-\frac{1}{2}+\frac{1}{4x^{4}}. \quad \text{(A1)}$$Add $1$:加$1$：
$$1+\left(\frac{dy}{dx}\right)^{2} = \frac{x^{4}}{4}+\frac{1}{2}+\frac{1}{4x^{4}} = \left(\frac{x^{2}}{2}+\frac{1}{2x^{2}}\right)^{2}. \quad \text{(A1)}$$This is a perfect square because the cross term $+\tfrac{1}{2}$ (from $2\cdot\tfrac{x^2}{2}\cdot\tfrac{1}{2x^2}$) exactly restores the $1$ that was added. (A1)这是完全平方式，因为交叉项$+\tfrac{1}{2}$（来自$2\cdot\tfrac{x^2}{2}\cdot\tfrac{1}{2x^2}$）恰好补足了所加的$1$。(A1)
Since $\dfrac{x^{2}}{2}+\dfrac{1}{2x^{2}}>0$ on $[1,3]$, the square root is simply that expression: (M1)由于$\dfrac{x^{2}}{2}+\dfrac{1}{2x^{2}}>0$在$[1,3]$上成立，平方根即为该表达式：(M1)
$$L = \int_{1}^{3}\!\left(\frac{x^{2}}{2}+\frac{1}{2x^{2}}\right)dx = \left[\frac{x^{3}}{6}-\frac{1}{2x}\right]_{1}^{3}. \quad \text{(A1)}$$ $$= \left(\frac{27}{6}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{2}\right) = \frac{26}{6}-\left(-\frac{2}{6}\right) = \frac{28}{6} = \frac{14}{3}. \quad \text{(A1)}$$Write $f'(x)=P(x)-Q(x)$ where $P=\dfrac{x^{2}}{2}$ and $Q=\dfrac{1}{2x^{2}}$. Then (R1)令$f'(x)=P(x)-Q(x)$，其中$P=\dfrac{x^{2}}{2}$，$Q=\dfrac{1}{2x^{2}}$。则(R1)
$$1+(P-Q)^{2} = 1+P^{2}-2PQ+Q^{2}.$$For this to equal $(P+Q)^{2}=P^{2}+2PQ+Q^{2}$, we need $1-2PQ=2PQ$, i.e. $PQ=\tfrac{1}{4}$. (A1)要使其等于$(P+Q)^{2}=P^{2}+2PQ+Q^{2}$，需要$1-2PQ=2PQ$，即$PQ=\tfrac{1}{4}$。(A1)
Check: $P\cdot Q=\dfrac{x^{2}}{2}\cdot\dfrac{1}{2x^{2}}=\dfrac{1}{4}$. ✓ The structural property is that the derivative splits into two parts whose product is the constant $\tfrac{1}{4}$; textbook arc-length problems are designed by choosing a function whose derivative has this property. (A1)验证：$P\cdot Q=\dfrac{x^{2}}{2}\cdot\dfrac{1}{2x^{2}}=\dfrac{1}{4}$。✓ 结构性质在于：导数可分解为两部分，其乘积为常数$\tfrac{1}{4}$；教材弧长题目正是通过选取导数具有该性质的函数来设计的。(A1)
Region $R$: $y=4-x^{2}$, $y=0$, $-2\le x\le 2$. (a) Shell method about $x=3$. (b) Why the washer method is harder here.区域$R$：$y=4-x^{2}$，$y=0$，$-2\le x\le 2$。(a) 绕$x=3$的柱壳法。(b) 为何垫圈法在此处更困难。
For a vertical shell at position $x\in[-2,2]$: the axis is $x=3$, which lies to the right of the entire region, so the shell radius is the distance from $x$ to the axis: (M1)对于位置$x\in[-2,2]$处的竖直壳：旋转轴为$x=3$，位于整个区域的右侧，故壳半径为$x$到旋转轴的距离：(M1)
$$r(x) = 3-x \quad (\ge 1 > 0 \text{ on } [-2,2]).$$The shell height is $h(x)=4-x^{2}$ (the parabola above the $x$-axis). (M1)壳高为$h(x)=4-x^{2}$（$x$轴上方的抛物线）。(M1)
$$V = 2\pi\int_{-2}^{2}(3-x)(4-x^{2})\,dx.$$Expand the integrand: $(3-x)(4-x^{2})=12-3x^{2}-4x+x^{3}$. Split into even and odd parts: $12-3x^{2}$ is even and $-4x+x^{3}$ is odd. Since the interval is symmetric about $x=0$, the odd part integrates to zero: (A1)展开被积函数：$(3-x)(4-x^{2})=12-3x^{2}-4x+x^{3}$。分解为奇偶部分：$12-3x^{2}$为偶函数，$-4x+x^{3}$为奇函数。由于积分区间关于$x=0$对称，奇函数部分的积分为零：(A1)
$$V = 2\pi\int_{-2}^{2}(12-3x^{2})\,dx = 2\pi\cdot 2\int_{0}^{2}(12-3x^{2})\,dx = 4\pi\!\left[12x-x^{3}\right]_{0}^{2}$$ $$= 4\pi(24-8) = 4\pi\cdot 16 = 64\pi. \quad \text{(A1·A1)}$$The washer method would integrate with respect to $y$. At a given height $y\in[0,4]$, the parabola $y=4-x^{2}$ contributes two $x$-values: $x=\pm\sqrt{4-y}$. To revolve about $x=3$, the outer washer radius is $3-(-\sqrt{4-y})=3+\sqrt{4-y}$ (distance from $x=3$ to the left branch) and the inner radius is $3-\sqrt{4-y}$ (distance to the right branch). (R1·A1)垫圈法需关于$y$积分。在给定高度$y\in[0,4]$处，抛物线$y=4-x^{2}$给出两个$x$值：$x=\pm\sqrt{4-y}$。绕$x=3$旋转时，垫圈外半径为$3-(-\sqrt{4-y})=3+\sqrt{4-y}$（从$x=3$到左支的距离），内半径为$3-\sqrt{4-y}$（到右支的距离）。(R1·A1)
Setting up the difference of squares in $y$, we get $R^2 - r^2 = (3+\sqrt{4-y})^2-(3-\sqrt{4-y})^2 = 12\sqrt{4-y}$, and the resulting integral $\pi\int_0^4 12\sqrt{4-y}\,dy$ is straightforward but requires recognising and tracking both branches of the parabola simultaneously. For a non-symmetric region about a shifted axis, failure to identify both branches produces an incorrect washer. The shell method avoids this entirely by working in $x$. (A1)在$y$中建立平方差，得$R^2 - r^2 = (3+\sqrt{4-y})^2-(3-\sqrt{4-y})^2 = 12\sqrt{4-y}$，由此得到的积分$\pi\int_0^4 12\sqrt{4-y}\,dy$虽可计算，但需要同时识别并追踪抛物线的两个分支。对于关于偏移轴的非对称区域，若未能正确识别两个分支，将得到错误的垫圈。柱壳法通过关于$x$积分完全规避了这一问题。(A1)