Unit 8: Applications of Integration
单元 8：积分的应用
This unit turns the definite integral into a problem-solving tool — average values, net change, geometric area, volume from cross sections, solids of revolution, and arc length. The unifying idea is simple: an integral accumulates many tiny pieces into one total quantity.
本单元把定积分（definite integral）当作解题工具——平均值、净变化、几何面积、由横截面（cross-section）得到的体积、旋转体（solid of revolution）以及弧长。统一思想很简单：积分把无数微小片段累加成一个整体量。
Finding the Average Value of a Function on an Interval
在区间上求函数的平均值
The average value of a continuous function is the constant height that gives the same area as the original curve over the same interval. It is a “one-number summary” of a varying function.
连续函数的平均值（average value of a function）是这样一个常数高度：它在同一区间上所围矩形面积与原曲线下的面积相等。它是一个变化函数（function）的“单值概括”。
Geometrically, this is the height of a rectangle with width b − a whose area matches the signed area under the graph of f on [a,b].
几何上，这就是一个宽度为 b − a 的矩形高度，使其面积等于 f 在 [a,b] 上图形下方的带号面积。
Average value is not average rate of change. Average value measures a typical output level; average rate of change measures how much the output changes per unit input.
函数的平均值不是平均变化率（average rate of change）。平均值衡量典型的输出水平；平均变化率衡量每单位输入对应输出的变化量。
Worked Example 1
例题 1
Find the average value of $f(x)=x^2+1$ on $[0,2]$.
求 $f(x)=x^2+1$ 在 $[0,2]$ 上的平均值。
- Write the average-value formula:写出平均值公式： $$f_{\text{avg}}=\frac{1}{2-0}\int_0^2 (x^2+1)\,dx$$
- Evaluate the integral:计算积分（
integral）： $$\int_0^2 (x^2+1)\,dx=\left[\frac{x^3}{3}+x\right]_0^2=\frac{8}{3}+2=\frac{14}{3}$$ - Divide by the interval length:除以区间长度： $$f_{\text{avg}}=\frac{1}{2}\cdot\frac{14}{3}=\frac{7}{3}$$
The curve’s average height on [0,2] is $\frac73$. A rectangle of width 2 and height $\frac73$ has the same area as the region under $y=x^2+1$ from 0 to 2.
曲线在 [0,2] 上的平均高度为 $\frac73$。宽 2、高 $\frac73$ 的矩形与 $y=x^2+1$ 在 0 到 2 之间的区域面积相同。
Interactive Example — Average Value as an Equal-Area Rectangle
交互示例——把函数平均值看作等面积矩形
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Connecting Position, Velocity, and Acceleration Using Integrals
用积分串联位置、速度与加速度
If velocity is the rate of change of position, then integrating velocity accumulates displacement. If acceleration is the rate of change of velocity, then integrating acceleration accumulates the change in velocity.
若速度是位置的变化率（rate of change），则对速度积分就累积出位移。若加速度是速度的变化率，则对加速度积分就累积出速度的变化量。
Displacement keeps track of direction. Distance ignores direction and is always nonnegative. If velocity changes sign, you must account for that before computing distance.
位移（displacement）会保留方向信息。路程（distance）忽略方向，且恒为非负。如果速度变号，在计算路程前必须先处理这一点。
Worked Example — Velocity Changes Sign
例题——速度变号
A particle moves along the $x$-axis with velocity $v(t) = t^{2} - 4t + 3$ for $0 \le t \le 4$ (in m/s). Find (a) the displacement and (b) the total distance traveled on $[0, 4]$.
一质点沿 $x$ 轴运动，速度为 $v(t) = t^{2} - 4t + 3$，其中 $0 \le t \le 4$（单位 m/s）。求 (a) 位移；(b) 在 $[0, 4]$ 上的总路程。
- Find where $v(t)$ changes sign. Factor: $v(t) = (t-1)(t-3)$, so $v(t) = 0$ at $t = 1$ and $t = 3$. Sign chart on $[0, 4]$: $v(0.5) = 1.25 > 0$, $v(2) = -1 < 0$, $v(3.5) = 1.25 > 0$. So $v > 0$ on $(0, 1)$ and $(3, 4)$, and $v < 0$ on $(1, 3)$.
- 找出 $v(t)$ 的变号点。 因式分解：$v(t) = (t-1)(t-3)$，所以 $v(t) = 0$ 在 $t = 1$ 与 $t = 3$。在 $[0, 4]$ 上的符号表：$v(0.5) = 1.25 > 0$，$v(2) = -1 < 0$，$v(3.5) = 1.25 > 0$。所以 $v > 0$ 在 $(0, 1)$ 和 $(3, 4)$ 上，$v < 0$ 在 $(1, 3)$ 上。
- (a) Displacement ignores direction — just integrate $v$ straight through: $$\int_0^{4}(t^{2} - 4t + 3)\,dt \;=\; \Bigl[\tfrac{t^{3}}{3} - 2t^{2} + 3t\Bigr]_0^{4} \;=\; \tfrac{64}{3} - 32 + 12 \;=\; \tfrac{4}{3}\text{ m}.$$
- (a) 位移忽略方向——直接对 $v$ 在整个区间积分： $$\int_0^{4}(t^{2} - 4t + 3)\,dt \;=\; \Bigl[\tfrac{t^{3}}{3} - 2t^{2} + 3t\Bigr]_0^{4} \;=\; \tfrac{64}{3} - 32 + 12 \;=\; \tfrac{4}{3}\text{ m}.$$
- (b) Total distance uses $|v(t)|$. Split at each sign change so absolute value resolves into a definite sign:
$$\int_0^{4}|v(t)|\,dt \;=\; \int_0^{1}\!v\,dt \;+\; \int_1^{3}\!(-v)\,dt \;+\; \int_3^{4}\!v\,dt.$$
Each piece evaluates separately. Using the same antiderivative $\tfrac{t^{3}}{3} - 2t^{2} + 3t$:
- $\int_0^{1} v \,dt = \tfrac{1}{3} - 2 + 3 = \tfrac{4}{3}$.
- $\int_1^{3} v \,dt = (9 - 18 + 9) - (\tfrac{1}{3} - 2 + 3) = 0 - \tfrac{4}{3} = -\tfrac{4}{3}$. Take absolute value: $\tfrac{4}{3}$.
- $\int_3^{4} v \,dt = \tfrac{4}{3} - 0 = \tfrac{4}{3}$.
- (b) 总路程使用 $|v(t)|$。在每个变号点处分段，让绝对值落到确定的符号：
$$\int_0^{4}|v(t)|\,dt \;=\; \int_0^{1}\!v\,dt \;+\; \int_1^{3}\!(-v)\,dt \;+\; \int_3^{4}\!v\,dt.$$
每段分别计算。使用同一个原函数（
antiderivative） $\tfrac{t^{3}}{3} - 2t^{2} + 3t$：- $\int_0^{1} v \,dt = \tfrac{1}{3} - 2 + 3 = \tfrac{4}{3}$。
- $\int_1^{3} v \,dt = (9 - 18 + 9) - (\tfrac{1}{3} - 2 + 3) = 0 - \tfrac{4}{3} = -\tfrac{4}{3}$。 取绝对值：$\tfrac{4}{3}$。
- $\int_3^{4} v \,dt = \tfrac{4}{3} - 0 = \tfrac{4}{3}$。
Computing $\int_0^4 v\,dt$ and reporting that as the total distance. That gives the displacement $\tfrac{4}{3}$ m — the cancellation across the sign change has thrown away $\tfrac{8}{3}$ m of "back-and-forth" motion. Always split at sign changes before computing distance.
把 $\int_0^4 v\,dt$ 当成总路程。这只算出位移 $\tfrac{4}{3}$ m——变号处的正负相消已经抹掉了 $\tfrac{8}{3}$ m 的“来回”运动。计算路程前务必在变号点处分段。
Using Accumulation Functions and Definite Integrals in Applied Contexts
在应用情境中使用累积函数与定积分
An accumulation function records how much of a changing quantity has built up from a starting point to a variable endpoint. In AP Calculus, this usually appears in the form $F(x)=\int_a^x f(t)\,dt$.
累积函数（accumulation function）记录某个变化量从起点到可变终点之间累积了多少。在 AP 微积分中，它通常写成 $F(x)=\int_a^x f(t)\,dt$ 的形式。
In words: the integral of a rate gives net change in the underlying quantity. This is why words like accumulation, flows in, net increase, and amount added usually point to integration.
用文字描述：对一个变化率（rate）积分得到的是底层量的净变化（net change）。这就是为什么累积、流入、净增加、新增量这类词通常都指向积分。
If the problem gives a rate and asks for an amount, integrate. If the problem gives an amount and asks how fast it changes, differentiate.
若题目给出变化率而求总量，则积分。若题目给出总量而问其变化速率，则求导（derivative）。
Worked Example — Tank Filling with a Variable-Rate Inflow
例题——变速率注水的水箱
Water flows into a tank at the rate $R(t) = 6 + 4t$ gal/min for $0 \le t \le 5$ min. The tank holds $W(0) = 20$ gal at $t = 0$.
水以速率 $R(t) = 6 + 4t$ gal/min 流入水箱，$0 \le t \le 5$ 分钟。$t = 0$ 时水箱中的水量为 $W(0) = 20$ gal。
(a) How much water enters the tank during the interval $[0, 5]$?
(a) 在 $[0, 5]$ 区间内有多少水流入水箱？
(b) How much water is in the tank at $t = 5$?
(b) $t = 5$ 时水箱里有多少水？
- Identify the integral. The amount that enters during $[0, 5]$ is the accumulation of the inflow rate: $$\text{Net inflow} \;=\; \int_0^{5} R(t)\,dt \;=\; \int_0^{5}(6 + 4t)\,dt.$$
- 识别积分。 在 $[0, 5]$ 区间内流入的水量（净流入量）就是流入速率的累积： $$\int_0^{5} R(t)\,dt \;=\; \int_0^{5}(6 + 4t)\,dt.$$
- Evaluate. $$\int_0^{5}(6 + 4t)\,dt \;=\; \bigl[\,6t + 2t^{2}\,\bigr]_0^{5} \;=\; 30 + 50 \;=\; 80\text{ gal}.$$
- 计算。 $$\int_0^{5}(6 + 4t)\,dt \;=\; \bigl[\,6t + 2t^{2}\,\bigr]_0^{5} \;=\; 30 + 50 \;=\; 80\text{ gal}.$$
- (b) Add to the initial amount. Net change formula: $W(5) = W(0) + \int_0^{5} R(t)\,dt = 20 + 80 = 100$ gal.
- (b) 加上初始量。 由净变化公式：$W(5) = W(0) + \int_0^{5} R(t)\,dt = 20 + 80 = 100$ gal。
The Net Change Theorem ($Q(b) - Q(a) = \int_a^b Q'(t)\,dt$) is the bridge between rates and amounts. We do not need an explicit formula for $W(t)$ — the rate $R$ is $W'$, and integrating it directly gives the change.
净变化定理 ($Q(b) - Q(a) = \int_a^b Q'(t)\,dt$) 是连接变化率与总量的桥梁。我们无需 $W(t)$ 的显式表达——速率 $R$ 就是 $W'$，直接积分就得到变化量。
If water also drains at rate $D(t)$, the net rate of change of the tank is $W'(t) = R(t) - D(t)$, and the same formula applies: $W(b) = W(a) + \int_a^b\!\bigl[\,R(t) - D(t)\,\bigr]\,dt$. AP free-response often pairs an inflow rate with an outflow rate and asks for the amount at a specific time.
若同时以速率 $D(t)$ 排水，则水箱的净变化率为 $W'(t) = R(t) - D(t)$，同一公式仍然适用：$W(b) = W(a) + \int_a^b\!\bigl[\,R(t) - D(t)\,\bigr]\,dt$。AP 自由响应题（FRQ）常把流入率与流出率配对出现，并要求某个时刻的总量。
Finding the Area Between Curves Expressed as Functions of x
求以 x 为自变量的两条曲线之间的面积
When curves are written as y = f(x) and y = g(x), the area between them is found by integrating the vertical distance between the top and bottom curves.
当曲线写成 y = f(x) 和 y = g(x) 时，两条曲线之间的面积（area between two curves）就是对上下两条曲线之间的垂直距离作积分。
Worked Example 2
例题 2
Find the area enclosed by $y=x$ and $y=x^2$ on $[0,1]$.
求 $y=x$ 与 $y=x^2$ 在 $[0,1]$ 上所围成的面积。
- Determine which graph is on top on [0,1]. Since $x \ge x^2$ there, the top curve is $y=x$ and the bottom curve is $y=x^2$.
- 先判断在 [0,1] 上哪条在上。因为该区间内 $x \ge x^2$，所以上曲线是 $y=x$，下曲线是 $y=x^2$。
- Set up the integral: $$A=\int_0^1 (x-x^2)\,dx$$
- 建立积分： $$A=\int_0^1 (x-x^2)\,dx$$
- Integrate: $$A=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac12-\frac13=\frac16$$
- 积分： $$A=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac12-\frac13=\frac16$$
Interactive Example — Area Between Two Curves
交互示例——两条曲线之间的面积
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As the slope changes, the “top minus bottom” gap changes point-by-point, so the total area changes. This is exactly what the definite integral is measuring.
当斜率变化时，“上减下”的间距逐点变化，因此总面积也随之改变。这正是定积分所衡量的东西。
Finding the Area Between Curves Expressed as Functions of y
求关于 y 轴的两曲线间面积
When the left and right boundaries are easier to describe than the top and bottom boundaries, integrate with respect to y.
若左右边界比上下边界更容易描述，则关于 y 积分（area between two curves with respect to y）。
Use vertical slices when the region is naturally “top minus bottom.” Use horizontal slices when the region is naturally “right minus left.” Choose the variable that produces one clean integral whenever possible.
当区域自然呈现“上减下”形式时，用垂直切片；当区域自然呈现“右减左”形式时，用水平切片。尽可能选择能写出一个干净积分的变量。
Finding the Area Between Curves That Intersect at More Than Two Points
求多次相交的两曲线之间的面积
If the top and bottom curves switch positions, a single integral will no longer work unless you use absolute value. Usually the safest method is to split the interval at every intersection.
如果上下曲线发生位置互换，一个不带绝对值的积分就不能直接用了。最稳妥的做法通常是在每一个交点处把区间拆开。
Students often forget to split at an intersection where the graphs cross. That causes cancellation and gives signed area instead of actual area.
学生常常忘记在两条曲线相交处拆分区间，导致正负相消，得到的是带号面积而不是实际面积。
Worked Example — $\displaystyle\int_{-1}^{1} |x^{3} - x|\,dx$
例题——$\displaystyle\int_{-1}^{1} |x^{3} - x|\,dx$
Find the area enclosed between $y = x^{3}$ and $y = x$ on $[-1, 1]$. The graphs cross at $x = -1, 0, 1$, so the "top minus bottom" identity flips at $x = 0$.
求 $y = x^{3}$ 与 $y = x$ 在 $[-1, 1]$ 上所围的面积。两图象在 $x = -1, 0, 1$ 处相交，所以“上减下”的关系在 $x = 0$ 处发生翻转。
- Locate the zeros of the difference. $x^{3} - x = x(x-1)(x+1) = 0$ at $x = -1, 0, 1$. These split $[-1, 1]$ into two subintervals.
- 找出差函数的零点。 $x^{3} - x = x(x-1)(x+1) = 0$，零点为 $x = -1, 0, 1$。它们把 $[-1, 1]$ 切成两段。
- Sign chart on each subinterval.
- On $(-1, 0)$, test $x = -\tfrac{1}{2}$: $(-\tfrac{1}{2})^{3} - (-\tfrac{1}{2}) = -\tfrac{1}{8} + \tfrac{1}{2} = \tfrac{3}{8} > 0$. So $x^{3} \ge x$ — cubic is on top.
- On $(0, 1)$, test $x = \tfrac{1}{2}$: $\tfrac{1}{8} - \tfrac{1}{2} = -\tfrac{3}{8} < 0$. So $x \ge x^{3}$ — line is on top.
- 对每段做符号表。
- 在 $(-1, 0)$ 上取 $x = -\tfrac{1}{2}$ 试： $(-\tfrac{1}{2})^{3} - (-\tfrac{1}{2}) = -\tfrac{1}{8} + \tfrac{1}{2} = \tfrac{3}{8} > 0$。所以 $x^{3} \ge x$——三次函数在上。
- 在 $(0, 1)$ 上取 $x = \tfrac{1}{2}$ 试： $\tfrac{1}{8} - \tfrac{1}{2} = -\tfrac{3}{8} < 0$。所以 $x \ge x^{3}$——直线在上。
- Split the absolute-value integral by sign. $$\int_{-1}^{1}|x^{3} - x|\,dx \;=\; \int_{-1}^{0}(x^{3} - x)\,dx \;+\; \int_{0}^{1}(x - x^{3})\,dx.$$
- 按符号拆分绝对值积分。 $$\int_{-1}^{1}|x^{3} - x|\,dx \;=\; \int_{-1}^{0}(x^{3} - x)\,dx \;+\; \int_{0}^{1}(x - x^{3})\,dx.$$
- Evaluate each piece. Use antiderivative $\tfrac{x^{4}}{4} - \tfrac{x^{2}}{2}$: $$\int_{-1}^{0}(x^{3} - x)\,dx \;=\; \Bigl[\tfrac{x^{4}}{4} - \tfrac{x^{2}}{2}\Bigr]_{-1}^{0} \;=\; 0 - \!\left(\tfrac{1}{4} - \tfrac{1}{2}\right) \;=\; \tfrac{1}{4}.$$ $$\int_{0}^{1}(x - x^{3})\,dx \;=\; \Bigl[\tfrac{x^{2}}{2} - \tfrac{x^{4}}{4}\Bigr]_{0}^{1} \;=\; \tfrac{1}{2} - \tfrac{1}{4} \;=\; \tfrac{1}{4}.$$
- 分别计算每一段。 使用原函数 $\tfrac{x^{4}}{4} - \tfrac{x^{2}}{2}$： $$\int_{-1}^{0}(x^{3} - x)\,dx \;=\; \Bigl[\tfrac{x^{4}}{4} - \tfrac{x^{2}}{2}\Bigr]_{-1}^{0} \;=\; 0 - \!\left(\tfrac{1}{4} - \tfrac{1}{2}\right) \;=\; \tfrac{1}{4}.$$ $$\int_{0}^{1}(x - x^{3})\,dx \;=\; \Bigl[\tfrac{x^{2}}{2} - \tfrac{x^{4}}{4}\Bigr]_{0}^{1} \;=\; \tfrac{1}{2} - \tfrac{1}{4} \;=\; \tfrac{1}{4}.$$
- Sum. Total area $= \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2}$.
- 求和。 总面积 $= \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2}$。
$|x^{3} - x|$ is an even function (the absolute value kills the odd-symmetry sign flip), so $\int_{-1}^{1}|x^{3} - x|\,dx = 2\!\int_0^1 |x^{3} - x|\,dx = 2 \cdot \tfrac{1}{4} = \tfrac{1}{2}$. Same answer.
$|x^{3} - x|$ 是偶函数（绝对值消去了奇函数的符号反转），所以 $\int_{-1}^{1}|x^{3} - x|\,dx = 2\!\int_0^1 |x^{3} - x|\,dx = 2 \cdot \tfrac{1}{4} = \tfrac{1}{2}$。答案相同。
Computing $\int_{-1}^{1}(x^{3} - x)\,dx$ without the absolute value gives $0$, because $x^{3} - x$ is odd and the contributions on $(-1,0)$ and $(0,1)$ cancel. That zero is signed area — definitely not the geometric area between the curves.
如果不加绝对值直接算 $\int_{-1}^{1}(x^{3} - x)\,dx$，结果为 $0$，因为 $x^{3} - x$ 是奇函数，$(-1,0)$ 和 $(0,1)$ 两段的贡献相互抵消。这个 $0$ 是带号面积——绝对不是两曲线之间的几何面积。
Volume Methods Roadmap
体积方法路线图
Every Unit 8 volume problem reduces to the same three moves:
- Pick a typical slice of the solid (perpendicular to one axis).
- Compute the area of that slice as a function of position.
- Integrate the area along the axis.
Topics 8.7 – 8.12 differ only in the shape of that slice. The 4-step decision tree below tells you which.
Unit 8 的每一道体积（volume）题都归结为同样的三步：
- 取实体的一个典型切片（垂直于某条轴）。
- 把该切片的面积表示为位置的函数。
- 沿该轴积分这个面积。
小节 8.7 – 8.12 之间的差别仅在切片的形状。下面的四步决策树告诉你该用哪种。
If the problem describes a solid built from specified cross-sectional shapes (squares, equilateral triangles, semicircles, …) sitting over a planar region, use the Cross-Section Method (Topics 8.7 – 8.8). Otherwise, if a planar region is revolved about a line, go to Step 2.
若题目描述的是由给定形状的横截面（cross-section）（正方形、等边三角形、半圆…）落在某平面区域上构成的实体，则使用已知横截面的体积法（volume by known cross-sections，小节 8.7 – 8.8）。否则，如果是把平面区域绕某条直线旋转得到旋转体，则进入第 2 步。
Sketch the region together with the axis of revolution and ask:
"Does the region share at least one boundary with the axis?"
- Yes — region's edge sits ON the axis → Disc Method. Each cross-section is a solid disc (no hole). Topics 8.9 (axis is $x$- or $y$-axis) and 8.10 (axis is $y=k$ or $x=k$).
- No — there is a gap between the region and the axis → Washer Method. Each cross-section is annular (hole in the middle). Topics 8.11 (standard axis) and 8.12 (shifted axis).
把区域与旋转轴（axis of revolution）一起画出来，自问：
“区域是否至少有一条边界落在旋转轴上？”
- 是——区域边缘落在轴上 → 圆盘法（
disk method）。每个横截面都是无孔的实心圆盘。涉及小节 8.9（轴为 $x$ 或 $y$ 轴）和 8.10（轴为 $y=k$ 或 $x=k$）。 - 否——区域与旋转轴之间有间隙 → 圆环法（
washer method）。每个横截面都是环形（中间有孔）。涉及小节 8.11（标准轴）和 8.12（平移过的轴）。
Slices must be perpendicular to the axis of revolution.
- Axis is horizontal (the $x$-axis, or any $y=k$) → vertical slices → integrate dx.
- Axis is vertical (the $y$-axis, or any $x=k$) → horizontal slices → integrate dy. (You will likely need to rewrite curves as $x=g(y)$.)
切片必须垂直于旋转轴。
- 旋转轴水平（$x$ 轴，或任意 $y=k$）→ 垂直切片 → 对 dx 积分。
- 旋转轴竖直（$y$ 轴，或任意 $x=k$）→ 水平切片 → 对 dy 积分（这时通常需要把曲线改写为 $x=g(y)$）。
A radius is the perpendicular distance from a boundary point of the region to the axis of revolution — not automatically the function value.
Disc (region edge sits on the axis):
- Axis is the $x$-axis: $R = |f(x)|$. Squaring kills the sign, so writing $R = f(x)$ in the integrand is fine.
- Axis is $y = k$: $R = |f(x) - k|$, which simplifies to $k - f(x)$ when the axis lies above the curve and $f(x) - k$ when below.
Washer (gap between region and axis): a region bounded by two curves, both on the same side of the axis. The slice is an annulus.
- $R$ = outer radius = distance from the farther boundary curve to the axis.
- $r$ = inner radius = distance from the nearer boundary curve to the axis.
- Critical: "outer" and "inner" depend on the axis location, not on which curve is called "$f$" or "$g$". Move the axis and the labels can swap. Always re-identify outer/inner from your sketch.
半径是区域边界上的一点到旋转轴的垂直距离——不一定等于函数值本身。
圆盘（区域边缘落在轴上）：
- 旋转轴为 $x$ 轴：$R = |f(x)|$。平方会消去符号，所以在被积式中写成 $R = f(x)$ 也可以。
- 旋转轴为 $y = k$：$R = |f(x) - k|$；轴在曲线上方时简化为 $k - f(x)$，轴在曲线下方时为 $f(x) - k$。
圆环（区域与轴之间有间隙）：由两条曲线围成的区域，两条曲线都在轴的同侧。切片是一个环形。
- $R$ = 外半径 = 较远的边界曲线到旋转轴的距离。
- $r$ = 内半径 = 较近的边界曲线到旋转轴的距离。
- 关键：“外”和“内”取决于旋转轴的位置，与哪条曲线被命名为 “$f$” 或 “$g$” 无关。轴一移动，标签就可能互换。每次都要根据自己的草图重新判定外/内。
| Method | When | Integrand | Topic |
|---|---|---|---|
| Cross Section | Solid built from named cross-sectional shapes | $A(x)\,dx$ (or $A(y)\,dy$) | 8.7 – 8.8 |
| Disc, axis = $x$ or $y$ | Revolved; region's edge sits on axis | $\pi\,R^2\,dx$ (or $dy$) | 8.9 |
| Disc, axis = $y=k$ or $x=k$ | Revolved; region's edge sits on the line | $\pi\,[\,\text{dist}(f, \text{axis})\,]^{\,2}\,dx$ | 8.10 |
| Washer, axis = $x$ or $y$ | Revolved; gap between region and axis | $\pi\,(R^2 - r^2)\,dx$ | 8.11 |
| Washer, axis = $y=k$ or $x=k$ | Revolved; gap between region and the line | $\pi\,(R^2 - r^2)\,dx$ where $R, r$ = distances to the axis | 8.12 |
| 方法 | 何时使用 | 被积式 | 小节 |
|---|---|---|---|
| 横截面法 | 实体由命名的横截面形状构成 | $A(x)\,dx$ （或 $A(y)\,dy$） | 8.7 – 8.8 |
| 圆盘法，轴 = $x$ 或 $y$ | 旋转体；区域边缘在轴上 | $\pi\,R^2\,dx$ （或 $dy$） | 8.9 |
| 圆盘法，轴 = $y=k$ 或 $x=k$ | 旋转体；区域边缘落在该直线上 | $\pi\,[\,\text{dist}(f, \text{axis})\,]^{\,2}\,dx$ | 8.10 |
| 圆环法，轴 = $x$ 或 $y$ | 旋转体；区域与轴之间有间隙 | $\pi\,(R^2 - r^2)\,dx$ | 8.11 |
| 圆环法，轴 = $y=k$ 或 $x=k$ | 旋转体；区域与该直线之间有间隙 | $\pi\,(R^2 - r^2)\,dx$ 其中 $R, r$ 为到轴的距离 | 8.12 |
Numerical check: with $R = 3$ and $r = 1$, $R^2 - r^2 = 9 - 1 = 8$, but $(R - r)^2 = 2^2 = 4$. They are not equal — and the washer formula uses the first.
数值验证：取 $R = 3$，$r = 1$， $R^2 - r^2 = 9 - 1 = 8$， 但 $(R - r)^2 = 2^2 = 4$。两者不等——圆环公式用的是前者。
Same Region, Three Axes — How the Setup Changes
同一区域，三个旋转轴——设置如何变化
To see the decision tree in action, take one fixed region and revolve it about three different axes. The region: bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$. (A half-parabola sitting on the $x$-axis, with maximum height $\sqrt{4} = 2$.)
为了看决策树如何运行，取一个固定区域，把它绕三个不同的旋转轴旋转。该区域由 $y = \sqrt{x}$、 $y = 0$、 $x = 4$ 围成。（一段坐在 $x$ 轴上的半抛物线，最大高度 $\sqrt{4} = 2$。）
| Axis of revolution | Method & reasoning | Radii | Volume integral |
|---|---|---|---|
| $y = 0$ (the $x$-axis) | Disc. Region's lower boundary is the axis. | $R = \sqrt{x}$ | $\displaystyle V = \pi\!\int_0^4 (\sqrt{x})^{\,2}\,dx = \pi\!\int_0^4 x\,dx = 8\pi$ |
| $y = 3$ (line above region) | Washer. Region's max height is $2$, so axis sits $1$ unit above the top — gap. | $R = 3 - 0 = 3$ (axis to the lower edge $y=0$, the farther boundary) $r = 3 - \sqrt{x}$ (axis to upper curve, the nearer boundary) |
$\displaystyle V = \pi\!\int_0^4 \!\!\left[\,9 - (3 - \sqrt{x})^{\,2}\,\right]dx$ |
| $y = -1$ (line below region) | Washer. Region's bottom is at $y=0$; gap of $1$ unit below. | $R = \sqrt{x} - (-1) = \sqrt{x} + 1$ (axis to upper curve, the farther boundary) $r = 0 - (-1) = 1$ (axis to lower edge, the nearer boundary) |
$\displaystyle V = \pi\!\int_0^4 \!\!\left[\,(\sqrt{x}+1)^{\,2} - 1\,\right]dx$ |
| 旋转轴 | 方法与判断依据 | 半径 | 体积积分 |
|---|---|---|---|
| $y = 0$ （$x$ 轴） | 圆盘法。 区域的下边界就是旋转轴。 | $R = \sqrt{x}$ | $\displaystyle V = \pi\!\int_0^4 (\sqrt{x})^{\,2}\,dx = \pi\!\int_0^4 x\,dx = 8\pi$ |
| $y = 3$ （区域上方的直线） | 圆环法。 区域最大高度为 $2$，所以轴位于顶部上方 $1$ 个单位——有间隙。 | $R = 3 - 0 = 3$ （轴到下边界 $y=0$，更远的边界） $r = 3 - \sqrt{x}$ （轴到上曲线，更近的边界） |
$\displaystyle V = \pi\!\int_0^4 \!\!\left[\,9 - (3 - \sqrt{x})^{\,2}\,\right]dx$ |
| $y = -1$ （区域下方的直线） | 圆环法。 区域底部在 $y=0$；轴在其下方 $1$ 个单位，有间隙。 | $R = \sqrt{x} - (-1) = \sqrt{x} + 1$ （轴到上曲线，更远的边界） $r = 0 - (-1) = 1$ （轴到下边界，更近的边界） |
$\displaystyle V = \pi\!\int_0^4 \!\!\left[\,(\sqrt{x}+1)^{\,2} - 1\,\right]dx$ |
Same region every time. The axis moved, and that flipped which boundary is farther. With $y=3$ above, the lower edge $y=0$ is the outer curve. With $y=-1$ below, the upper curve $y=\sqrt{x}$ is the outer curve. Mechanical reuse of a previous problem's $R$ and $r$ assignments will give a wrong answer here. Always re-derive from a fresh sketch.
每次都是同一区域。轴的位置改变了，从而翻转了哪条边界更远。当 $y=3$ 在上方时，下边界 $y=0$ 是外曲线。当 $y=-1$ 在下方时，上曲线 $y=\sqrt{x}$ 是外曲线。直接照搬上一道题里的 $R$、$r$ 在这里会得出错误答案。每次都要根据新草图重新推导。
Before writing any integral: (1) draw the region; (2) draw the axis of revolution as a dashed line; (3) draw one representative slice perpendicular to that axis; (4) on the slice, mark the boundary points and label the perpendicular distance to the axis as $R$ (and $r$ if there's a hole). If your sketch is right, the integrand reads itself off the page.
写积分前先做：(1) 画出区域；(2) 把旋转轴画成虚线；(3) 在该轴的垂直方向上画一片代表性切片；(4) 在切片上标出边界点，并把它到轴的垂直距离标为 $R$（如果有孔，再标一个 $r$）。如果草图画对了，被积式就会自己从图上显现出来。
Volumes with Cross Sections: Squares and Rectangles
已知横截面的体积：正方形与矩形
For solids built from specified cross-sectional shapes (no rotation), the volume reduces to a single question: what is the area of a typical slice as a function of position? Integrate that area along the axis.
对于由给定横截面形状构成的实体（非旋转体），求体积归结为一个问题：典型切片的面积如何随位置变化？把该面积沿轴积分即可。
When cross-sections are squares with side equal to the vertical distance between two curves, $A(x)=[f(x)-g(x)]^2$. For rectangles, replace $A(x)$ with the correct length × width matching the problem statement.
当横截面是正方形（square cross-section），且边长等于两曲线之间的垂直距离时，$A(x)=[f(x)-g(x)]^2$。对于矩形，把 $A(x)$ 替换为题目所述的正确“长 × 宽”。
Worked Example — Square Cross-Sections
例题——正方形横截面
The base of a solid is the region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$. Cross-sections perpendicular to the $x$-axis are squares with one side in the base. Find the volume.
实体的底面是由 $y=\sqrt{x}$、 $y=0$、 $x=4$ 围成的区域。垂直于 $x$ 轴的横截面是正方形，且其一边位于底面上。求体积。
- Read the side length off the base. A vertical slice of the base at position $x$ runs from $y=0$ to $y=\sqrt{x}$, so the square's side is $s = \sqrt{x}$.
- 从底面读出边长。 在位置 $x$ 处，底面的垂直切片从 $y=0$ 到 $y=\sqrt{x}$，所以正方形的边长为 $s = \sqrt{x}$。
- Cross-sectional area. $A(x) = s^2 = (\sqrt{x})^2 = x$.
- 横截面面积。 $A(x) = s^2 = (\sqrt{x})^2 = x$。
- Integrate. $\displaystyle V = \int_0^4 x\,dx = \tfrac{x^2}{2}\Big|_0^4 = 8.$
- 积分。 $\displaystyle V = \int_0^4 x\,dx = \tfrac{x^2}{2}\Big|_0^4 = 8.$
Volumes with Cross Sections: Triangles and Semicircles
已知横截面的体积：三角形与半圆
These problems use the same structure as Topic 8.7, but the cross-sectional area now depends on a geometry formula.
这类问题与 8.7 结构相同，只是横截面面积现在要用几何公式计算。
| Cross Section | Area in terms of side / diameter s |
|---|---|
| Square | $A=s^2$ |
| Equilateral triangle | $A=\frac{\sqrt3}{4}s^2$ |
| Isosceles right triangle (leg = s) | $A=\frac12 s^2$ |
| Semicircle (diameter = s) | $A=\frac{\pi}{8}s^2$ |
| 横截面 | 用边长 / 直径 s 表示的面积 |
|---|---|
正方形（square cross-section） | $A=s^2$ |
等边三角形横截面（equilateral triangle cross-section） | $A=\frac{\sqrt3}{4}s^2$ |
| 等腰直角三角形（直角边 = s） | $A=\frac12 s^2$ |
半圆横截面（semicircle cross-section，直径 = s） | $A=\frac{\pi}{8}s^2$ |
If the region gives you the diameter, do not use it as the radius. Radius is half the diameter.
如果题目给出的是直径，不要把它当作半径用。半径等于直径的一半。
Volume with Disc Method: Revolving Around the x- or y-Axis
圆盘法求体积：绕 x 轴或 y 轴旋转
Use the disc method when the region's boundary sits on the axis of revolution — every cross-section is a solid disc with no hole. (See the Roadmap above for the disc-vs-washer decision.)
当区域边界落在旋转轴上时使用圆盘法（disk method）——每个横截面都是无孔的实心圆盘。（圆盘 vs. 圆环的判别见上方路线图。）
$R$ is the perpendicular distance from the curve to the axis. About the $x$-axis: $R = f(x)$. About the $y$-axis (slicing in $y$): $R = g(y)$ where $x = g(y)$.
$R$ 是曲线到旋转轴的垂直距离。绕 $x$ 轴旋转（rotating about x-axis）：$R = f(x)$。绕 $y$ 轴旋转（沿 $y$ 切片，rotating about y-axis）：$R = g(y)$，其中 $x = g(y)$。
Worked Example — Revolve $y = x^2$ on $[0,1]$ about the $x$-axis
例题——将 $y = x^2$ 在 $[0,1]$ 上绕 $x$ 轴旋转
- Region touches axis? Yes — the region under $y=x^2$ from $0$ to $1$ has its lower boundary on $y=0$. Disc applies.
- 区域是否与轴相接？ 是——$y=x^2$ 在 $0$ 到 $1$ 之间下方的区域，下边界落在 $y=0$ 上。可用圆盘法。
- Slice direction. Axis is horizontal, so slice vertically and integrate $dx$.
- 切片方向。 轴水平，故作垂直切片，对 $dx$ 积分。
- Radius. $R(x) = x^2$ (distance from curve to $x$-axis).
- 半径。 $R(x) = x^2$（曲线到 $x$ 轴的距离）。
- Set up and evaluate. $$V = \pi\!\int_0^1 (x^2)^{\,2}\,dx = \pi\!\int_0^1 x^4\,dx = \pi \cdot \tfrac{1}{5} = \tfrac{\pi}{5}.$$
- 建立并计算。 $$V = \pi\!\int_0^1 (x^2)^{\,2}\,dx = \pi\!\int_0^1 x^4\,dx = \pi \cdot \tfrac{1}{5} = \tfrac{\pi}{5}.$$
Volume with Disc Method: Revolving Around Other Axes
圆盘法求体积：绕其它轴旋转
When the axis is $y=k$ or $x=k$, the formula is the same — only the radius changes. The radius is now the perpendicular distance from the curve to that line, not to the standard axis.
当旋转轴为 $y=k$ 或 $x=k$ 时，公式不变——只是半径变了。半径现在是曲线到该直线的垂直距离，而不是到标准坐标轴的距离。
If the axis lies above the curve, swap to $\,k - f(x)\,$. Either way, $|f(x)-k|$ is the perpendicular distance — and squaring it makes the sign moot.
如果轴在曲线上方，则改写为 $\,k - f(x)\,$。无论如何，$|f(x)-k|$ 才是垂直距离——平方后符号无关紧要。
Worked Example — Translation Invariance
例题——平移不变性
Revolve $y = x + 3$ on $[0, 2]$ about the line $y = 3$. The region touches the axis at $x = 0$ (where $y = 3$).
将 $y = x + 3$ 在 $[0, 2]$ 上绕直线 $y = 3$ 旋转。区域在 $x = 0$ 处（此时 $y = 3$）与轴相接。
- Radius. $R(x) = (x + 3) - 3 = x$.
- 半径。 $R(x) = (x + 3) - 3 = x$。
- Volume. $$V = \pi\!\int_0^2 x^2\,dx = \pi\cdot \tfrac{8}{3} = \tfrac{8\pi}{3}.$$
- 体积。 $$V = \pi\!\int_0^2 x^2\,dx = \pi\cdot \tfrac{8}{3} = \tfrac{8\pi}{3}.$$
- Sanity check. The same line $y = x$ on $[0, 2]$ revolved about the $x$-axis gives $\pi\!\int_0^2 x^2\,dx = \tfrac{8\pi}{3}$. Translating the region and the axis together by the same amount does not change the volume — a useful self-check.
- 验证。 同一条直线 $y = x$ 在 $[0, 2]$ 上绕 $x$ 轴旋转得到 $\pi\!\int_0^2 x^2\,dx = \tfrac{8\pi}{3}$。把区域与旋转轴一起平移同样的距离不会改变体积——这是一个有用的自检手段。
Draw one representative disc on your sketch and label the radius on the picture. The picture catches "is the axis above or below the curve" before you write a single integral.
在草图上画一个代表性的圆盘，并在图上标出半径。图像在你写下一个积分之前就能告诉你“轴在曲线上方还是下方”。
Volume with Washer Method: Revolving Around the x- or y-Axis
圆环法求体积：绕 x 轴或 y 轴旋转
The washer method is used when rotating a region that leaves a hole in the middle. Each cross section is an outer disc with an inner disc removed.
当旋转一个会在中间留下孔的区域时使用圆环法（washer method）。每个横截面都是从外圆盘中扣掉一个内圆盘后的环。
Worked Example 3
例题 3
Find the volume generated by revolving the region between $y=x$ and $y=x^2$ on $[0,1]$ around the x-axis.
求 $y=x$ 与 $y=x^2$ 在 $[0,1]$ 上之间的区域绕 $x$ 轴旋转所得的体积。
- On [0,1], the outer radius is $R(x)=x$ and the inner radius is $r(x)=x^2$.
- 在 [0,1] 上，外半径为 $R(x)=x$，内半径为 $r(x)=x^2$。
- Set up the washer integral: $$V=\pi\int_0^1 (x^2-x^4)\,dx$$
- 写出圆环积分： $$V=\pi\int_0^1 (x^2-x^4)\,dx$$
- Evaluate: $$V=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=\pi\left(\frac13-\frac15\right)=\frac{2\pi}{15}$$
- 计算： $$V=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=\pi\left(\frac13-\frac15\right)=\frac{2\pi}{15}$$
Compute R² − r², not (R − r)². Those are different expressions.
计算的是 R² − r²，不是 (R − r)²。两者不是同一个表达式。
Volume with Washer Method: Revolving Around Other Axes
圆环法求体积：绕其它轴旋转
Same structure as Topic 8.11. The only change: $R$ and $r$ are now perpendicular distances to a shifted axis $y=k$ or $x=k$.
结构与 8.11 完全相同。唯一的变化：$R$ 与 $r$ 现在是到平移过的轴 $y=k$ 或 $x=k$ 的垂直距离。
where $R, r$ are distances from the farther and nearer boundary curves to the axis, respectively.
其中 $R$、$r$ 分别为较远和较近的边界曲线到旋转轴的距离。
"Outer" depends on the axis location, not on which boundary you happened to call $f$. With axis above the region, the lower boundary is farther (outer); with axis below the region, the upper boundary is farther. Re-identify on every problem from a fresh sketch.
“外”取决于旋转轴的位置，而不是你恰好把哪条边界叫作 $f$。轴在区域上方时，下边界更远（外）；轴在区域下方时，上边界更远。每道题都要根据新草图重新判别。
Worked Example — Region $\{y=x, y=x^2\}$ on $[0,1]$ Revolved About $y = -1$
例题——把 $\{y=x, y=x^2\}$ 在 $[0,1]$ 上的区域绕 $y = -1$ 旋转
- Region touches axis? No — the region sits above $y=0$, while the axis is at $y=-1$. Gap of at least 1 unit. Washer.
- 区域是否接轴？ 否——区域位于 $y=0$ 上方，而轴在 $y=-1$。至少有 1 个单位的间隙。用圆环法。
- Identify outer and inner. Both boundaries are above the axis. The line $y=x$ is farther from $y=-1$ than the parabola $y=x^2$ is, so $y=x$ is outer.
- 判别外/内。 两条边界都在轴上方。直线 $y=x$ 到 $y=-1$ 的距离比抛物线 $y=x^2$ 大，所以 $y=x$ 是外曲线。
- Radii.
- $R = x - (-1) = x + 1$
- $r = x^2 - (-1) = x^2 + 1$
- 半径。
- $R = x - (-1) = x + 1$
- $r = x^2 - (-1) = x^2 + 1$
- Set up and evaluate. $$\begin{aligned} V &= \pi\!\int_0^1 \!\bigl[(x+1)^2 - (x^2+1)^2\bigr]\,dx \\[4pt] &= \pi\!\int_0^1 \!\bigl[(x^2 + 2x + 1) - (x^4 + 2x^2 + 1)\bigr]\,dx \\[4pt] &= \pi\!\int_0^1 \!\bigl(2x - x^2 - x^4\bigr)\,dx \\[4pt] &= \pi\Bigl[\,x^2 - \tfrac{x^3}{3} - \tfrac{x^5}{5}\,\Bigr]_0^1 \\[4pt] &= \pi\Bigl(1 - \tfrac{1}{3} - \tfrac{1}{5}\Bigr) = \tfrac{7\pi}{15}. \end{aligned}$$
- 建立并计算。 $$\begin{aligned} V &= \pi\!\int_0^1 \!\bigl[(x+1)^2 - (x^2+1)^2\bigr]\,dx \\[4pt] &= \pi\!\int_0^1 \!\bigl[(x^2 + 2x + 1) - (x^4 + 2x^2 + 1)\bigr]\,dx \\[4pt] &= \pi\!\int_0^1 \!\bigl(2x - x^2 - x^4\bigr)\,dx \\[4pt] &= \pi\Bigl[\,x^2 - \tfrac{x^3}{3} - \tfrac{x^5}{5}\,\Bigr]_0^1 \\[4pt] &= \pi\Bigl(1 - \tfrac{1}{3} - \tfrac{1}{5}\Bigr) = \tfrac{7\pi}{15}. \end{aligned}$$
- Compare to Topic 8.11. The same region revolved about the $x$-axis gives $\tfrac{2\pi}{15}$. Shifting the axis from $y=0$ down to $y=-1$ enlarged the volume to $\tfrac{7\pi}{15}$ — each radius grew by 1, but $R^2 - r^2$ changes in a non-trivial way, so always rewrite from scratch.
- 与 8.11 对比。 同一区域绕 $x$ 轴旋转得到 $\tfrac{2\pi}{15}$。把轴从 $y=0$ 向下移到 $y=-1$，体积扩大到 $\tfrac{7\pi}{15}$——每个半径都增加了 1，但 $R^2 - r^2$ 的变化并非线性，所以每次都要从头重新写。
The Arc Length of a Smooth, Planar Curve and Distance Traveled
光滑平面曲线的弧长与路程
Arc length measures the accumulated length of tiny tangent-like segments along a curve.
弧长（arc length）衡量沿曲线累积起来的、近似切向的小段长度之和。
For parametric motion in the plane, the total distance traveled is:
对于平面参数运动，总路程为：
Derivation Sketch — From Polygonal Approximation to the Integral推导提纲——从折线近似到积分
The idea: approximate a smooth curve by a chain of straight segments, sum their lengths, then refine.
思路：用一串直线段去逼近光滑曲线，把它们的长度加起来，再不断加密。
- Partition. Split $[a,b]$ into $n$ equal subintervals of width $\Delta x$, with sample points $x_0,x_1,\ldots,x_n$. The curve passes through $\bigl(x_i,\,f(x_i)\bigr)$.
- 分划。 把 $[a,b]$ 等分成 $n$ 个宽度为 $\Delta x$ 的子区间，分点为 $x_0,x_1,\ldots,x_n$。曲线经过 $\bigl(x_i,\,f(x_i)\bigr)$。
- Approximate each piece by a line segment. The straight segment from $\bigl(x_{i-1},f(x_{i-1})\bigr)$ to $\bigl(x_i,f(x_i)\bigr)$ has length, by the distance formula, $$ \Delta L_i \;=\; \sqrt{(\Delta x)^2 + (\Delta y_i)^2}, \quad \text{where } \Delta y_i = f(x_i)-f(x_{i-1}). $$
- 用线段近似每一小段。 由距离公式，从 $\bigl(x_{i-1},f(x_{i-1})\bigr)$ 到 $\bigl(x_i,f(x_i)\bigr)$ 的线段长度为 $$ \Delta L_i \;=\; \sqrt{(\Delta x)^2 + (\Delta y_i)^2}, \quad \Delta y_i = f(x_i)-f(x_{i-1}). $$
- Factor out $\Delta x$. $$ \Delta L_i \;=\; \sqrt{1 + \!\left(\dfrac{\Delta y_i}{\Delta x}\right)^{\!2}}\;\Delta x. $$
- 提出 $\Delta x$。 $$ \Delta L_i \;=\; \sqrt{1 + \!\left(\dfrac{\Delta y_i}{\Delta x}\right)^{\!2}}\;\Delta x. $$
- Mean Value Theorem. Since $f$ is differentiable, MVT guarantees a point $x_i^{*}\in(x_{i-1},x_i)$ with $\dfrac{\Delta y_i}{\Delta x}=f'(x_i^{*})$. Substitute: $$ \Delta L_i \;=\; \sqrt{1 + [f'(x_i^{*})]^2}\;\Delta x. $$
- 中值定理（MVT）。 因为 $f$ 可微，MVT 保证存在一点 $x_i^{*}\in(x_{i-1},x_i)$，使 $\dfrac{\Delta y_i}{\Delta x}=f'(x_i^{*})$。代入： $$ \Delta L_i \;=\; \sqrt{1 + [f'(x_i^{*})]^2}\;\Delta x. $$
- Sum and take the limit. The total polygonal length is $\displaystyle\sum_{i=1}^{n}\sqrt{1+[f'(x_i^{*})]^2}\,\Delta x$, a Riemann sum. As $n\to\infty$ (so $\Delta x\to 0$), $$ L \;=\; \lim_{n\to\infty}\sum_{i=1}^{n}\sqrt{1+[f'(x_i^{*})]^2}\,\Delta x \;=\; \int_a^b \sqrt{1+[f'(x)]^2}\,dx. $$
- 求和并取极限。 折线总长度为 $\displaystyle\sum_{i=1}^{n}\sqrt{1+[f'(x_i^{*})]^2}\,\Delta x$，这是一个黎曼和（
Riemann sum）。当 $n\to\infty$（$\Delta x\to 0$）时， $$ L \;=\; \lim_{n\to\infty}\sum_{i=1}^{n}\sqrt{1+[f'(x_i^{*})]^2}\,\Delta x \;=\; \int_a^b \sqrt{1+[f'(x)]^2}\,dx. $$
Geometrically: if you walk a horizontal step of length $1$ along a curve, you also rise by $\tfrac{dy}{dx}$ vertically, so the actual arc distance is the hypotenuse $\sqrt{1+(dy/dx)^2}$. The integral just accumulates this "speed-along-the-curve" over $[a,b]$.
几何上：沿曲线水平方向走 $1$ 单位，竖直方向同时上升 $\tfrac{dy}{dx}$ 单位，因此真正的弧线距离是斜边 $\sqrt{1+(dy/dx)^2}$。积分就是把这个“沿曲线的速度”在 $[a,b]$ 上累加。
Repeat the derivation with the segment from $\bigl(x(t_{i-1}),y(t_{i-1})\bigr)$ to $\bigl(x(t_i),y(t_i)\bigr)$. Its length is $\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(\Delta x/\Delta t)^2+(\Delta y/\Delta t)^2}\,\Delta t$. MVT on each component yields the parametric arc-length integral.
把推导应用到从 $\bigl(x(t_{i-1}),y(t_{i-1})\bigr)$ 到 $\bigl(x(t_i),y(t_i)\bigr)$ 的线段上。其长度为 $\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(\Delta x/\Delta t)^2+(\Delta y/\Delta t)^2}\,\Delta t$。对每个分量分别用 MVT，就得到参数形式的弧长积分。
The formula uses $[f'(x)]^2$, not $|f'(x)|$ or $f'(x)$. Because of the square, the sign of $f'$ doesn't matter — but you must square before integrating, never after. A common student error is writing $\int f'(x)\sqrt{\cdots}\,dx$.
公式使用的是 $[f'(x)]^2$，不是 $|f'(x)|$，也不是 $f'(x)$。由于平方，$f'$ 的符号无所谓——但必须在积分前就平方，绝不能积分后再平方。常见学生错误是写成 $\int f'(x)\sqrt{\cdots}\,dx$。
Exam Strategy
考试策略
Identify the quantity
Area, volume, net change, average value, and arc length all use different integrands even though they share the same accumulation idea.
先识别所求量
面积、体积、净变化、平均值与弧长虽然都源自同一“累积”思想，但被积式各不相同。
Match slice direction
Vertical slices lead to dx. Horizontal slices lead to dy. For disc and washer problems, slices must be perpendicular to the axis of revolution.
切片方向要对
垂直切片对 dx 积分。水平切片对 dy 积分。圆盘和圆环问题中，切片必须垂直于旋转轴。
State the setup first
On free response, a correct integral with correct bounds earns meaningful credit even before evaluation.
先写出建立式
在自由响应题（FRQ）中，写对积分与上下限本身就能拿到相当一部分分数，甚至不必算出最终数值。
Common Mistakes
常见错误
One uses an integral divided by interval length. The other uses endpoint outputs divided by endpoint inputs.
一个是积分除以区间长度，另一个是端点输出之差除以端点输入之差。
$\int v(t)\,dt$ is displacement. For total distance, use $\int |v(t)|\,dt$.
$\int v(t)\,dt$ 是位移。求总路程要用 $\int |v(t)|\,dt$。
Radius is distance to the axis, not automatically the function itself.
半径是到旋转轴的距离，并不一定就是函数本身。
When graphs cross, one “top minus bottom” rule will not work over the full interval.
两图象相交时，单一的“上减下”规则不能贯穿整个区间。
Washer method is $R^2-r^2$, not $(R-r)^2$.
圆环法是 $R^2-r^2$，不是 $(R-r)^2$。
Flashcards
闪卡
Click a card to reveal the answer.
点击卡片查看答案。
Unit Quiz
单元小测
Select an answer for each question.
为每题选择一个答案。
Readiness Checklist
备考清单
Click each item you've mastered. Aim for 100% before exam day.
点击你已经掌握的每一项。考前争取 100%。
- Compute the average value of a function on a closed interval在闭区间上计算函数的平均值
- Relate position, velocity, and acceleration via integration通过积分把位置、速度、加速度联系起来
- Distinguish displacement from total distance traveled区分位移与总路程
- Set up accumulation / net-change integrals from a rate function由变化率函数建立累积 / 净变化积分
- Find area between curves integrating with respect to $x$对 $x$ 积分求两曲线之间的面积
- Find area between curves integrating with respect to $y$对 $y$ 积分求两曲线之间的面积
- Split the integral when curves cross or intersect multiple times在曲线相交或多次相交时拆分积分区间
- Compute volume by known cross sections (squares, triangles, semicircles)用已知横截面（正方形、三角形、半圆）计算体积
- Apply the Disc Method about the $x$- or $y$-axis用圆盘法绕 $x$ 或 $y$ 轴旋转
- Apply the Disc Method about a horizontal or vertical line用圆盘法绕水平或竖直直线旋转
- Apply the Washer Method for volumes of revolution用圆环法求旋转体体积
- Choose outer/inner radii correctly for non-axis axes of rotation在非标准旋转轴情形下正确选择外/内半径
- Compute arc length with $\int \sqrt{1 + (f'(x))^2}\,dx$ (BC)用 $\int \sqrt{1 + (f'(x))^2}\,dx$ 计算弧长（BC）
- Sketch the region and axis of rotation before setting up the integral写积分前先画出区域和旋转轴
- Select the appropriate method (disc, washer, cross-section) per problem针对每道题选择合适方法（圆盘、圆环、横截面）
- Use proper notation and units in contextual applications在情境应用题中使用正确的记号与单位
AP-Style Practice Problems
AP 风格练习题
Exam-level practice for this unit — multiple-choice plus extended-response items modeled on the AP rubric. Built for top-score prep; go here after you've worked through the notes and the in-page quizzes above.
面向本单元的考试级练习——按 AP 评分标准设计的选择题加扩展回答题。专为冲击高分准备；在做完上面的讲解与页内小测后，再到这里训练。