📚 Exercise 18B.2: Volumes of Revolution | 练习18B.2: 旋转体体积
In IB Mathematics, one of the most visual and practical applications of definite integration is computing the volume of a solid formed by rotating a curve around an axis. Exercise 18B.2 focuses precisely on this technique, offering a range of problems that require you to set up and evaluate volume integrals using the disk method. Whether you are working with polynomials, roots, or composite functions, mastering this set of exercises builds a strong bridge between abstract integration and three‑dimensional geometry.
在IB数学中，定积分最直观、最实际的应用之一就是计算由曲线绕坐标轴旋转所形成的立体体积。练习18B.2正是围绕这一技巧展开，提供了多种需要运用圆盘法建立并计算体积积分的问题。无论面对的是多项式、根式还是复合函数，掌握这一组练习都能在抽象积分和三维几何之间架起牢固的桥梁。
1. Understanding the Disk Method | 理解圆盘法
The disk method arises naturally when a plane region is revolved about a line, sweeping out a solid composed of infinitely many thin circular disks. Each disk’s radius is given by the value of the function at that point, and its thickness is an infinitesimal change in the independent variable. Summing the volumes of all disks via integration yields the total volume.
圆盘法源于平面区域绕某一直线旋转生成由无数薄圆盘组成的立体。每个圆盘的半径由该点处的函数值决定，其厚度为自变量的无穷小增量。通过积分将所有圆盘的体积累加起来，就得到了总体积。
In Exercise 18B.2, you are typically asked to rotate a region bounded by a curve, the x‑axis, and two vertical lines about the x‑axis or y‑axis. The method remains the same: identify the radius function, square it, multiply by π, and integrate over the given interval.
在练习18B.2中，通常要求将曲线、x轴及两条垂线围成的区域绕x轴或y轴旋转。方法始终保持一致：确定半径函数，将其平方，乘以π，然后在指定区间上积分。
2. The Formula for Rotation about the x‑axis | 绕x轴旋转的公式
a region under the curve y = f(x), from x = a to x = b, is rotated completely about the x‑axis, the volume V is given by the well‑known formula. It assumes the disks are perpendicular to the axis of rotation.
当曲线 y = f(x) 下方、从 x = a 到 x = b 的区域绕x轴完整旋转时，体积 V 由熟悉的公式给出。该公式假设圆盘与旋转轴垂直。
V = π ∫ₐᵇ [f(x)]² dx
This formula is the heart of Exercise 18B.2. Pay close attention to the limits a and b: they are the x‑coordinates of the intersection points of the boundaries, not simply the endpoints of the function’s domain unless the problem specifies the region exactly.
这个公式是练习18B.2的核心。需要特别留意积分限 a 和 b：它们是区域边界交点的x坐标，而不仅仅是函数定义域的端点，除非题目明确指定了区域。
If the curve sometimes dips below the x‑axis, you still square f(x), so the radius is always taken as a positive length. That is why absolute values are not needed inside the square.
即使曲线有时位于x轴下方，仍需对 f(x) 平方，因此半径总是取正值长度。这就是平方内不需要加绝对值的原因。
3. Step‑by‑Step Guide to Exercise 18B.2 | 攻克练习18B.2的分步指南
A systematic approach is essential for the varied functions you will encounter. Begin by sketching the region, noting the axis of rotation clearly. Identify the upper and lower boundaries with respect to that axis, then express the radius as a function of the appropriate variable.
面对各种函数，一套系统的解题方法至关重要。首先画出区域的草图，明确标出旋转轴。确定该轴对应的上下边界，然后用合适的变量表示半径。
Next, write down the volume integral exactly as per the disk formula. Check that the limits match the direction of rotation. Finally, carry out the integration, which may require polynomial expansion, substitution, or trigonometric identities. Always simplify your answer and include the factor of π.
接着，严格按照圆盘公式写出体积积分。检查积分限是否与旋转方向匹配。最后进行积分运算，可能涉及多项式展开、代换法或三角恒等式。始终化简答案并保留π因子。
4. Example 1: Simple Polynomial | 例题1: 简单多项式
Consider a typical problem from Exercise 18B.2: the region bounded by y = x², the x‑axis, x = 0 and x = 2 is rotated about the x‑axis. Find the volume. Apply the disk formula directly with f(x) = x².
考虑练习18B.2中一道典型题目：由 y = x²，x轴，x = 0 和 x = 2 围成的区域绕x轴旋转。求体积。直接应用圆盘公式，取 f(x) = x²。
V = π ∫₀² (x²)² dx = π ∫₀² x⁴ dx
Integrate x⁴ to get (1/5)x⁵. Evaluate from 0 to 2, giving (32/5). Multiply by π to obtain the final volume.
对 x⁴ 积分得到 (1/5)x⁵。从0到2求值得 (32/5)。乘以π即得最终体积。
V = (32π)/5 cubic units
This straightforward example reinforces the basic pattern. Exercise 18B.2 gradually introduces more complicated functions, but the underlying logic never changes.
这个简单的例子巩固了基本模式。练习18B.2会逐步引入更复杂的函数，但根本的逻辑始终不变。
5. Handling Boundaries and Limits | 处理边界与积分限
nded by vertical lines, you must find where the curve meets the axis of rotation. For an x‑axis rotation, set f(x) = 0 to solve for x‑intercepts that become limits.
一个常见陷阱是错误识别积分限 a 和 b。如果区域并非由明确的垂线界定，就必须找到曲线与旋转轴的交点。对于绕x轴旋转，令 f(x)=0 解出x截距，作为积分限。
Sometimes the region is between two curves, and the volume is generated by revolving the enclosed area. In that case, the ‘washer method’ – an extension of the disk method – subtracts the inner radius squared. Exercise 18B.2 focuses primarily on the simple disk case, but being comfortable with limits sets a strong foundation for washers later.
有时区域介于两条曲线之间，旋转封闭区域即生成体积。此时需使用“垫圈法”——圆盘法的扩展——减去内半径的平方。练习18B.2主要集中于简单圆盘情形，但熟练处理积分限为后续的垫圈法奠定了坚实基础。
Always double‑check whether the axis of rotation coincides with one boundary. If the region is above the x‑axis and rotated about the x‑axis, the radius is just y. If rotated about the y‑axis, the roles swap, which we address separately.
务必检查旋转轴是否与某一边界重合。若区域在x轴上方且绕x轴旋转，半径就是y。如果绕y轴旋转，则角色互换，我们将单独讨论。
6. Rotation about the y‑axis | 绕y轴旋转
When the axis of rotation is the y‑ st be bounded by a curve of the form x = g(y), horizontal lines y = c and y = d, and the y‑axis itself. The volume formula then becomes:
当旋转轴为y轴时，圆盘法要求半径用y表示。区域须由形如 x = g(y) 的曲线、水平线 y = c 和 y = d 以及y轴本身围成。体积公式则变为：
V = π ∫_cᵈ [g(y)]² dy
quare roots or inverse trigonometric functions appear. Always solve for x in terms of y, and ensure you use the correct branch if the function is not one‑to‑one over the entire domain.
这种反转是练习18B.2测试的一项关键技能。许多学生觉得将给定方程 y = f(x) 改写为 x = g(y) 很棘手，特别是出现平方根或反三角函数时。务必解出x用y表示，并确保在函数不是整个定义域上一一对应时使用正确的分支。
7. Reversing the Roles of x and y | 交换x和y的角色
To illustrate, suppose the region under y = √x from x = 0 to x = 4 is rotated about the y‑ on y go from 0 to 2 because when x = 4, y = 2. The volume is π ∫₀² (y²)² dy = π ∫₀² y⁴ dy.
举例说明，若 y = √x 下方从 x=0 到 x=4 的区域绕y轴旋转。此时，曲线 x = y²（y ≥ 0）是相关的表示。y的积分限从0到2，因为当 x=4 时 y=2。体积为 π ∫₀² (y²)² dy = π ∫₀² y⁴ dy。
Evaluating yields (32π/5), which interestingly matches the previous x‑axis volume for a different region. Such symmetries occur often in Exercise 18B.2 and can serve as a partial check.
求值得到 (32π/5)，有趣的是这与之前不同区域绕x轴旋转的体积相同。这类对称性在练习18B.2中经常出现，可作为部分检验。
If the same region is bounded by the x‑axis and rotated about the y‑axis, you must be careful: the region is defined by vertical strips, so integration with respect to y is more natural. Always sketch and decide which variable makes the radius expression simpler.
如果同一区域由x轴界定且绕y轴旋转，则需小心：区域由垂直条带定义，因此关于y积分更自然。始终画图并决定哪个变量能使半径表达式更简单。
8. Example 2: Piecewise and Trigonometric Functions | 例题2: 分段与三角函数
Exercise 18B.2 might present a piecewise‑d function, such as a region bounded by y = sin x from x = 0 to x = π, rotated about the x‑ radius is sin x, and the square is sin² x. Integrate sin² x using the identity sin² x = (1 – cos 2x)/2.
练习18B.2可能出现分段定义的函数，例如由 y = sin x 从 x=0 到 x=π 围成的区域绕x轴旋转。半径为 sin x，平方为 sin² x。使用恒等式 sin² x = (1 – cos 2x)/2 进行积分。
V = π ∫₀^π sin² x dx = π ∫₀^π (1 – cos 2x)/2 dx
ghlights the need for trigonometric integration skills, which are assumed in IB Mathematics and regularly appear in this exercise set.
完成积分运算得到 V = (π²/2) 立方单位。该例题突出了对三角积分技巧的需求，这是IB数学所要求的，并经常出现在这组练习中。
For piecewise functions, break the integral at the points where the definition changes. Calculate the volume for each piece separately and sum the results, always respecting the original bounds.
对于分段函数，在定义改变的点处拆分积分。分别计算每一段的体积并求和，并始终遵循原始边界。
9. Combining with Other Integration Techniques | 结合其他积分技巧
Many problems in Exercise 18B.2 test your ability to integrate functions that require more than simple power rules. You may need u‑substitution when the function inside the square is composite, or integration by parts if the radius involves a product like x·eˣ.
练习18B.2中的许多问题考验你对需要超越简单幂法则的函数进行积分的能力。当平方内的函数是复合函数时，可能需要u代换；若半径包含如 x·eˣ 的乘积，则可能需要分部积分。
For instance, rotating the region under y = ln x between x = 1 and x = e about the x‑ lifying, and consider whether it is more efficient to rotate a different way, such as using shells, though Exercise 18B.2 sticks to disks.
例如，将 y = ln x 在 x=1 到 x=e 之间的区域绕x轴旋转，将得到 ∫ (ln x)² dx。这需要两次分部积分或识别一个标准结果。始终在化简前写出完整积分，并考虑是否采用不同方式旋转更有效，例如使用柱壳法，但练习18B.2坚持使用圆盘。
When a function involves roots like ∛x, express it with a fractional exponent before squaring. This small step saves time and prevents algebraic errors.
当函数包含如 ∛x 的根式时，在平方前先将它用分数指数表示。这个小步骤能节省时间并避免代数错误。
10. Common Mistakes to Avoid | 常见错误
One frequent error is forgetting to square the function. The volume formula contains [f(x)]², not f(x). Integrating π f(x) dx would compute an area scaled by π, not a volume. Always write the square explicitly before integrating.
一个常犯错误是忘记对函数平方。体积公式中含有 [f(x)]²，而不是 f(x)。对 π f(x) dx 积分会得到一个倍乘π的面积，而非体积。一定要在积分前明确写出平方。
te the limits accordingly. Also, check whether the region extends on both sides of the axis; if a part lies below the axis, squaring automatically handles it, but the integral must still cover the entire interval.
另一个错误是在变量代换后使用了错误的积分限。如果设 u 为 f(x) 内部的某表达式，则相应更新积分限。同时，检查区域是否延伸至轴的两侧；若有一部分位于轴下方，平方会自动处理，但积分仍须覆盖整个区间。
11. Practice Problem from Exercise 18B.2 | 练习18B.2中的一道自测题
Try this problem: The region bounded by y = 3x – x² and the x‑axis is revolved about the x‑axis. Find the volume. First, find the x‑intercepts by solving 3x – x² = 0 → x = 0 and x = 3. The function is above the axis on [0,3], so radius = 3x – x².
试做此题：由 y = 3x – x² 和x轴围成的区域绕x轴旋转。求体积。首先解 3x – x² = 0 → x=0 和 x=3 得到x截距。函数在 [0,3] 上位于轴上方，因此半径 = 3x – x²。
V = π ∫₀³ (3x – x²)² dx = π ∫₀³ (9x² – 6x³ + x⁴) dx
Integrate term by term: 9·(x³/3) = 3x³, –6·(x⁴/4) = –(3/2)x⁴, and (x⁵/5). Evaluate from 0 to 3 to get (81 – 243/2 + 243/5) = 81 – 121.5 + 48.6 = 8.1? Let's compute exactly: common denominator 10: 3x³ from 0 to 3 is 81, –(3/2)x⁴ = –(3/2)·81 = –121.5, + x⁵/5 = 243/5 = 48.6. Sum = 81 - 121.5 + 48.6 = 8.1. So V = (8.1)π = 81π/10.
逐项积分：9·(x³/3) = 3x³，–6·(x⁴/4) = –(3/2)x⁴，以及 (x⁵/5)。从0到3求值：3x³ 为 81，–(3/2)x⁴ = –121.5，x⁵/5 = 48.6。总和 = 8.1。因此 V = 8.1π = 81π/10 立方单位。
This result illustrates how accurate expansion and fraction arithmetic are vital. Double‑check signs and coefficients; one small slip can alter the entire volume.
这一结果表明准确的展开和分数运算至关重要。仔细检查符号和系数，一个小失误就可能改变整个体积。
12. Summary and Beyond Exercise 18B.2 | 总结与练习18B.2之后
Exercise 18B.2 is designed to solidify your grasp of the disk method for volumes of revolution. By working through polynomial, radical, trigonometric, and inverse functions, you develop the integration versatility needed for more advanced topics such as arc length and surface area in later chapters. Each problem reinforces the critical connection between geometry and calculus.
练习18B.2旨在巩固你对旋转体体积圆盘法的掌握。通过练习多项式、根式、三角函数和反函数等问题，你将培养出积分灵活性，为后续章节中弧长和表面积等更高级的主题做好准备。每道题都强化了几何与微积分之间的关键联系。
Once comfortable with disks, you can explore the washer method for regions between curves and the shell method for alternative axis orientations. But remember, the foundation built here – setting up the radius, squaring it, and integrating – remains the core skill.
一旦熟练掌握圆盘法，你就可以探索用于曲线间区域的垫圈法，以及用于不同轴方向的柱壳法。但请记住，这里打下的基础——设定半径、平方、积分——始终是核心技能。
As you complete the exercises, always reflect on the units: volumes are in cubic units, and the factor of π is a constant companion. With consistent practice, these problems will become both intuitive and rewarding.
在完成这些练习时，始终思考单位：体积以立方单位计，π因子始终相伴。通过持续练习，这些问题将变得直观而富有成就感。
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