Exercise 15H.2: Volumes of Revolution | 习题15H.2: 旋转体体积

📚 Exercise 15H.2: Volumes of Revolution | 习题15H.2: 旋转体体积

In IB Mathematics Analysis and Approaches (AA) Higher Level, integration is not only about finding areas under curves—it extends to calculating volumes of solids of revolution. Exercise 15H.2 focuses on using definite integrals to compute the volume generated when a plane region is rotated about a coordinate axis. This article provides a thorough walkthrough of the key concepts, step-by-step solutions for typical problems found in this exercise, and practical tips to avoid common pitfalls and excel in the exam.

在IB数学分析与方法（AA）高级课程中，积分不仅用于求曲线下的面积，还扩展到计算旋转体的体积。习题15H.2重点在于利用定积分计算平面区域绕坐标轴旋转所生成的体积。本文将从核心概念、典型习题的分步解答到常见错误避免和应试技巧，进行全面讲解。

1. Understanding Volumes of Revolution | 理解旋转体体积

A solid of revolution is created when a two‑dimensional region is rotated about a straight line called the axis of revolution. The most common axes in IB problems are the x‑axis and the y‑axis. To find the volume, we imagine slicing the solid perpendicular to the axis into infinitely many thin disks (or washers) and summing their volumes using integration. The disk method is the primary tool for Exercise 15H.2.

旋转体是将平面区域绕着一条称为旋转轴的直线旋转一周所形成的立体。IB考题中最常见的轴是x轴和y轴。为求体积，我们设想垂直于轴将立体切成无穷多个薄圆盘（或垫圈），再通过积分累加其体积。圆盘法是习题15H.2的核心方法。

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2. Formula for Revolution Around the x‑axis | 绕x轴旋转的公式

When the region bounded by y = f(x), the x‑axis, and the vertical lines x = a and x = b is rotated completely about the x‑axis, the resulting volume V is given by the disk formula:

V = π ∫_a^b [f(x)]² dx

This arises because each thin slice perpendicular to the x‑axis is a disk of radius |f(x)| and thickness Δx, so its volume is approximately π[f(x)]² Δx. Integration adds these infinitesimal contributions.

当由曲线 y = f(x)、x轴及直线 x = a 和 x = b 围成的区域绕x轴旋转一周时，所得体积V的圆盘公式为：

V = π ∫_a^b [f(x)]² dx

这是因为垂直于x轴的薄片是一个半径为|f(x)|、厚度为Δx的圆盘，其体积近似为π[f(x)]² Δx，积分累加这些微元即得总体积。

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3. Formula for Revolution Around the y‑axis | 绕y轴旋转的公式

If the region is bounded by x = g(y), the y‑axis, and the horizontal lines y = c and y = d, rotation about the y‑axis yields:

V = π ∫_c^d [g(y)]² dy

Notice that the roles of x and y are swapped: the radius is given by the horizontal distance from the y‑axis, which is |g(y)|, and integration is with respect to y. This symmetric extension is frequently tested alongside x‑axis revolutions.

若区域由曲线 x = g(y)、y轴及直线 y = c 和 y = d 围成，绕y轴旋转则得到：

V = π ∫_c^d [g(y)]² dy

注意x与y的角色互换：半径为由y轴出发的水平距离|g(y)|，积分变量为y。这种对称扩展常与绕x轴旋转一同考查。

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4. Step‑by‑Step Solution to Exercise 15H.2(a) | 习题15H.2(a)分步解答

Problem: The region R is bounded by y = x², the x‑axis, x = 1 and x = 2. Find the volume generated when R is rotated through 360° about the x‑axis.

Using the formula V = π ∫₁² (x²)² dx = π ∫₁² x⁴ dx. Integrate: π [x⁵/5]₁² = π (32/5 − 1/5) = 31π/5. The volume is 31π/5 cubic units.

题目：区域R由 y = x²、x轴、x = 1 和 x = 2 围成。将R绕x轴旋转360°，求生成的体积。

应用公式 V = π ∫₁² (x²)² dx = π ∫₁² x⁴ dx。积分得 π [x⁵/5]₁² = π (32/5 − 1/5) = 31π/5。体积为 31π/5 立方单位。

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5. Step‑by‑Step Solution to Exercise 15H.2(b) | 习题15H.2(b)分步解答

Problem: The region S is bounded by y = √x, the y‑axis, y = 1 and y = 2. Find the volume generated when S is rotated through 360° about the y‑axis.

First express x in terms of y: from y = √x we obtain x = y². The limits are y = 1 to y = 2. The volume is V = π ∫₁² (y²)² dy = π ∫₁² y⁴ dy = π [y⁵/5]₁² = π (32/5 − 1/5) = 31π/5. Interestingly, this yields the same numerical result, demonstrating the symmetry of the two problems.

题目：区域S由 y = √x、y轴、y = 1 和 y = 2 围成。将S绕y轴旋转360°，求生成的体积。

首先将x表示为y的函数：由 y = √x 得 x = y²。积分限为 y = 1 到 y = 2。体积 V = π ∫₁² (y²)² dy = π ∫₁² y⁴ dy = π [y⁵/5]₁² = π (32/5 − 1/5) = 31π/5。有趣的是，此题得到同样的数值结果，展示了两题的对称性。

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6. Handling Limits and Definite Integration | 处理积分限和定积分

When setting up the integral, always ensure the limits match the variable of integration. For x‑axis rotation, the limits are x‑values; for y‑axis rotation, limits are y‑values. If a region is bounded by both curves, carefully identify the correct upper and lower boundaries. Never guess the limits—sketch the region or solve intersection points algebraically. Also, remember to square the function before integrating. A common error is to integrate f(x) and then square the result; that is incorrect.

构建积分时，务必使积分限与积分变量一致。绕x轴旋转时限为x值；绕y轴旋转时限为y值。若区域由两条曲线围成，应仔细确定上下边界。不要凭空猜测积分限——可绘制草图或通过解方程求交点。另外，务必先将函数平方再积分。常见错误是先积分f(x)再对结果平方，这是完全错误的。

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7. Common Mistakes and How to Avoid Them | 常见错误与避免方法

• Forgetting the square: Volume integrals require [f(x)]², not f(x). Always write the square explicitly inside the integral. Tip: write (function)² immediately after setting up the formula.
• Confusing axes: When rotating about the y‑axis, you must rewrite the equation as x = g(y) and integrate with respect to y. Double‑check the variable in the limits and the differential.
• Misidentifying limits: For regions between curves, subtraction may be needed, and the outer and inner radii must be found. Practice with washers before moving to more complex shapes.
• Ignoring units: While IB often leaves answers in terms of π, always state “cubic units” to show understanding of volume.

• 忘记平方：体积公式需要 [f(x)]²，而非 f(x)。应在积分内明确写出平方。技巧：列式后立刻写出 (函数)²。
• 混淆坐标轴：绕y轴旋转时，必须将方程改写为 x = g(y) 并对y积分。仔细检查积分变量和上下限。
• 积分限判断错误：对于曲线间区域，可能需用垫圈法并确定外半径和内半径。先练好圆盘法再处理复杂形状。
• 忽略单位：尽管IB常允许答案保留π，但最好注明“立方单位”以体现对体积的理解。

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8. Graphical Interpretation | 图形解释

Visualising the solid helps avoid algebraic mistakes. For Exercise 15H.2(a), imagine revolving the parabolic strip between x=1 and x=2 about the x‑axis. Each cross‑section perpendicular to the x‑axis is a disk of radius x². As x increases, the radius grows, creating a flared solid resembling a trumpet. For 15H.2(b), revolving the region under y=√x (or x=y²) about the y‑axis produces a solid whose cross‑section perpendicular to the y‑axis is a disk of radius y², giving a similar flared shape. Sketching the region and a typical disk solidifies the concept of “summing disks.”

将立体可视化有助于避免代数错误。对于习题15H.2(a)，想象将x=1到x=2之间的抛物线带状区域绕x轴旋转。每个垂直于x轴的截面都是半径为x²的圆盘。随着x增大，半径增加，形成一个向外扩大的类似喇叭的立体。15H.2(b)中，将y=√x（即x=y²）下的区域绕y轴旋转，产生一个垂直于y轴的截面半径为y²的圆盘构成的立体，形状同样向外展开。绘制区域和一个典型圆盘能强化“圆盘求和”的概念。

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9. Using Technology: GDC Tips | 使用技术：图形计算器技巧

In IB exams, you may use a graphical display calculator (GDC) to evaluate definite integrals. After setting up the integral analytically (e.g., π ∫₁² x⁴ dx), you can compute it directly using the integration function. However, always show the setup on paper—the GDC is only a calculation tool. Be aware that the calculator gives a decimal approximation when the exact answer involves π; to obtain the precise value, input the integral and check that the result matches your manual calculation (e.g., 31π/5 ≈ 19.48). Use the GDC to verify your work, not to replace understanding.

在IB考试中，可用图形计算器（GDC）计算定积分。在建立解析式（如 π ∫₁² x⁴ dx）后，可直接用积分功能求值。但必须在卷面上展示式子构建过程——GDC仅是计算工具。注意当精确值含π时，计算器仅给出小数近似；为获得准确值，可输入积分并核对与手算结果是否一致（如31π/5 ≈ 19.48）。用GDC验证工作，而非取代理解。

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10. Practice Variations | 练习变化

To deepen understanding, try modifying Exercise 15H.2:

• Rotate the region between y = x² and y = x about the x‑axis between x=0 and x=1 (requires washers).
• Rotate the same region about the y‑axis using the shell method (if covered) or by expressing bounds in terms of y.
• Find the volume when the region bounded by y = sin x, x=0, x=π/2 and the x‑axis is rotated about the x‑axis. This leads to π ∫₀^π/2 sin²x dx, requiring a trigonometric identity.

Such variations test the ability to adapt the core method and are excellent exam preparation.

为加深理解，可尝试以下变式：

• 将 y = x² 和 y = x 之间的区域绕x轴旋转（x从0到1），需用垫圈法。
• 将同一区域绕y轴旋转，可利用壳层法（如已学习）或以y表示边界。
• 求由 y = sin x、x=0、x=π/2 和x轴围成的区域绕x轴旋转的体积。这将导出 π ∫₀^π/2 sin²x dx，需用三角恒等式。

此类变式考查核心方法的灵活运用能力，是极佳的备考训练。

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11. Linking to IB Exam Questions | 链接IB考题

Typical IB AA HL Paper 1 or Paper 2 questions may ask: “The region enclosed by the curves y = x³ and y = √x is rotated about the x‑axis. Find the volume of the solid generated.” Here you must first find intersection points (0,0) and (1,1), then set up the difference of squares: V = π ∫₀¹ [(√x)² − (x³)²] dx = π ∫₀¹ (x − x⁶) dx. The exercise 15H.2 methodology of squaring, integrating, and applying limits transfers directly. Always look for opportunities to link back to the disk or washer method.

典型的IB AA HL试卷一或试卷二可能给出：“求由曲线 y = x³ 和 y = √x 围成的区域绕x轴旋转所得立体的体积。”此时首先求交点(0,0)和(1,1)，然后建立平方差：V = π ∫₀¹ [(√x)² − (x³)²] dx = π ∫₀¹ (x − x⁶) dx。习题15H.2中的平方后积分、代入限值的方法可以直接迁移。解题时始终寻找与圆盘或垫圈法的联系。

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12. Summary and Key Takeaways | 总结与关键点

Exercise 15H.2 solidifies the disk method for volumes of revolution. Remember: Identify the axis → express the radius in terms of the integration variable → square the radius → set up the definite integral → integrate → evaluate at limits. Pay meticulous attention to squaring and limit selection. Regular practice with both x‑axis and y‑axis rotations, alongside graphical visualisation and GDC verification, will build the fluency needed for top marks in the IB exam.

习题15H.2巩固了旋转体体积的圆盘法。牢记：确定旋转轴 → 用积分变量表示半径 → 平方半径 → 建立定积分 → 积分 → 代入上下限求值。特别注意平方和积分限的选择。通过绕x轴和y轴旋转的充分练习，结合图形可视化与GDC验证，将培养出IB考试夺高分所需的熟练度。

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