Example 5.5.4 – Application of Integration for Volumes of Revolution | 例5.5.4

📚 Example 5.5.4 – Application of Integration for Volumes of Revolution | 例5.5.4 – 旋转体体积的积分应用

In this worked example, we explore a classic AQA A-Level Mathematics problem involving the volume of a solid generated by rotating a curve about an axis. Example 5.5.4 demonstrates how to set up a definite integral, evaluate it using standard techniques, and interpret the result geometrically.

在本例题中，我们研究一个典型的AQA A-Level数学问题：将一条曲线绕坐标轴旋转所得旋转体的体积。例5.5.4展示了如何建立定积分、用标准方法计算积分，并从几何上解释结果。

1. Problem Statement | 题目陈述

Consider the curve \( y = x^2 \) from \( x = 0 \) to \( x = 1 \). The region bounded by this curve, the x-axis, and the line \( x = 1 \) is rotated through \( 360^\circ \) about the x-axis. Find the volume of the solid of revolution formed.

考虑曲线 \( y = x^2 \) 从 \( x = 0 \) 到 \( x = 1 \) 的部分。由这条曲线、x轴和直线 \( x = 1 \) 所围成的区域绕x轴旋转 \( 360^\circ \)。求所形成的旋转体的体积。

2. Method: Volume about the x-axis | 方法：绕x轴旋转的体积

For a curve \( y = f(x) \) rotated about the x-axis between \( x = a \) and \( x = b \), the volume is given by the integral formula:

对于曲线 \( y = f(x) \) 在 \( x = a \) 和 \( x = b \) 之间绕x轴旋转，体积由以下积分公式给出：

V = π ∫ₐᵇ y² dx

This formula arises from summing infinitesimally thin discs of radius \( y \) and thickness \( dx \). Each disc has volume \( \pi y² dx \).

该公式来源于对半径 \( y \)、厚度 \( dx \) 的无限薄圆盘求和。每个圆盘的体积为 \( \pi y² dx \)。

3. Setting Up the Integral | 建立积分

Given \( y = x^2 \), we have \( y² = x^4 \). The limits of integration are \( x = 0 \) and \( x = 1 \). Therefore, the volume is:

给定 \( y = x^2 \)，则 \( y² = x^4 \)。积分上下限为 \( x = 0 \) 和 \( x = 1 \)。因此体积为：

V = π ∫₀¹ (x²)² dx = π ∫₀¹ x⁴ dx

We have successfully converted the geometry problem into a definite integral. The next step is to evaluate this integral.

我们已成功将几何问题转化为定积分。下一步是计算该积分。

4. Evaluating the Integral | 计算积分

Using the power rule for integration:

使用幂函数积分法则：

∫ x⁴ dx = x⁵ / 5 + C

Applying the limits \( 0 \) and \( 1 \):

代入上下限 \( 0 \) 和 \( 1 \)：

V = π [ x⁵ / 5 ]₀¹ = π ( 1⁵ / 5 – 0⁵ / 5 ) = π / 5

Thus, the exact volume of the solid is \( \pi / 5 \) cubic units.

因此，该旋转体的精确体积为 \( \pi / 5 \) 立方单位。

5. Geometric Interpretation | 几何解释

The solid formed has a parabolic profile and resembles a smooth, pointed dome. The maximum radius is 1 at \( x = 1 \), and the radius decreases to 0 at \( x = 0 \). The volume \( \pi / 5 \) is less than the volume of a cone of the same height and base radius, which would be \( \pi / 3 \), because the parabola lies below the line \( y = x \).

所生成的固体具有抛物线轮廓，类似于一个平滑的尖顶圆顶。最大半径为 \( x = 1 \) 处的1，半径在 \( x = 0 \) 处减小到0。体积 \( \pi / 5 \) 小于同高度同底面半径的圆锥体积 \( \pi / 3 \)，因为抛物线位于直线 \( y = x \) 下方。

6. Verifying with the Washer Method (Alternative) | 用垫圈法验证（替代方法）

If instead we rotate the region about the y-axis, we would use the washer method, but in this example we only rotate about the x-axis. However, we can verify our result by approximating the volume numerically using a midpoint Riemann sum.

如果我们改为绕y轴旋转，则需要使用垫圈法，但本例仅绕x轴旋转。不过，我们可以通过中点黎曼和进行数值近似来验证结果。

Divide \( [0,1] \) into 10 equal subintervals of width \( \Delta x = 0.1 \). Use midpoints \( x = 0.05, 0.15, …, 0.95 \). The approximate volume is:

将 \( [0,1] \) 分成10个等宽子区间，宽度 \( \Delta x = 0.1 \)。取中点 \( x = 0.05, 0.15, …, 0.95 \)。近似体积为：

V ≈ π Σ (xᵢ)⁴ Δx = π × 0.1 × Σ (xᵢ)⁴

Computing the sum of the fourth powers of the midpoints gives approximately 0.1998, so \( V ≈ 0.1998\pi ≈ 0.6276 \). The exact value \( \pi/5 ≈ 0.6283 \), confirming our result.

计算中点四次方的和约为0.1998，所以 \( V ≈ 0.1998\pi ≈ 0.6276 \)。精确值 \( \pi/5 ≈ 0.6283 \)，验证了我们的结果。

7. Common Mistakes to Avoid | 常见错误避坑

Forgetting to square the function. The formula requires \( y² \), not \( y \). Using \( y \) instead of \( y² \) gives \( \pi/2 \) instead of \( \pi/5 \).

忘记对函数平方。公式要求 \( y² \)，而不是 \( y \)。使用 \( y \) 而非 \( y² \) 会得到 \( \pi/2 \) 而不是 \( \pi/5 \)。
Incorrect limits. Always identify the correct x-interval from the problem statement. Here, the region is bounded by \( x = 0 \) and \( x = 1 \).

上下限错误。始终从题目中识别正确的x区间。这里区域由 \( x = 0 \) 和 \( x = 1 \) 界定。
Forgetting \( \pi \). The volume formula includes the factor \( \pi \); omitting it leads to an incorrect numerical answer.

忘记 \( \pi \)。体积公式包含因子 \( \pi \)；漏掉它会导致数值答案错误。
Not simplifying the integrand. \( (x²)² \) must be simplified to \( x⁴ \) before integrating.

未化简被积函数。必须先化简 \( (x²)² \) 为 \( x⁴ \)，再积分。

8. Extension: Rotating about the y-axis | 扩展：绕y轴旋转

If the same region were rotated about the y-axis, we would express \( x \) as a function of \( y \): since \( y = x² \), \( x = √y \). The volume would then be:

如果同一区域绕y轴旋转，我们需要将 \( x \) 表示为 \( y \) 的函数：因为 \( y = x² \)，所以 \( x = √y \)。体积则为：

V = π ∫₀¹ (√y)² dy = π ∫₀¹ y dy = π [ y² / 2 ]₀¹ = π / 2

This is a different solid, and its volume \( \pi/2 \) is larger than \( \pi/5 \). Understanding which axis to rotate about is crucial in exam questions.

这是一个不同的旋转体，其体积 \( \pi/2 \) 大于 \( \pi/5 \)。理解绕哪个轴旋转在考试题目中至关重要。

9. Connection to Exam Questions | 与考试题目的联系

Example 5.5.4 is representative of AQA A-Level questions that ask students to “find the volume of the solid formed when the region bounded by … is rotated through \( 360^\circ \) about the x-axis.” These questions often involve curves like \( y = x² \), \( y = √x \), or trigonometric functions.

例5.5.4代表了AQA A-Level中常见的题型：“求由……围成的区域绕x轴旋转 \( 360^\circ \) 所形成的旋转体体积”。这些问题常涉及曲线如 \( y = x² \)、\( y = √x \) 或三角函数。

In the exam, you should:

在考试中，你应当：

Sketch the curve and shade the region to visualise the solid.

画出曲线并标出区域，以可视化旋转体。
Write down the formula \( V = \pi \int y² dx \) or \( V = \pi \int x² dy \) depending on the axis.

根据旋转轴写出公式 \( V = \pi \int y² dx \) 或 \( V = \pi \int x² dy \)。
Evaluate the integral carefully, showing all working.

仔细计算积分，并展示所有步骤。

10. Further Practice Problems | 更多练习题目

Try these similar questions to master the technique:

尝试以下类似题目以掌握该技巧：

1. Rotate the region bounded by \( y = √x \), \( x = 0 \), \( x = 4 \), and the x-axis about the x-axis. Find the volume.

1. 将 \( y = √x \)、\( x = 0 \)、\( x = 4 \) 和x轴所围区域绕x轴旋转，求体积。
2. Rotate the region bounded by \( y = \sin x \), \( x = 0 \), \( x = \pi \), and the x-axis about the x-axis. Find the volume.

2. 将 \( y = \sin x \)、\( x = 0 \)、\( x = \pi \) 和x轴所围区域绕x轴旋转，求体积。
3. Rotate the region bounded by \( y = x³ \), \( x = 0 \), \( x = 2 \), and the x-axis about the x-axis. Find the volume.

3. 将 \( y = x³ \)、\( x = 0 \)、\( x = 2 \) 和x轴所围区域绕x轴旋转，求体积。

Answers: 1. \( 8\pi \) 2. \( \pi² / 2 \) 3. \( 128\pi / 7 \)

11. Summary of Key Steps | 关键步骤总结

Identify the axis of rotation and the limits of integration.

确定旋转轴和积分上下限。
Express the radius of the cross-sectional disc in terms of the variable of integration.

用积分变量表示横截面圆盘的半径。
Apply the volume formula \( V = \pi \int r² \, dx \) or \( V = \pi \int r² \, dy \).

应用体积公式 \( V = \pi \int r² \, dx \) 或 \( V = \pi \int r² \, dy \)。
Integrate and evaluate using the limits to obtain the exact volume.

积分并根据上下限计算，得到精确体积。
Check the reasonableness of your answer using numerical approximation or a sketch.

通过数值近似或草图检查答案的合理性。

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Example 5.5.4 – Application of Integration for Volumes of Revolution | 例5.5.4 – 旋转体体积的积分应用