📚 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 = x2 from x = 0 to x = 1 . The region bounded by this curve, the x-axis, and the line x = 1 is rotated through 360circ about the x-axis. Find the volume of the solid of revolution formed.
考虑曲线 y = x2 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 360circ 。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 π 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 π y² dx 。
3. Setting Up the Integral | 建立积分
Given y = x2 , we have y² = x4 . The limits of integration are x = 0 and x = 1 . Therefore, the volume is:
给定 y = x2 ，则 y² = x4 。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 π / 5 cubic units.
因此，该旋转体的精确体积为 π / 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 π / 5 is less than the volume of a cone of the same height and base radius, which would be π / 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 π / 5 小于同高度同底面半径的圆锥体积 π / 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 Δ 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 Δ 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 ≈ 0.1998π ≈ 0.6276 . The exact value π/5 ≈ 0.6283 , confirming our result.
计算中点四次方的和约为0.1998，所以 V ≈ 0.1998π ≈ 0.6276 。精确值 π/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 π/2 instead of π/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 π/2 而不是 π/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 π . The volume formula includes the factor π ; omitting it leads to an incorrect numerical answer.
忘记 π 。体积公式包含因子 π ；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 π/2 is larger than π/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 π/2 大于 π/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 360circ 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 360circ 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 = π ∫ y² dx or V = π ∫ x² dy depending on the axis.
根据旋转轴写出公式 V = π ∫ y² dx 或 V = π ∫ 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 = π , and the x-axis about the x-axis. Find the volume.
2. 将 y = sin x 、 x = 0 、 x = π 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. 8π 2. π² / 2 3. 128π / 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 = π ∫ r² dx or V = π ∫ r² dy .
应用体积公式 V = π ∫ r² dx 或 V = π ∫ 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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