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

📚 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。

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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。

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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 )。

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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。

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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。

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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。

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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 ≈ 0.1998pi ≈ 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 ≈ 0.1998pi ≈ 0.6276 )。精确值 ( pi/5 ≈ 0.6283 )，验证了我们的结果。

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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。

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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。

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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。

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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. ( 8pi ) 2. ( pi² / 2 ) 3. ( 128pi / 7 )

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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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