📚 Volume of Solids of Revolution | 旋转体体积的计算方法
In A-level Mathematics, calculating the volume of a solid formed by rotating a curve around an axis is a key application of integration. This article explains the standard methods, formulas, and common pitfalls to help you master this topic.
在 A-level 数学中，利用积分计算曲线绕坐标轴旋转所形成立体的体积，是积分应用的核心考点。本文将系统讲解标准方法、公式以及常见易错点，帮助你全面掌握这一知识点。
1. The Basic Idea | 基本思想
When a region under a curve y = f(x) between x = a and x = b is rotated through 360° about the x-axis, it sweeps out a three-dimensional solid. To find its volume, we imagine slicing the solid into thin discs perpendicular to the x-axis.
当曲线 y = f(x) 在 x = a 到 x = b 之间的区域绕 x 轴旋转 360° 时，会扫出一个三维立体。为了求其体积，我们想象将该立体沿 x 轴方向切成许多薄圆盘。
Each disc has radius y = f(x) and thickness δx. Its volume is approximately πy²δx. Summing these volumes and taking the limit as δx → 0 gives the integral formula.
每个薄圆盘的半径为 y = f(x)，厚度为 δx，其体积近似为 πy²δx。将这些体积求和并令 δx → 0 取极限，便得到积分公式。
V = π ∫ₐᵇ y² dx
This is the disc method for rotation about the x-axis.
这就是绕 x 轴旋转时的圆盘法公式。
2. Rotation about the x-axis | 绕 x 轴旋转
If the region bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is rotated about the x-axis, the volume is given by:
若由曲线 y = f(x)、x 轴以及直线 x = a 与 x = b 所围成的区域绕 x 轴旋转，其体积公式为：
V = π ∫ₐᵇ [f(x)]² dx
It is essential that f(x) is non-negative on [a, b], or you must use the absolute value squared, which is the same as squaring.
注意 f(x) 在 [a, b] 上必须非负，或者直接使用平方，因为平方后符号不影响结果。
Example: Find the volume generated by rotating y = √x from x = 0 to x = 4 about the x-axis.
例：求曲线 y = √x 从 x = 0 到 x = 4 绕 x 轴旋转所得立体的体积。
Solution: V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = 8π.
解：V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = 8π。
3. Rotation about the y-axis | 绕 y 轴旋转
When the region under a curve is rotated about the y-axis, we use horizontal discs. Express x as a function of y, x = g(y), with y ranging from c to d.
当曲线下的区域绕 y 轴旋转时，我们使用水平薄圆盘。将 x 表示为 y 的函数 x = g(y)，y 的取值范围为 c 到 d。
V = π ∫꜀ᵈ x² dy
Alternatively, this can be written as V = π ∫꜀ᵈ [g(y)]² dy.
等价地，可写成 V = π ∫꜀ᵈ [g(y)]² dy。
Example: Rotate the region bounded by y = x², the y-axis, and y = 4 about the y-axis. Since y = x², we have x = √y for y ∈ [0,4].
例：求由 y = x²、y 轴和 y = 4 围成的区域绕 y 轴旋转所得体积。由 y = x² 得 x = √y，y ∈ [0,4]。
V = π ∫₀⁴ (√y)² dy = π ∫₀⁴ y dy = 8π.
V = π ∫₀⁴ (√y)² dy = π ∫₀⁴ y dy = 8π。
4. Rotation about a Horizontal Line | 绕水平线旋转
If the axis of rotation is a horizontal line y = k, the radius of each disc is the vertical distance from the curve to the line, i.e. |f(x) − k|. The volume is:
如果旋转轴是水平线 y = k，则每个圆盘的半径是曲线上点到该线的竖直距离，即 |f(x) − k|。体积公式为：
V = π ∫ₐᵇ [f(x) − k]² dx
Make sure you square the whole expression, not each term separately.
务必对整个表达式平方，而不是分别平方各项。
For example, rotating y = x² from x = 0 to x = 2 about y = −1 gives:
例如，将 y = x² 从 x = 0 到 x = 2 绕 y = −1 旋转，得到：
V = π ∫₀² (x² + 1)² dx = π ∫₀² (x⁴ + 2x² + 1) dx = π [x⁵/5 + 2x³/3 + x]₀² = π (32/5 + 16/3 + 2).
V = π ∫₀² (x² + 1)² dx = π ∫₀² (x⁴ + 2x² + 1) dx = π [x⁵/5 + 2x³/3 + x]₀² = π (32/5 + 16/3 + 2)。
5. Rotation about a Vertical Line | 绕垂直线旋转
If the axis is a vertical line x = h, integrate with respect to y, using the horizontal distance from the curve to the line, |x − h|, as the radius.
若旋转轴为垂直线 x = h，则对 y 积分，以曲线上点到该线的水平距离 |x − h| 作为半径。
V = π ∫꜀ᵈ [g(y) − h]² dy
Again, square the whole bracket. Pay attention to which side of the axis the curve lies.
同样，对整个括号平方。注意曲线位于轴的哪一侧。
Example: Rotate the region bounded by x = y², y = 0, y = 2 about x = 3. Here x ranges from 0 to 4, but for y between 0 and 2, x = y² ≤ 4 < 3? Actually y² can be > 3, so the distance is |y² − 3|.
例：求由 x = y²、y = 0、y = 2 围成的区域绕 x = 3 旋转所得体积。这里 x 范围是 0 到 4，但 y 在 0 到 2 之间时，x = y² 可能大于 3，所以距离为 |y² − 3|。
V = π ∫₀² (y² − 3)² dy = π ∫₀² (y⁴ − 6y² + 9) dy = π [y⁵/5 − 2y³ + 9y]₀² = π (32/5 − 16 + 18) = π (32/5 + 2) = 42π/5.
V = π ∫₀² (y² − 3)² dy = π ∫₀² (y⁴ − 6y² + 9) dy = π [y⁵/5 − 2y³ + 9y]₀² = π (32/5 − 16 + 18) = π (32/5 + 2) = 42π/5。
6. The Shell Method | 壳层法
For rotation about the y-axis, an alternative to horizontal discs is the cylindrical shell method. A vertical strip of width δx at distance x from the y-axis, with height y = f(x), rotates to form a thin shell of circumference 2πx, height y, and thickness δx.
对于绕 y 轴旋转的情况，除了水平圆盘法，还可以用圆柱壳层法。距 y 轴为 x、宽度为 δx、高度为 y = f(x) 的竖直条带旋转后形成一个薄壳，其周长为 2πx，高度为 y，厚度为 δx。
Volume of shell ≈ 2πx · y · δx, so the total volume is:
壳的体积近似为 2πx · y · δx，因此总体积为：
V = 2π ∫ₐᵇ x f(x) dx
This method is particularly useful when solving for x in terms of y is difficult.
当难以将 x 表示为 y 的函数时，此方法特别有用。
Example: Use the shell method to find the volume of the solid obtained by rotating the region under y = x² from x = 0 to x = 2 about the y-axis.
例：用壳层法求曲线 y = x² 在 x = 0 到 x = 2 之间的区域绕 y 轴旋转所得立体的体积。
V = 2π ∫₀² x · x² dx = 2π ∫₀² x³ dx = 2π [x⁴/4]₀² = 2π (16/4) = 8π.
V = 2π ∫₀² x · x² dx = 2π ∫₀² x³ dx = 2π [x⁴/4]₀² = 2π (16/4) = 8π。
7. Shell Method for the x-axis | 绕 x 轴的壳层法
Similarly, to rotate about the x-axis, use horizontal strips. A horizontal strip of width δy at distance y from the x-axis, with length x = g(y), generates a shell of volume 2πy · x · δy.
类似地，绕 x 轴旋转时使用水平条带。距 x 轴为 y、宽度为 δy、长度为 x = g(y) 的水平条带生成体积为 2πy · x · δy 的壳。
V = 2π ∫꜀ᵈ y g(y) dy
This is equivalent to the disc method applied to the inverse function.
这等价于对反函数应用圆盘法。
When choosing between methods, pick the one that avoids solving for the inverse function or splitting the region.
在选择方法时，应优先选择能避免求反函数或分割区域的方法。
8. Volume between Two Curves | 两曲线之间的旋转体体积
When a region between two curves is rotated about an axis, the volume is the difference of the volumes generated by the outer and inner curves.
当两条曲线之间的区域绕轴旋转时，体积等于外曲线与内曲线分别生成的体积之差。
V = π ∫ₐᵇ ([f(x)]² − [g(x)]²) dx
where f(x) is the upper (outer) curve and g(x) is the lower (inner) curve, assuming rotation about the x-axis.
其中 f(x) 是上方（外侧）曲线，g(x) 是下方（内侧）曲线，适用于绕 x 轴旋转。
Example: Find the volume generated by rotating the region between y = x and y = x² about the x-axis. The curves intersect at x = 0 and x = 1.
例：求由 y = x 与 y = x² 围成的区域绕 x 轴旋转所得体积。两曲线交于 x = 0 和 x = 1。
For 0 < x < 1, x > x², so f(x) = x and g(x) = x².
在 0 < x < 1 内，x > x²，因此 f(x) = x，g(x) = x²。
V = π ∫₀¹ [x² − (x²)²] dx = π ∫₀¹ (x² − x⁴) dx = π [x³/3 − x⁵/5]₀¹ = π (1/3 − 1/5) = 2π/15.
V = π ∫₀¹ [x² − (x²)²] dx = π ∫₀¹ (x² − x⁴) dx = π [x³/3 − x⁵/5]₀¹ = π (1/3 − 1/5) = 2π/15。
9. Using Parametric Equations | 参数方程的情形
If a curve is given parametrically as x = x(t), y = y(t), then dx = x′(t) dt, and the volume of revolution about the x-axis becomes:
若曲线由参数方程 x = x(t)、y = y(t) 给出，则 dx = x′(t) dt，绕 x 轴旋转的体积变为：
V = π ∫ₜ₁ᵗ² [y(t)]² x′(t) dt
Be careful with the direction of integration: the limits t₁ and t₂ correspond to the original x-limits.
注意积分方向：t₁ 和 t₂ 对应原 x 的上下限。
Similarly, for rotation about the y-axis, use V = π ∫ [x(t)]² y′(t) dt.
类似地，绕 y 轴旋转时，使用 V = π ∫ [x(t)]² y′(t) dt。
10. Worked Exam Example | 真题演算示例
Question: The region enclosed by the curve y = eˣ, the x-axis, and the lines x = 1 and x = 2 is rotated through 2π radians about the x-axis. Find the volume of the solid generated.
题目：由曲线 y = eˣ、x 轴以及直线 x = 1 和 x = 2 所围成的区域绕 x 轴旋转 2π 弧度，求所得立体的体积。
Solution:
解：
V = π ∫₁² (eˣ)² dx = π ∫₁² e²ˣ dx
V = π ∫₁² (eˣ)² dx = π ∫₁² e²ˣ dx
= π [e²ˣ/2]₁² = π/2 (e⁴ − e²)
= π [e²ˣ/2]₁² = π/2 (e⁴ − e²)
Thus the volume is π/2 (e⁴ − e²) cubic units.
因此体积为 π/2 (e⁴ − e²) 立方单位。
11. Common Mistakes and Tips | 常见错误与技巧
Common mistakes include:
常见错误包括：
- Forgetting to square the function before integrating.
- Forgetting to multiply by π.
- Using the wrong variable of integration or wrong limits.
- Subtracting volumes instead of subtracting squares of radii.
- Ignoring discontinuities or negative regions.
- 积分前忘记对函数平方。
- 忘记乘以 π。
- 用错积分变量或上下限。
- 错误地相减体积而不是相减半径的平方。
- 忽略不连续点或负值区域。
Tips:
技巧：
- Always sketch the curve and the axis of rotation first.
- Identify whether the disc or shell method is simpler.
- If the axis is not x or y, shift coordinates or use the distance formula.
- Check that your final volume is positive and reasonable.
- 先画出曲线与旋转轴。
- 判断圆盘法还是壳层法更简单。
- 若轴不是 x 或 y 轴，可平移坐标或用距离公式。
- 检查最终体积为正且数值合理。
12. Summary | 总结
The volume of a solid of revolution can be found using the disc method or the shell method. For rotation about the x-axis, use V = π ∫ y² dx or V = 2π ∫ y x dy depending on the orientation. The key is to correctly identify the radius and the variable of integration.
旋转体体积可以通过圆盘法或壳层法求解。绕 x 轴旋转时，根据情况使用 V = π ∫ y² dx 或 V = 2π ∫ y x dy。关键是正确识别半径与积分变量。
Remember to square the entire radius expression and always include π. Practice with both methods to build confidence.
记住对整个半径表达式平方，并始终包含 π。通过两种方法的练习来增强信心。
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