Volumes | 体积

📚 Volumes | 体积

Volumes are a central topic in AQA A-Level Mathematics, appearing in pure integration, geometry, and applied rates-of-change problems. This revision guide covers every volume-related skill you need for the exam: solids of revolution, standard formulae, composite regions, parametric forms, similarity, and connected rates.

体积是 AQA A-Level 数学的核心考点，出现在纯数积分、几何以及应用题中的变化率问题里。本复习指南涵盖考试所需的全部体积技能：旋转体、标准公式、组合区域、参数形式、相似形以及关联变化率。

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1. Volume of Revolution about the x-axis | 绕 x 轴旋转体的体积

When a curve y = f(x) is rotated through 360° about the x-axis between x = a and x = b, the volume of the solid formed is given by integrating the area of circular cross-sections. Each disc has radius y and thickness dx, so its volume is πy² dx.

当曲线 y = f(x) 在 x = a 与 x = b 之间绕 x 轴旋转 360° 时，所形成立体的体积通过对圆形截面面积积分得到。每个圆盘的半径为 y、厚度为 dx，因此其体积为 πy² dx。

V = π∫ₐᵇ y² dx

This formula is the most frequently tested volume result in AQA Paper 1. You must square the function completely before integrating – a common slip is to integrate y and then square the result, which is incorrect.

这是 AQA Paper 1 中考查频率最高的体积公式。你必须先将函数完全平方再积分——常见错误是先对 y 积分再平方，这样做是不对的。

Worked Example: Find the volume of the solid formed when y = √x is rotated about the x-axis from x = 1 to x = 4.

例：求曲线 y = √x 从 x = 1 到 x = 4 绕 x 轴旋转一周所得立体的体积。

V = π∫₁⁴ (√x)² dx = π∫₁⁴ x dx = π[x²/2]₁⁴ = π(16/2 − 1/2) = 15π/2

V = 15π/2 cubic units

The x-axis itself forms the axis of symmetry of the solid. If the curve dips below the x-axis, y² is still positive, so the formula automatically handles negative regions.

x 轴本身构成立体的对称轴。若曲线位于 x 轴下方，y² 仍为正，因此公式自动处理负值区域。

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2. Volume of Revolution about the y-axis | 绕 y 轴旋转体的体积

When a curve is rotated about the y-axis, the roles of x and y are swapped. You must express x as a function of y, giving cross-sections of radius x and thickness dy.

当曲线绕 y 轴旋转时，x 与 y 的角色互换。你必须将 x 表示为 y 的函数，得到半径为 x、厚度为 dy 的截面。

V = π∫ₐᵇ x² dy

For example, the curve y = x³ from y = 0 to y = 8 is rearranged as x = y^(1/3). The volume about the y-axis is then:

例如，曲线 y = x³ 从 y = 0 到 y = 8，改写为 x = y^(1/3)，绕 y 轴旋转的体积为：

V = π∫₀⁸ y^(2/3) dy = π[(3/5)y^(5/3)]₀⁸ = π × (3/5) × 32 = 96π/5

Note that when rotating about the y-axis, the integration limits are y-values, not x-values. Always check which axis is the axis of rotation before setting up the integral.

注意绕 y 轴旋转时，积分限是 y 值，而非 x 值。建立积分前务必先确认旋转轴是哪一条。

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3. Volume of Revolution with a Gap: The Washer Method | 有间隙的旋转体：垫圈法

If the region between two curves y₁ and y₂ is rotated about the x-axis, we subtract the inner volume from the outer volume. This is called the washer or annulus method.

若两条曲线 y₁ 与 y₂ 之间的区域绕 x 轴旋转，则用外侧体积减去内侧体积。这称为垫圈法或环形法。

V = π∫ₐᵇ (y₁² − y₂²) dx

Here y₁ is the upper curve and y₂ is the lower curve. It is essential to identify which curve is outermost before squaring – swapping them gives a negative volume, which signals an error.

其中 y₁ 是上方曲线，y₂ 是下方曲线。平方前必须确认哪条曲线在外侧——颠倒两者会得到负体积，表明出错。

Example: The region between y = 2x and y = x² from x = 0 to x = 2 is rotated about the x-axis. Find the volume.

例：区域介于 y = 2x 与 y = x² 之间，从 x = 0 到 x = 2，绕 x 轴旋转。求体积。

V = π∫₀² [(2x)² − (x²)²] dx = π∫₀² (4x² − x⁴) dx

V = π[4x³/3 − x⁵/5]₀² = π(32/3 − 32/5) = 64π/15

This method extends naturally to rotation about the y-axis by substituting x₁ and x₂.

此法可自然推广到绕 y 轴旋转，只需代入 x₁ 与 x₂。

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4. Volume of Revolution with Parametric Equations | 参数方程旋转体体积

When a curve is given parametrically as x = x(t) and y = y(t), the volume about the x-axis is found using the chain rule: replace dx with (dx/dt) dt.

当曲线以参数方程 x = x(t)、y = y(t) 给出时，绕 x 轴旋转的体积利用链式法则：将 dx 替换为 (dx/dt) dt。

V = π∫ₐᵇ y² (dx/dt) dt

The limits a and b are now t-values corresponding to the start and end of the curve. You must convert the x-limits into parameter values before integrating.

此时上下限 a、b 是对应曲线起点与终点的 t 值。积分前必须将 x 的限转换为参数值。

Example: A curve is defined by x = t², y = 2t for t from 0 to 2. Find the volume generated by rotating about the x-axis.

例：曲线由 x = t²、y = 2t 定义，t 从 0 到 2。求绕 x 轴旋转所得体积。

V = π∫₀² (2t)² × (2t) dt = π∫₀² 8t³ dt = π[2t⁴]₀² = 32π

For rotation about the y-axis, the formula becomes V = π∫ x² (dy/dt) dt. Since dx/dt may be negative for some curves, take the absolute value if the curve is traced in the negative direction.

绕 y 轴旋转时，公式变为 V = π∫ x² (dy/dt) dt。由于某些曲线在反向绘制时 dx/dt 可能为负，若曲线沿负方向追踪，请取绝对值。

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5. Standard Volume Formulae for Solids | 常见立体体积公式

AQA expects you to recall and apply the standard volume formulae for common solids. These are not provided in the formula booklet, so you must memorise them.

AQA 要求你熟记并应用常见立体的标准体积公式。这些公式不在公式册中提供，必须牢记。

Solid	Volume
Prism / Cylinder	V = Ah (area of cross-section × length)
Cone	V = ⅓πr²h
Sphere	V = ⁴⁄₃πr³
Pyramid	V = ⅓ × base area × height
Frustum of a cone	V = ⅓πh(R² + Rr + r²)

For a pyramid, the base may be any polygon – triangle, rectangle, or regular polygon. The factor of ⅓ always applies.

棱锥的底面可以是任意多边形——三角形、矩形或正多边形。系数 ⅓ 始终适用。

A frustum is formed by cutting a cone by a plane parallel to its base. The formula above requires the height h of the frustum itself, not the original cone, and R and r are the radii of the larger and smaller bases respectively.

圆台由平行于底面的平面截圆锥得到。上述公式中的 h 是圆台自身的高度，而非原圆锥的高度，R 和 r 分别为大底与小底半径。

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6. Composite Volumes and Subtracting Regions | 组合体积与区域相减

Many exam questions combine multiple solids. The strategy is to add or subtract known volumes rather than integrate from scratch. For example, a cylindrical container with a hemispherical top has volume V = πr²h + ⅔πr³.

许多考试题组合多个立体。策略是加减已知体积，而非从头积分。例如，一个圆柱容器带半球形顶部，其体积为 V = πr²h + ⅔πr³。

For a hollow solid, subtract the inner volume from the outer volume. A pipe of outer radius R, inner radius r and length L has volume V = π(R² − r²)L.

对于空心立体，用外体积减去内体积。外半径 R、内半径 r、长度 L 的管道，其体积为 V = π(R² − r²)L。

When integrating volumes of revolution for composite regions, split the integral at the x-coordinate where the bounding curve changes. This preserves the correct radii for each segment.

对组合区域做旋转体体积积分时，应在边界曲线改变的 x 坐标处拆分积分。这样能保持每一段的正确半径。

Example: A solid is made by rotating y = x² from x = 0 to x = 1 and then y = 1 − x from x = 1 to x = 2 about the x-axis. Compute it as two separate integrals and add.

例：一个立体由 y = x²（x 从 0 到 1）和 y = 1 − x（x 从 1 到 2）分别绕 x 轴旋转构成。分成两个积分再相加。

V = π∫₀¹ x⁴ dx + π∫₁² (1 − x)² dx = π/5 + π/3 = 8π/15

Always verify that the curve segments meet at the boundary x-value; otherwise the solid has a gap or an overlap that must be accounted for.

务必验证曲线段在边界 x 值处相接；否则立体将存在间隙或重叠，必须予以考虑。

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7. Similar Shapes and Volume Scale Factors | 相似形与体积比例因子

For two mathematically similar solids, every corresponding length is scaled by a linear scale factor k. Since volume has three dimensions, the volume scale factor is k³.

对于两个数学相似的立体，所有对应长度按线性比例因子 k 缩放。由于体积有三个维度，体积比例因子为 k³。

V₂ / V₁ = (L₂ / L₁)³

This relationship is tested in both pure and applied contexts. If the surface area scale factor is k², then the volume scale factor is (√(area ratio))³. Always convert carefully between length, area, and volume ratios.

此关系在纯数与应用题中均有考查。若表面积比例因子为 k²，则体积比例因子为 (√(面积比))³。在长度、面积与体积比之间转换时务必小心。

Example: Two similar cones have heights 6 cm and 15 cm. If the smaller cone has volume 40 cm³, find the larger volume.

例：两个相似圆锥的高分别为 6 cm 与 15 cm。若小圆锥体积为 40 cm³，求大圆锥体积。

k = 15/6 = 2.5, V₂ = 40 × 2.5³ = 40 × 15.625 = 625 cm³

A common mistake is to apply the linear factor to the volume directly. Remember: volumes change by the cube of the length ratio, so doubling dimensions multiplies volume by 8.

常见错误是直接将线性因子应用于体积。记住：体积按长度比的立方变化，因此尺寸加倍会使体积变为 8 倍。

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8. Connected Rates of Change Involving Volume | 涉及体积的关联变化率

AQA Paper 2 and Paper 3 frequently test differential equations that connect dV/dt with other rates. The general approach uses the chain rule to link the volume to a changing dimension.

AQA Paper 2 与 Paper 3 经常考查将 dV/dt 与其他变化率相联系的微分方程。通常方法是用链式法则将体积与变化的尺寸关联。

dV/dt = dV/dr × dr/dt

For a sphere filling with water, V = ⁴⁄₃πr³ so dV/dr = 4πr², which equals the surface area. If water enters at a constant rate, you can solve for dr/dt as a function of r.

对于充水的球体，V = ⁴⁄₃πr³，因此 dV/dr = 4πr²，恰好等于表面积。若水以恒定速率注入，就可解出 dr/dt 关于 r 的函数。

Worked Example: A spherical balloon is inflated at 100 cm³/s. Find dr/dt when the radius is 5 cm.

例：一个球形气球以 100 cm³/s 的速度充气。求半径 r = 5 cm 时的 dr/dt。

dV/dt = 4πr² (dr/dt) → 100 = 4π(25)(dr/dt)

dr/dt = 100/(100π) = 1/π cm/s

For a cone, the radius and height are linked by similar triangles, so you must eliminate one variable before differentiating. If r/h is constant, substitute r = (r/h)h into V = ⅓πr²h first.

对于圆锥，半径与高度通过相似三角形关联，因此微分前必须消去一个变量。若 r/h 恒定，先将 r = (r/h)h 代入 V = ⅓πr²h。

Pay attention to units. If dV/dt is given in cm³/s but lengths are in metres, convert to a consistent unit system before applying the chain rule.

注意单位。若 dV/dt 以 cm³/s 给出而长度以米为单位，应用链式法则前请转换为一致的单位系统。

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9. Volume of Revolution: Integrating with Respect to y | 旋转体体积：对 y 积分

Some questions require rotation about the y-axis of a region bounded by y = f(x). You must rearrange to x = g(y) and determine the y-limits carefully.

某些题目要求区域绕 y 轴旋转，区域由 y = f(x) 界定。你必须改写为 x = g(y) 并仔细确定 y 的限。

The area between x = 0 and x = 2 under y = x², rotated about the y-axis, produces a bowl shape. The y-limits are 0 and 4, and x = √y, giving:

y = x² 下方从 x = 0 到 x = 2 的区域绕 y 轴旋转，形成碗状。y 的限为 0 与 4，且 x = √y，于是：

V = π∫₀⁴ (√y)² dy = π∫₀⁴ y dy = π[y²/2]₀⁴ = 8π

Notice that this bowl has a different volume from the solid obtained by rotating the same region about the x-axis. Always identify the axis before writing the integral.

注意，这个碗状体的体积与同一区域绕 x 轴旋转所得立体不同。写出积分前务必先确定旋转轴。

In AQA, the command word “hence” after a volume-of-revolution question usually signals that the previous integration result should be reused, saving time and reducing arithmetic errors.

在 AQA 中，旋转体体积题后的 “hence”（因此）通常表示应复用前面的积分结果，这样既省时又减少计算错误。

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10. Common Exam Traps and Revision Summary | 常见考试陷阱与复习总结

Below are the most frequent mistakes students make in AQA volume questions, together with the correct approach.

以下是学生在 AQA 体积题目中最常犯的错误及正确处理方法。

• Forgetting to square the function before integrating: the integrand is always y² or x², never y or x.

  忘记先平方再积分：被积函数永远是 y² 或 x²，绝不是 y 或 x。

• Using x-limits for a y-axis rotation: limits must match the variable of integration.

  绕 y 轴旋转时误用 x 限：积分限必须与积分变量一致。

• Writing π∫ y² dx − π∫ z² dx as π∫ (y − z)² dx: the correct form is π∫ (y² − z²) dx.

  将 π∫ y² dx − π∫ z² dx 写成 π∫ (y − z)² dx：正确形式应为 π∫ (y² − z²) dx。

• Applying the linear scale factor instead of the cube when comparing similar solids.

  比较相似立体时误用线性比例因子而不是立方。

• Ignoring units in rates-of-change problems, especially when mixing cm and m.

  在变化率题目中忽视单位，尤其是混用 cm 与 m 时。

To revise effectively, practise a full mixture: one revolution about x, one about y, one parametric question, one composite solid, and one connected-rates problem. Timing yourself under exam conditions builds confidence for the real paper.

有效复习的方法是练习完整组合：一道绕 x 轴旋转、一道绕 y 轴旋转、一道参数题、一道组合立体、一道关联变化率题。在考试条件下计时练习，能为真实考试建立信心。

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