Area of Surface of Revolution | 旋转体表面积

📚 Area of Surface of Revolution | 旋转体表面积

A surface of revolution is formed when a curve is rotated 360° about an axis. The area of such a surface is a natural extension of arc length and is widely used in geometry, physics and engineering. In this revision guide, we will derive the formulas, understand the key ideas, and work through typical exam-style questions.

当一条平面曲线绕某条轴旋转 360° 时，就形成一个旋转曲面。该曲面的面积是弧长概念的自然推广，在几何、物理和工程中都有广泛用途。在本复习指南中，我们将推导公式、理解核心思想，并练习典型的考试题型。

1. Understanding the Surface of Revolution | 理解旋转曲面

Imagine taking the graph of a function y = f(x) between x = a and x = b, and rotating it around the x-axis. Every point on the curve sweeps out a circle perpendicular to the x-axis. The union of all these circles forms a smooth surface. The total area of that surface is what we call the area of revolution.

想象函数 y = f(x) 在 x = a 到 x = b 之间的图像，将其绕 x 轴旋转。曲线上的每一点都会扫出一个垂直于 x 轴的圆。所有这些圆共同构成一个光滑曲面，该曲面的总面积就是我们所说的旋转体表面积。

If instead the curve is rotated about the y-axis, the surface is generated by the same curve but the circles now lie in horizontal planes. Both cases lead to different formulas, and choosing the correct one is a key skill in exams.

如果曲线绕 y 轴旋转，则同样由该曲线生成曲面，但这些圆位于水平面内。两种情况对应不同的公式，在考试中正确选择公式是关键技能。

2. The Arc Length Element | 弧长微元

To compute the surface area, we first need the arc length element ds. For a curve y = f(x), the arc length of a tiny segment can be written as

要计算旋转体表面积，我们首先需要弧长微元 ds。对于曲线 y = f(x)，一小段弧长可以写成

ds = √(1 + (dy/dx)²) dx

Equivalently, if we treat y as the independent variable, we have ds = √(1 + (dx/dy)²) dy. This small piece of arc length, when rotated, sweeps out a thin band on the surface.

等价地，如果把 y 看作自变量，则 ds = √(1 + (dx/dy)²) dy。这段微小的弧长旋转后，会扫过曲面上一条细窄的带状区域。

3. Formula for Revolution about the x-axis | 绕 x 轴旋转的表面积公式

When the curve y = f(x) from x = a to x = b is rotated about the x-axis, the surface area is

当曲线 y = f(x) 从 x = a 到 x = b 绕 x 轴旋转时，表面积为

S = 2π ∫_a^b y √(1 + (dy/dx)²) dx

Here y is the radius of the circle swept out by the point (x, y), and the factor 2πy is the circumference of that circle. The square root term is the arc length element ds.

其中 y 是点 (x, y) 所扫出圆的半径，因子 2πy 是该圆的周长，根号项就是弧长微元 ds。

This formula can also be remembered as S = ∫ 2π y ds, where ds is the arc length element along the original curve.

这个公式也可以记忆为 S = ∫ 2π y ds，其中 ds 是原曲线上的弧长微元。

4. Why the 2πy Factor? | 为什么会有因子 2πy？

Consider a tiny segment of the curve of length ds. When rotated around the x-axis, it sweeps out a frustum (a sliced cone) of slant height ds. The curved surface area of a frustum is the average circumference multiplied by the slant height. For an infinitesimal frustum, the average circumference is simply 2πy, because the change in y across the tiny segment is negligible.

考虑曲线上长度为 ds 的微小线段。它绕 x 轴旋转后，扫过一个斜高为 ds 的圆台（被切开的圆锥）。圆台的侧面积等于平均周长乘以斜高。对于无穷小的圆台，平均周长就是 2πy，因为在该微小线段上 y 的变化可以忽略不计。

Thus the contribution to the total area is dS = 2πy ds. Summing these contributions with an integral gives the surface area formula.

因此，该微元对总面积的贡献是 dS = 2πy ds。用积分把这些贡献累加起来，就得到了表面积公式。

5. Formula for Revolution about the y-axis | 绕 y 轴旋转的表面积公式

If the same curve y = f(x) is rotated about the y-axis, the radius of the swept circle is x rather than y. The surface area is then

如果同一条曲线 y = f(x) 绕 y 轴旋转，那么扫出圆的半径是 x 而不是 y。此时表面积为

S = 2π ∫_a^b x √(1 + (dy/dx)²) dx

In this formula the integration is still carried out with respect to x. This version is often easier when the curve is given as y = f(x) and the x-limits are straightforward.

在此公式中，积分仍然对 x 进行。当曲线以 y = f(x) 给出且 x 的上下限比较直接时，这个形式往往更方便。

Alternatively, if we write x = g(y), then the surface area about the y-axis is S = 2π ∫_c^d x √(1 + (dx/dy)²) dy, where y runs from c to d.

另一种写法是，如果 x = g(y)，则绕 y 轴的表面积为 S = 2π ∫_c^d x √(1 + (dx/dy)²) dy，其中 y 从 c 到 d。

6. Choosing the Correct Variable | 选择正确的变量

Many students struggle with which variable to integrate with respect to. The guiding principle is simple: the radius in the factor 2πr must be measured perpendicular to the axis of rotation.

许多学生难以决定对哪个变量积分。指导原则很简单：因子 2πr 中的半径 r 必须是从旋转轴垂直测量得到的距离。

For revolution about the x-axis, the radius is y, so use S = 2π ∫ y ds.

绕 x 轴旋转时，半径是 y，因此使用 S = 2π ∫ y ds。
For revolution about the y-axis, the radius is x, so use S = 2π ∫ x ds.

绕 y 轴旋转时，半径是 x，因此使用 S = 2π ∫ x ds。

You must also decide whether to write ds in terms of dx or dy. If the curve is y = f(x), ds = √(1 + (dy/dx)²) dx is usually the simplest. If the curve is x = g(y), use ds = √(1 + (dx/dy)²) dy.

你还必须决定把 ds 写成关于 dx 还是 dy。若曲线为 y = f(x)，通常使用 ds = √(1 + (dy/dx)²) dx 最简单。若曲线为 x = g(y)，则使用 ds = √(1 + (dx/dy)²) dy。

7. Parametric Forms | 参数方程形式

When a curve is given parametrically as x = x(t), y = y(t), t₁ ≤ t ≤ t₂, the arc length element is

当曲线由参数方程 x = x(t)，y = y(t)，t₁ ≤ t ≤ t₂ 给出时，弧长微元为

ds = √((dx/dt)² + (dy/dt)²) dt

The surface area formulas become:

表面积公式变为：

About x-axis: S = 2π ∫_t₁^t₂ y √((dx/dt)² + (dy/dt)²) dt

About y-axis: S = 2π ∫_t₁^t₂ x √((dx/dt)² + (dy/dt)²) dt

These formulas are especially useful for curves that are not functions, such as circles or ellipses.

这些公式特别适用于不是函数的曲线，例如圆或椭圆。

8. Polar Forms | 极坐标形式

If a curve is described in polar coordinates r = r(θ), θ₁ ≤ θ ≤ θ₂, the arc length element is

如果曲线用极坐标 r = r(θ)，θ₁ ≤ θ ≤ θ₂ 描述，则弧长微元为

ds = √(r² + (dr/dθ)²) dθ

For revolution about the polar axis (the initial line, equivalent to the x-axis), the surface area is

对于绕极轴（即初始线，等价于 x 轴）的旋转，表面积为

S = 2π ∫_θ₁^θ₂ r sin θ √(r² + (dr/dθ)²) dθ

This is an advanced extension, but some Further Mathematics syllabuses include it. Always check the axis definition carefully.

这是一个进阶扩展内容，部分 Further Mathematics 考纲会包含它。务必仔细确认轴的定义。

9. Step-by-Step Solution Strategy | 分步解题策略

Follow these steps to solve surface area problems reliably:

按照以下步骤可以可靠地求解表面积问题：

Draw or visualise the curve and identify the axis of rotation.

画出或想象曲线，并确定旋转轴。
Write down the appropriate formula: S = 2π ∫ (radius) ds.

写出适当的公式：S = 2π ∫ (半径) ds。
Find dy/dx or dx/dy as needed, and substitute into the arc length element ds.

按需计算 dy/dx 或 dx/dy，并代入弧长微元 ds。
Simplify the integrand algebraically before integrating.

在积分之前先对根号内的表达式进行代数化简。
Apply the correct limits of integration, which must match the chosen variable.

应用正确的积分上下限，上下限必须与所选择的变量一致。
Evaluate the integral and present the final answer, often involving exact forms.

计算积分并写出最终答案，通常需要保留精确形式。

10. Common Pitfalls | 常见易错点

There are several traps that can cost marks in exams:

考试中有几个陷阱会让我们丢分：

Using y when rotating about the y-axis, or x when rotating about the x-axis. Remember: the radius is the perpendicular distance to the axis.

绕 y 轴时误用 y，或绕 x 轴时误用 x。记住：半径是到旋转轴的垂直距离。
Forgetting the square root in ds. The surface area is not just 2π ∫ y dx; that would give the volume of a cylinder, not the surface area of the revolution.

忘记 ds 中的根号。表面积不是简单的 2π ∫ y dx，否则得到的是圆柱体积而不是旋转曲面面积。
Using the wrong limits. If you convert from x to y, you must change the limits to y-values.

用错上下限。如果从 x 换成 y 积分，必须把上下限也换成 y 值。
Neglecting absolute values when simplifying square roots, for example √(y²) = |y|.

化简根号时忽略绝对值，例如 √(y²) = |y|。

11. Worked Example 1 (x-axis) | 示例1（绕 x 轴）

Find the surface area generated by rotating the curve y = √x from x = 0 to x = 4 about the x-axis.

求曲线 y = √x 从 x = 0 到 x = 4 绕 x 轴旋转所得曲面的面积。

Solution: Here dy/dx = 1/(2√x), so

解：这里 dy/dx = 1/(2√x)，因此

ds = √(1 + 1/(4x)) dx = √((4x + 1)/(4x)) dx

The surface area is

表面积为

S = 2π ∫₀⁴ √x · √((4x + 1)/(4x)) dx = 2π ∫₀⁴ √x · √(4x + 1)/(2√x) dx

The √x cancels, leaving

√x 约去，得到

S = π ∫₀⁴ √(4x + 1) dx

Let u = 4x + 1, so du = 4 dx. When x = 0, u = 1; when x = 4, u = 17. Then

令 u = 4x + 1，则 du = 4 dx。当 x = 0 时 u = 1；当 x = 4 时 u = 17。于是

S = π/4 ∫₁¹⁷ u^1/2 du = π/4 · [ (2/3)u^3/2 ]₁¹⁷

S = π/6 (17√17 − 1)

This is the exact surface area.

这就是精确的表面积值。

12. Worked Example 2 (y-axis) | 示例2（绕 y 轴）

Find the surface area when the curve y = x² from x = 0 to x = 1 is rotated about the y-axis.

求曲线 y = x² 从 x = 0 到 x = 1 绕 y 轴旋转所得曲面的面积。

Solution: Since we rotate about the y-axis, the radius is x. We have dy/dx = 2x, so

解：由于绕 y 轴旋转，半径为 x。已知 dy/dx = 2x，所以

ds = √(1 + 4x²) dx

Thus

因此

S = 2π ∫₀¹ x √(1 + 4x²) dx

Let u = 1 + 4x², so du = 8x dx. When x = 0, u = 1; when x = 1, u = 5. The integral becomes

令 u = 1 + 4x²，则 du = 8x dx。当 x = 0 时 u = 1；当 x = 1 时 u = 5。积分变为

S = 2π · (1/8) ∫₁⁵ u^1/2 du = π/4 · [ (2/3)u^3/2 ]₁⁵

S = π/6 (5√5 − 1)

Both examples demonstrate how simplification before integration is essential.

这两个例子都表明，在积分之前进行化简是至关重要的。

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Area of Surface of Revolution | 旋转体表面积