IB Math: Line Integrals over Closed Paths | IB数学：闭合路径上的线积分

📚 IB Math: Line Integrals over Closed Paths | IB数学：闭合路径上的线积分

Line integrals over closed paths appear frequently in IB Mathematics Higher Level and Further Mathematics. They connect vector calculus, geometry and physics through ideas such as circulation, work and Green's Theorem.

在IB数学高级水平与进阶数学中，闭合路径上的线积分十分常见。它通过环量、功以及格林定理等概念，将向量微积分、几何与物理学联系在一起。

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1. What Is a Line Integral? | 什么是线积分？

esents the arc-length element. If C is parametrized by r(t) for a ≤ t ≤ b, the formula is:

对于标量函数 f(x,y)，沿曲线 C 的线积分记为 ∫_C f ds，其中 ds 表示弧长微元。若 C 由 r(t)（a ≤ t ≤ b）参数化，则有：

∫_C f ds = ∫_aᵇ f(r(t)) |r'(t)| dt

), the line integral along C is ∫_C F · dr. In two dimensions this equals ∫_C P dx + Q dy.

对于向量场 F = (P, Q)，沿曲线 C 的线积分为 ∫_C F · dr。在二维情形下，它等于 ∫_C P dx + Q dy。

When the curve C is closed, we write ∮ C. This special notation emphasises that the starting point and ending point are the same.

当曲线 C 为闭合曲线时，我们用 ∮_C 代替 ∫_C。这一特殊记号强调起点与终点相同。

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2. Parametrising a Closed Path | 闭合路径的参数化

. The orientation matters: counterclockwise is usually taken as positive, clockwise as negative.

一条闭合路径必须满足 r(a) = r(b)。方向很重要：通常取逆时针为正，顺时针为负。

Common examples include:

常见的例子包括：

• Circle: r(t) = (R cos t, R sin t), 0 ≤ t ≤ 2π.
• 圆： r(t) = (R cos t, R sin t)，0 ≤ t ≤ 2π。
• Ellipse: r(t) = (a cos t, b sin t), 0 ≤ t ≤ 2π.
• 椭圆： r(t) = (a cos t, b sin t)，0 ≤ t ≤ 2π。
• Triangle: parametrise each side separately, ensuring continuity at the vertices.
• 三角形： 逐边参数化，并确保在顶点处连续。

Always check that the parametrisation traces the entire curve exactly once. If the curve is traversed twice, the integral doubles.

始终检查参数化是否恰好沿曲线走过一次。若曲线被走两次，积分值也会加倍。

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3. Direct Evaluation of a Closed Line Integral | 闭合线积分的直接计算

To evaluate ∮_C F · dr directly, follow these steps.

要直接计算 ∮_C F · dr，请按以下步骤进行。

1. Parametrise C as r(t), with the correct orientation.
2. 求 C 的参数方程 r(t)，并注意方向。
3. Compute dr = r'(t) dt.
4. 计算 dr = r'(t) dt。
5. Form the dot product F(r(t)) · r'(t).
6. 写出点积 F(r(t)) · r'(t)。
7. Integrate with respect to t from a to b.
8. 对 t 从 a 到 b 积分。

Example: Let F = (-y, x) and let C be the circle x² + y² = R² oriented counterclockwise.

示例： 设 F = (-y, x)，C 为圆 x² + y² = R²，取逆时针方向。

Parametrise r(t) = (R cos t, R sin t). Then r'(t) = (-R sin t, R cos t).

令 r(t) = (R cos t, R sin t)，则 r'(t) = (-R sin t, R cos t)。

F(r(t)) = (-R sin t, R cos t).

F(r(t)) = (-R sin t, R cos t)。

F · r'(t) = (-R sin t)(-R sin t) + (R cos t)(R cos t) = R²(sin² t + cos² t) = R².

F · r'(t) = (-R sin t)(-R sin t) + (R cos t)(R cos t) = R²(sin² t + cos² t) = R²。

∮_C F · dr = ∫₀²π R² dt = 2πR²

This result is positive because the field tends to rotate around the origin in the counterclockwise direction.

结果为正值，因为该向量场倾向于绕原点逆时针旋转。

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4. Work and Circulation | 功与环量

In physics, if F represents a force, then ∮ · dr is the net work done by the force as a particle travels once around the closed path.

在物理学中，若 F 表示力，则 ∮_C F · dr 是质点沿闭合路径运动一圈时力所做的净功。

For a velocity field F, the same integral represents the circulation around the curve. Stronger rotation inside the curve leads to larger circulation.

若 F 是速度场，则同一积分表示沿曲线的环量。曲线内部的旋转越强，环量越大。

If the integral is zero, the net rotation effect around the loop is zero, even if the field is locally rotating.

若积分为零，则绕该回路的净旋转效应为零，即使局部场存在旋转。

Circulation = ∮_C F · dr = ∮_C P dx + Q dy

This interpretation is especially useful when applying Green's Theorem in the next sections.

这一解释在接下来应用格林定理时尤其有用。

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5. Conservative Vector Fields | 保守向量场

A vector field F is conservative if F = ∇f for some scalar function f(x,y). In components:

若存在标量函数 f(x,y) 使得 F = ∇f，则向量场 F 是保守的。分量形式为：

P = ∂f/∂x,  Q = ∂f/∂y

For a conservative field, the line integral around any closed path is zero:

对于保守场，沿任何闭合路径的线积分为零：

∮_C F · dr = 0

Why? Since F = ∇ f along the path, and because the path is closed, the net change is f(r(a)) - f(r(a)) = 0.

原因如下：因为 F = ∇f，积分等于 f 沿路径的总变化量，而路径闭合，所以净变化为 f(r(a)) - f(r(a)) = 0。

In two dimensions, the condition for conservatism on a simply connected region is:

在二维中，单连通区域上保守的条件是：

∂P/∂y = ∂Q/∂x

o saying the curl, or the scalar quantity ∂Q/∂x - ∂P/∂y, is zero everywhere in the region.

这等价于旋度（即标量式 ∂Q/∂x - ∂P/∂y）在该区域内处处为零。

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6. Path Independence | 路径无关性

The following statements are equivalent for a vector field F on a simply connected region:

对于单连通区域上的向量场 F，以下命题等价：

Statement | 命题	Meaning | 含义
F is conservative.	F 是保守的。
F = ∇f for some f.	存在 f 使得 F = ∇f。
∮_C F · dr = 0 for every closed path C.	对所有闭合路径 C，∮_C F · dr = 0。
∂P/∂y = ∂Q/∂x throughout the region.	区域内处处 ∂P/∂y = ∂Q/∂x。

Therefore, if you must show a closed line integral is zero, it is often sufficient to check the partial derivative condition.

因此，若要证明某个闭合线积分为零，通常只需检查偏导数条件即可。

everywhere except at the origin, but its integral around the unit circle equals 2π, not 0.

然而，这种检验在带孔区域上会失效。例如，场 F = (-y/(x² + y²), x/(x² + y²)) 在除原点外处处满足 ∂Q/∂x = ∂P/∂y，但它绕单位圆的积分等于 2π，而不是 0。

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7. Green's Theorem | 格林定理

Green's Theorem gives a powerful way to evaluate closed line integrals by converting them into double integrals.

格林定理提供了一种强有力的方法，将闭合线积分转化为二重积分来计算。

Let C be a positively oriented, piecewise smooth, simple closed curve in the plane, and let D be the region bounded by C. If P and Q have continuous partial derivatives in an open region containing D, then:

设 C 为平面中一条正向、分段光滑的简单闭合曲线，D 为 C 所围成的区域。若 P 和 Q 在包含 D 的开区域内具有连续偏导数，则：

∮_C P dx + Q dy = ∬_D (∂Q/∂x - ∂P/∂y) dA

The term ∂Q/∂x - ∂P/∂y is often called the scalar curl of F.

∂Q/∂x - ∂P/∂y 通常被称为 F 的标量旋度。

For the earlier example F = (-y, x), we have P = -y and Q = x:

在前面的例子 F = (-y, x) 中，P = -y，Q = x：

∂Q/∂x - ∂P/∂y = 1 - (-1) = 2

Therefore:

因此：

∮_C F · dr = ∬_D 2 dA = 2 × Area(D) = 2πR²

This matches the direct calculation.

这与直接计算的结果一致。

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8. Using Green's Theorem with a Square | 用格林定理计算正方形上的积分

Example: Evaluate ∮ ) and C is the square with vertices (0,0), (1,0), (1,1), (0,1), oriented counterclockwise.

示例： 计算 ∮_C F · dr，其中 F = (x² - y, x + y²)，C 是顶点为 (0,0)、(1,0)、(1,1)、(0,1) 且逆时针方向的正方形。

Here P = x² - y and Q = x + y².

这里 P = x² - y，Q = x + y²。

Compute:

计算：

∂Q/∂x = 1,  ∂P/∂y = -1

Thus the integrand is 1 - (-1) = 2.

因此被积函数为 1 - (-1) = 2。

The region D is the unit square, so Area(D) = 1.

区域 D 是单位正方形，所以 Area(D) = 1。

∮_C F · dr = ∬_D 2 dA = 2

Without Green's Theorem, one would need to integrate along four separate edges. Green's Theorem reduces the problem to a single area calculation.

若不用格林定理，则需要沿四条边分别积分。而格林定理将问题简化为一次面积计算。

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9. Area from a Line Integral | 用线积分求面积

A special application of Green's Theorem is finding the area of a region D using a line integral over its boundary.

格林定理的一个特殊应用是，通过区域边界上的线积分求区域 D 的面积。

Choose P = 0 and Q = x, or P = -y and Q = 0, or take the symmetric version:

令 P = 0、Q = x，或令 P = -y、Q = 0，或取对称形式：

Area(D) = ½ ∮_C x dy - y dx

Example: For the ellipse x²/a² + y²/b² = 1, parametrise x = a cos t, y = b sin t, with 0 ≤ t ≤ 2π.

示例： 对于椭圆 x²/a² + y²/b² = 1，令 x = a cos t，y = b sin t，0 ≤ t ≤ 2π。

Then dy = b cos t dt and dx = -a sin t dt:

于是 dy = b cos t dt，dx = -a sin t dt：

½ ∮_C (x dy - y dx) = ½ ∫₀²π (ab cos² t + ab sin² t) dt = ½ ∫₀²π ab dt = πab

This recovers the familiar formula for the area of an ellipse.

这一结果重新得到了熟悉的椭圆面积公式。

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10. Common Exam Pitfalls | 常见考试误区

Students often make mistakes in the following areas:

学生常在以下几个方面出错：

• Forgetting the ∮ symbol: Use ∮_C only when the path is closed. Regular ∫_C is for open paths.
• 忘记 ∮ 符号： 只有闭合路径时才用 ∮_C。一般的 ∫_C 用于开放路径。
• Wrong orientation: Green's Theorem requires positive (counterclockwise) orientation. Clockwise introduces a negative sign.
• 方向错误： 格林定理要求正向（逆时针）。顺时针方向会引入负号。
• Ignoring the domain: The partial derivative condition for conservatism only works on simply connected regions.
• 忽略区域： 保守场的偏导数条件只在单连通区域上成立。
• Inconsistent parametrisation: The path must be traversed completely and exactly once.
• 参数化不一致： 路径必须完整且恰好被遍历一次。
• Confusing P and Q: In Green's Theorem, remember to use ∂Q/∂x - ∂P/∂y, not the reverse.
• 混淆 P 和 Q： 在格林定理中，要使用 ∂Q/∂x - ∂P/∂y，而不是反过来。

Always check whether the vector field is conservative before performing a long integration. If it is, the closed integral is immediately zero.

在进行冗长积分之前，先检查向量场是否保守。若保守，则闭合积分立即为零。

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11. Practice Questions | 练习题目

Question 1: Evaluate ∮_C (y dx + x dy), where C is the unit circle x² + y² = 1, counterclockwise.

练习 1： 计算 ∮_C (y dx + x dy)，其中 C 是单位圆 x² + y² = 1，逆时针方向。

Hint: Use Green' = y and Q = x. Since ∂Q/∂x = 1 and ∂P/∂y = 1, the integrand is 0. Thus the integral is 0.

提示： 用格林定理，P = y，Q = x。因为 ∂Q/∂x = 1，∂P/∂y = 1，被积函数为 0，所以积分为 0。

Question 2: Let F = (3x² + y, x - 2y). Show that ∮_C F · dr = 0 for any closed path in the plane.

练习 2： 设 F = (3x² + y, x - 2y)。证明对平面内任意闭合路径 C，∮_C F · dr = 0。

Solution: and ∂Q/∂x = 1, so the field is conservative. Therefore every closed line integral is zero.

解答： 这里 ∂P/∂y = 1，∂Q/∂x = 1，因此场保守。所以所有闭合线积分均为零。

Question 3: Use Green's Theorem to evaluate ∮_C (eˣ + y²) dx + (x³ - sin y) dy, where C is the triangle with vertices (0,0), (2,0), (0,2).

练习 3： 用格林定理计算 ∮_C (eˣ + y²) dx + (x³ - sin y) dy，其中 C 是顶点为 (0,0)、(2,0)、(0,2) 的三角形。

Solution: ², ∂P/∂y = 2y, so the integrand is 3x² - 2y. The region is 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 - x. Then:

解答： ∂Q/∂x = 3x²，∂P/∂y = 2y，所以被积函数为 3x² - 2y。区域为 0 ≤ x ≤ 2，0 ≤ y ≤ 2 - x。于是：

∫₀² ∫₀^2-x (3x² - 2y) dy dx = ∫₀² (6x² - 6x - x) dx? Let's compute carefully.

Actually, let's compute: ∫₀^2-x (3x² - 2y) dy = [3x² y - y²]₀^2-x = 3x²(2 - x) - (2 - x)².

实际上，计算：∫₀^2-x (3x² - 2y) dy = [3x² y - y²]₀^2-x = 3x²(2 - x) - (2 - x)²。

Simplify: 6x² - 3x³ - (4 - 4x + x²) = -3x³ + 5x² + 4x - 4.

化简：6x² - 3x³ - (4 - 4x + x²) = -3x³ + 5x² + 4x - 4。

Integrate from 0 to 2:

从 0 到 2 积分：

[-3/4 x⁴ + 5/3 x³ + 2x² - 4x]₀² = -12 + 40/3 + 8 - 8 = 4/3

Therefore the integral equals 4/3.

因此积分值为 4/3。

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12. Summary | 总结

Line integrals over closed paths are a central topic in IB vector calculus. The key ideas are direct parametrisation, recognition of conservative fields, and Green's Theorem as a shortcut.

闭合路径上的线积分是IB向量微积分中的核心主题。关键方法包括直接参数化、识别保守场，以及利用格林定理进行简化。

For any closed path, first ask: is the field conservative? If yes, the integral is zero. If not, decide whether to integrate directly or apply Green's Theorem.

对于任意闭合路径，首先应问：该场是否保守？若是，积分为零。若否，再决定是直接积分，还是应用格林定理。

Practising parametrisation and checking orientation will help you avoid the most common errors in exams.

练习参数化并检查方向，将帮助你在考试中避免最常见的错误。

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