Polar Curves: Finding the Area Enclosed | 极坐标曲线：求所围面积

📚 Polar Curves: Finding the Area Enclosed | 极坐标曲线：求所围面积

In A-Level Mathematics, one of the most elegant applications of integration is finding the area enclosed by a polar curve. Unlike Cartesian coordinates, where we integrate with respect to x or y, polar coordinates require a formula based on the angle θ and the radius r. This article explains the method step by step, with worked examples that match the AQA syllabus.

在 A-Level 数学中，积分最优雅的应用之一就是求极坐标曲线所围成的面积。与笛卡尔坐标中关于 x 或 y 积分不同，极坐标需要基于角度 θ 和半径 r 的公式。本文将逐步解释这种方法，并提供符合 AQA 考纲的例题。

1. The Polar Coordinate System | 极坐标系

A point P in polar coordinates is written as (r, θ), where r is the distance from the origin (called the pole) and θ is the angle measured anticlockwise from the positive x-axis. The relationship between polar and Cartesian coordinates is given by x = r cos θ and y = r sin θ.

极坐标中的点 P 写作 (r, θ)，其中 r 是到极点（原点）的距离，θ 是从 x 轴正方向逆时针测量的角度。极坐标与直角坐标的关系为 x = r cos θ，y = r sin θ。

A polar curve is defined by an equation of the form r = f(θ). For example, r = a is a circle of radius a, while r = aθ is a spiral. To find the area enclosed by such a curve, we cannot simply apply the Cartesian area formula; we must derive a specialised integral.

极坐标曲线由形如 r = f(θ) 的方程定义。例如，r = a 是半径为 a 的圆，而 r = aθ 是螺线。要求这类曲线所围的面积，我们不能直接套用直角坐标面积公式，而必须推导一个专门的积分式。

2. The Area Formula | 面积公式

For a polar curve r = f(θ), the area enclosed between two angles θ = α and θ = β is given by the formula:

对于极坐标曲线 r = f(θ)，在角度 θ = α 与 θ = β 之间所围成的面积为：

A = ½ ∫_α^β r² dθ

where r is a function of θ. The factor ½ arises because a small sector of angle dθ has area ½ r² dθ, just as in a circle the area of a sector with angle θ is ½ r² θ.

其中 r 是 θ 的函数。系数 ½ 来源于一个小扇形：角度为 dθ 的扇形面积为 ½ r² dθ。这正如圆中角度为 θ 的扇形面积为 ½ r² θ。

This formula is valid only when the curve is traced exactly once over the interval [α, β]. If the curve completes a full loop, you must choose the interval that corresponds to one full trace of the loop.

这个公式仅在曲线在区间 [α, β] 上恰好被描画一次时成立。如果曲线完成一个完整回路，你必须选择对应回路被完整描画一次的区间。

3. Derivation of the Formula | 公式的推导

To understand the formula, imagine splitting the region into many thin sectors, each with a small angle dθ. Each sector is approximately a triangle with two sides of length r and included angle dθ. The area of such a triangle is ½ r² sin(dθ). For very small dθ, sin(dθ) ≈ dθ, so the sector area is approximately ½ r² dθ.

为了理解这个公式，设想将区域分割成许多细小的扇形，每个扇形对应一个微小角度 dθ。每个扇形近似为一个两条边长为 r、夹角为 dθ 的三角形。该三角形面积为 ½ r² sin(dθ)。当 dθ 很小时，sin(dθ) ≈ dθ，所以扇形面积近似为 ½ r² dθ。

Summing all these sectors and taking the limit as dθ → 0 gives the integral formula. This derivation mirrors the Riemann sum approach used for Cartesian integration, but with the independent variable as the angle rather than x.

将所有扇形面积求和并令 dθ → 0 取极限，就得到积分公式。这个推导与直角坐标积分中的黎曼和类似，只是自变量是角度而不是 x。

A = lim_n→∞ Σ ½ [f(θᵢ)]² Δθ = ½ ∫_α^β [f(θ)]² dθ

This derivation also shows why it is essential to integrate r² rather than r: area depends on the square of the linear dimension, exactly as it does for a circle.

这个推导也说明了为什么必须对 r² 积分而不是对 r 积分：面积依赖于线性尺寸的平方，正如圆的面积一样。

4. Choosing the Limits of Integration | 选择积分限

The most challenging part of finding polar areas is determining the correct limits α and β. You should first sketch the curve or reason about its behaviour. Identify the range of θ for which the curve traces the boundary of the desired region exactly once.

求极坐标面积最困难的部分是确定正确的上下限 α 和 β。你应当先画出曲线草图，或分析其行为。找出 θ 的范围，使曲线在该范围内恰好描出所求区域的边界一次。

For a full circle r = a, one full trace occurs from θ = 0 to θ = 2π. For a circle centred at the origin, the area is simply πa². Substituting into the formula gives:

对于完整圆 r = a，从 θ = 0 到 θ = 2π 恰好描画一周。对于圆心在原点的圆，面积自然是 πa²。代入公式得：

A = ½ ∫₀^2π a² dθ = ½ a² [θ]₀^2π = πa²

For curves that are symmetric about the initial line, you may compute the area of half the region and double it. This often simplifies the integrals and reduces the risk of sign errors.

对于关于极轴对称的曲线，你可以计算一半区域的面积再乘以 2。这常常能简化积分，并降低符号错误的风险。

5. Worked Example 1: A Cardioid | 例题一：心形线

Find the area enclosed by the cardioid r = a(1 + cos θ), where a > 0.

求心形线 r = a(1 + cos θ)（a > 0）所围成的面积。

The cardioid is traced out completely as θ varies from 0 to 2π. The area is:

当 θ 从 0 变化到 2π 时，心形线被完整描画一次。面积为：

A = ½ ∫₀^2π a²(1 + cos θ)² dθ

Expand the integrand:

展开被积函数：

(1 + cos θ)² = 1 + 2 cos θ + cos² θ

Use the identity cos² θ = ½(1 + cos 2θ). Then:

利用恒等式 cos² θ = ½(1 + cos 2θ)，于是：

1 + 2 cos θ + cos² θ = 3/2 + 2 cos θ + ½ cos 2θ

Now integrate from 0 to 2π. The integrals of cos θ and cos 2θ over a full period are zero, so:

现在从 0 到 2π 积分。cos θ 与 cos 2θ 在一个完整周期上的积分为零，因此：

A = ½ a² × (3/2) × 2π = (3/2)πa²

This result is well known: the area of a cardioid is 1.5πa².

这个结果是众所周知的：心形线面积为 1.5πa²。

6. Worked Example 2: Three-Leaf Rose | 例题二：三叶玫瑰线

Find the area of one leaf of the rose curve r = a sin 3θ.

求玫瑰线 r = a sin 3θ 的一个叶子的面积。

The curve r = a sin 3θ has three leaves. One leaf is traced as θ varies from 0 to π/3, because sin 3θ is positive on this interval and returns to zero at θ = π/3.

曲线 r = a sin 3θ 有三个叶子。当 θ 从 0 变化到 π/3 时描出一个叶子，因为 sin 3θ 在这个区间上为正，并且在 θ = π/3 时回到零。

The area of one leaf is therefore:

因此一个叶子的面积为：

A_leaf = ½ ∫₀^π/3 a² sin²(3θ) dθ

Use sin²(3θ) = ½(1 − cos 6θ). Then:

使用 sin²(3θ) = ½(1 − cos 6θ)，则：

A_leaf = ½ a² × ½ ∫₀^π/3 (1 − cos 6θ) dθ

Integrating gives:

积分得到：

A_leaf = ¼ a² [θ − (sin 6θ)/6]₀^π/3 = ¼ a² × π/3 = πa²/12

Since the three leaves are equal, the total area is 3 × πa²/12 = πa²/4.

由于三个叶子相等，总面积约为 3 × πa²/12 = πa²/4。

7. Worked Example 3: Area Between Two Polar Curves | 例题三：两条极坐标曲线之间的面积

Sometimes you need the area between two polar curves r₁ = f(θ) and r₂ = g(θ). If r₁ ≥ r₂ for α ≤ θ ≤ β, then the area between them is:

有时需要求两条极坐标曲线 r₁ = f(θ) 与 r₂ = g(θ) 之间的面积。若在 α ≤ θ ≤ β 上 r₁ ≥ r₂，则两者之间的面积为：

A = ½ ∫_α^β (r₁² − r₂²) dθ

For example, find the area inside r = 2 + cos θ but outside r = 2. First find the intersection points by solving 2 + cos θ = 2, which gives cos θ = 0, so θ = π/2 and θ = 3π/2.

例如，求位于 r = 2 + cos θ 内部且位于 r = 2 外部的面积。首先解 2 + cos θ = 2 求得交点，即 cos θ = 0，因此 θ = π/2 和 θ = 3π/2。

Over the interval from π/2 to 3π/2, we have r₁ = 2 + cos θ ≤ 2? Actually, check: for θ between π/2 and 3π/2, cos θ ≤ 0, so 2 + cos θ ≤ 2. Thus the inner curve is r₁ = 2 + cos θ and the outer curve is r₂ = 2. The required area is:

在 π/2 到 3π/2 区间上，cos θ ≤ 0，因此 2 + cos θ ≤ 2。所以内曲线为 r₁ = 2 + cos θ，外曲线为 r₂ = 2。所求面积为：

A = ½ ∫_π/2^3π/2 [4 − (2 + cos θ)²] dθ

Expand (2 + cos θ)² = 4 + 4 cos θ + cos² θ. Then 4 − (2 + cos θ)² = −4 cos θ − cos² θ. So:

展开 (2 + cos θ)² = 4 + 4 cos θ + cos² θ。于是 4 − (2 + cos θ)² = −4 cos θ − cos² θ。因此：

A = ½ ∫_π/2^3π/2 (−4 cos θ − cos² θ) dθ

Using cos² θ = ½(1 + cos 2θ), integrate over the interval. The result is:

利用 cos² θ = ½(1 + cos 2θ)，在该区间上积分，结果为：

A = ½ [ −4 sin θ − θ/2 − (sin 2θ)/4 ]_π/2^3π/2 = ½ [0 − π/2] = −π/4

The negative sign indicates that our assumption about which curve is outer should be checked carefully. In fact, since the difference is negative, the outer curve is r = 2 and the inner is r = 2 + cos θ. Taking absolute value gives A = π/4.

负号说明我们对内外的判断需要仔细检查。事实上，由于差值为负，外曲线是 r = 2，内曲线是 r = 2 + cos θ。取绝对值得到 A = π/4。

8. Using Symmetry | 利用对称性

Many polar curves are symmetric. For example, r = a(1 + cos θ) is symmetric about the initial line (θ = 0). Instead of integrating from 0 to 2π, you can integrate from 0 to π and double the result.

许多极坐标曲线具有对称性。例如，r = a(1 + cos θ) 关于极轴（θ = 0）对称。你可以在 0 到 π 上积分，然后将结果乘以 2，而不必在 0 到 2π 上积分。

When using symmetry, be careful to ensure that the half-region you choose actually covers exactly half of the total area. For a rose curve with an even number of petals, symmetry can help reduce the integration interval even more.

使用对称性时，要确保所选的半区域恰好覆盖总面积的一半。对于花瓣数为偶数的玫瑰线，对称性可以进一步缩短积分区间。

A_total = 2 × ½ ∫₀^π r² dθ = ∫₀^π r² dθ

This trick is especially useful when the interval [0, π] yields a simpler integrand than [0, 2π].

这个技巧在区间 [0, π] 上的被积函数比 [0, 2π] 上更简单时特别有用。

9. Common Pitfalls | 常见错误

A frequent mistake is forgetting to square r before integrating. The formula uses r², not r. Another error is using the wrong limits, especially when the curve crosses the pole (r = 0) and the loop shape is not obvious.

一个常见错误是忘记在积分前将 r 平方。公式中使用的是 r²，而不是 r。另一个错误是使用错误的积分限，尤其是当曲线经过极点（r = 0）时，回路的形状不太明显。

Always check that r is positive (or define negative r carefully) on the interval you integrate.
Always check that r is positive (or define negative r carefully) on the interval you integrate.
如果曲线在不同的 θ 区间上 r 的正负发生变化，需要通过解 r = 0 来找到分割点。
算两条曲线之间的面积时，必须确定在整个区间上哪条曲线在外，哪条在内。

Another pitfall is integrating over a full period of sin or cos when the curve repeats. For r = sin 2θ (a four-leaf rose), one loop is traced from 0 to π/2, so integrating from 0 to 2π would give twice the total area.

另一个陷阱是对 sin 或 cos 的整个周期积分，而曲线已经重复。对于 r = sin 2θ（四叶玫瑰线），一个回路在 0 到 π/2 上描出，因此在 0 到 2π 上积分会得到总面积的二倍。

10. Practice Questions | 练习题目

Try these questions on your own before checking the answers.

请先独立尝试以下题目，再核对答案。

Question \| 题目	Solution Hint \| 提示
1. Find the area enclosed by r = 2(1 − cos θ).	Use symmetry from 0 to π, double the result. Answer: 6π.
2. Find the area of one petal of r = sin 3θ.	Integrate from 0 to π/3. Answer: π/12.
3. Find the area inside r = 1 + sin θ and outside r = 1.	Find intersections: sin θ = 0 gives θ = 0, π. Integrate r₁² − r₂² from 0 to π.

For question 3, the inner curve is r = 1 and the outer is r = 1 + sin θ on [0, π]. The area is ½ ∫₀^π [(1 + sin θ)² − 1] dθ = ½ ∫₀^π (2 sin θ + sin² θ) dθ = ½ × (4 + π/2) = 2 + π/4.

对于第 3 题，在 [0, π] 上内曲线为 r = 1，外曲线为 r = 1 + sin θ。面积为 ½ ∫₀^π [(1 + sin θ)² − 1] dθ = ½ ∫₀^π (2 sin θ + sin² θ) dθ = ½ × (4 + π/2) = 2 + π/4。

11. Exam-Style Advice | 考试建议

In AQA A-Level exams, polar area questions usually appear in the Further Maths or extension papers. Show all steps clearly: write down the formula, substitute the function, simplify the integrand, integrate, and evaluate the limits.

在 AQA A-Level 考试中，极坐标面积问题通常出现在进阶数学或扩展试卷中。请清晰写出所有步骤：写出公式、代入函数、化简被积函数、积分并代入上下限。

Always state the limits explicitly. If you use symmetry, write “by symmetry” explicitly so the examiner can follow your reasoning. Remember that the angle is measured in radians in all integration involving polar coordinates.

始终明确写出积分上下限。如果使用对称性，请明确写出“由对称性”字样，以便阅卷人理解你的思路。请记住，所有涉及极坐标的积分中角度单位都是弧度。

Check your final answer for reasonableness. Areas of polar curves are typically some multiple of π times a². If your answer contains no π, you have probably forgotten the ½ factor or used the wrong limits.

检查最终答案是否合理。极坐标曲线的面积通常是某个倍数乘以 πa²。如果你的答案中不含 π，那么很可能是漏了 ½ 因子或使用了错误的上下限。

12. Summary | 总结

The area enclosed by a polar curve r = f(θ) is found using A = ½ ∫_α^β r² dθ. The derivation comes from summing small sectors. Choosing the correct limits is crucial: you must find the interval of θ over which the curve traces the boundary exactly once.

极坐标曲线 r = f(θ) 所围面积由 A = ½ ∫_α^β r² dθ 求得。该公式源于对小扇形的求和。选择正确的积分限至关重要：你必须找到 θ 的区间，使曲线在该区间内恰好描出边界一次。

For multiple loops or rose curves, find the interval for one loop and multiply by the number of loops. For areas between two curves, subtract the squared radii. With careful limits and systematic simplification, polar area problems become straightforward.

对于多个回路或玫瑰线，先求一个回路的区间，再乘以回路的数量。对于两条曲线之间的面积，减去半径的平方。只要积分限选择准确，化简步骤系统化，极坐标面积问题就会变得简单直接。

Master this topic by practising plenty of past paper questions. Each question typically follows the same pattern: identify the curve, choose the limits, integrate r², and evaluate.

通过大量练习真题来掌握这一主题。每道题通常都遵循相同的模式：识别曲线、选择积分限、对 r² 积分并代入上下限求值。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Polar Curves: Finding the Area Enclosed | 极坐标曲线：求所围面积