IB Math: Polar Coordinate Transformations in Double Integrals | IB数学：二重积分中的极坐标变换

📚 IB Math: Polar Coordinate Transformations in Double Integrals | IB数学：二重积分中的极坐标变换

Polar coordinate transformations are one of the most powerful techniques in multivariable calculus. In IB Mathematics Analysis and Approaches HL, students are expected to evaluate double integrals over circular or sector-shaped regions using polar coordinates. This article provides a thorough, exam-focused guide to understanding and applying this transformation correctly.

极坐标变换是多变量微积分中最强大的技巧之一。在IB数学分析与方法HL课程中，学生需要学会利用极坐标计算圆形或扇形区域上的二重积分。本文提供一份系统、紧扣考点的指南，帮助你正确理解并应用这一变换。

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1. Why Use Polar Coordinates? | 为什么要用极坐标？

Some regions are naturally described in Cartesian coordinates, but many regions — such as circles, annuli, and sectors — become much simpler when expressed using polar coordinates. When the integrand contains expressions like \(x^2 + y^2\) or \(\sqrt{x^2 + y^2}\), switching to polar coordinates can turn a complicated integral into a manageable one.

有些区域天然适合用直角坐标描述，但许多区域——例如圆、圆环和扇形——用极坐标表达时会更简洁。当被积函数中含有形如 \(x^2 + y^2\) 或 \(\sqrt{x^2 + y^2}\) 的表达式时，转换为极坐标往往能化繁为简。

The fundamental relationship between Cartesian and polar coordinates is given by:

直角坐标与极坐标之间的基本关系如下：

x = r cos θ, y = r sin θ, x² + y² = r²

When converting a double integral from Cartesian to polar coordinates, we must also replace the area element dA. In Cartesian coordinates, dA = dx dy. In polar coordinates, the correct area element is:

在将二重积分从直角坐标转换为极坐标时，还必须替换面积元素 dA。在直角坐标中，dA = dx dy。在极坐标中，正确的面积元素是：

dA = r dr dθ

The extra factor of r is crucial. It arises because a small “rectangle” in polar coordinates has side lengths dr and r dθ, so its area is r dr dθ. Forgetting this factor is one of the most common errors in IB exams.

多出的因子 r 至关重要。它来源于极坐标中的微小“矩形”边长为 dr 和 r dθ，因此其面积为 r dr dθ。忘记这个因子是IB考试中最常见的错误之一。

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2. The General Transformation Formula | 一般变换公式

For a region R in the xy-plane, the double integral in polar coordinates takes the form:

对于xy平面上的区域R，极坐标下的二重积分形式为：

∬_R f(x, y) dA = ∫_θ=α^β ∫_r=g₁(θ)^g₂(θ) f(r cos θ, r sin θ) · r dr dθ

Here α and β are the angular limits, and g₁(θ) and g₂(θ) are the radial limits. The region R must be described in polar form: r between two functions of θ, and θ between two fixed angles.

其中 α 和 β 是角度界限，g₁(θ) 和 g₂(θ) 是径向界限。区域R必须以极坐标形式描述：r在θ的两个函数之间，θ在两个固定角度之间。

This formula is a direct consequence of the change of variables theorem. The Jacobian determinant of the transformation from (r, θ) to (x, y) is:

这个公式是变量替换定理的直接推论。从 (r, θ) 到 (x, y) 的变换的雅可比行列式为：

∂(x, y)/∂(r, θ) = r

This determinant measures how much the area is stretched or compressed by the transformation, which is why dA = r dr dθ.

该行列式度量了变换对面积的拉伸或压缩程度，这正是 dA = r dr dθ 的原因。

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3. Step-by-Step Procedure | 分步解题步骤

Follow these steps when solving a double integral using polar coordinates:

使用极坐标计算二重积分时，请遵循以下步骤：

• Step 1: Sketch the region — Identify the region R and determine its shape. Look for circles, semicircles, sectors, or annuli.
• Step 2: Convert the integrand — Replace x with r cos θ and y with r sin θ. Simplify expressions like x² + y² to r².
• Step 3: Determine the limits — Find the range of θ (from α to β) and the range of r for each θ (from g₁(θ) to g₂(θ)).
• Step 4: Multiply by r — Do not forget the Jacobian factor r in the integrand.
• Step 5: Evaluate — Integrate with respect to r first, then with respect to θ.

• 第一步：画出区域——识别区域R并判断其形状，寻找圆、半圆、扇形或圆环。
• 第二步：转换被积函数——将x替换为r cos θ，将y替换为r sin θ，并将x² + y²等表达式简化为r²。
• 第三步：确定界限——求出θ的范围（从α到β）以及每个θ对应的r的范围（从g₁(θ)到g₂(θ)）。
• 第四步：乘以r——不要忘记被积函数中的雅可比因子r。
• 第五步：求值——先对r积分，再对θ积分。

The inner integral is always with respect to r first, because the radial limits may depend on θ. The outer integral is with respect to θ, whose limits are constants.

内层积分总是先对r积分，因为径向界限可能依赖于θ。外层积分对θ积分，其界限是常数。

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4. Example 1: Integral over a Full Circle | 例1：整个圆上的积分

Evaluate the double integral of f(x, y) = x² + y² over the disk of radius a centred at the origin.

计算函数 f(x, y) = x² + y² 在以原点为中心、半径为a的圆盘上的二重积分。

In polar coordinates, the disk is described by 0 ≤ r ≤ a and 0 ≤ θ ≤ 2π. The integrand becomes r². Thus:

在极坐标中，该圆盘描述为 0 ≤ r ≤ a，0 ≤ θ ≤ 2π。被积函数变为r²。因此：

∬_R (x² + y²) dA = ∫₀^2π ∫₀^a r² · r dr dθ = ∫₀^2π ∫₀^a r³ dr dθ

First integrate with respect to r:

先对r积分：

∫₀^a r³ dr = [r⁴/4]₀^a = a⁴/4

Then integrate with respect to θ:

再对θ积分：

∫₀^2π (a⁴/4) dθ = (a⁴/4) · 2π = πa⁴/2

So the value of the integral is πa⁴/2. Notice how the polar transformation made the problem almost trivial.

因此，该积分的值为 πa⁴/2。可以看到，极坐标变换使问题变得几乎微不足道。

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5. Example 2: Integral over a Semicircle | 例2：半圆上的积分

Evaluate ∬_R y dA, where R is the upper half of the circle x² + y² = 4.

计算 ∬_R y dA，其中R是圆 x² + y² = 4 的上半部分。

The region is described by 0 ≤ θ ≤ π and 0 ≤ r ≤ 2. The integrand y = r sin θ, and the Jacobian factor is r. Therefore:

该区域描述为 0 ≤ θ ≤ π，0 ≤ r ≤ 2。被积函数 y = r sin θ，雅可比因子为r。因此：

∬_R y dA = ∫₀^π ∫₀² (r sin θ) · r dr dθ = ∫₀^π ∫₀² r² sin θ dr dθ

Integrate with respect to r first:

先对r积分：

∫₀² r² dr = [r³/3]₀² = 8/3

Then integrate with respect to θ:

再对θ积分：

∫₀^π (8/3) sin θ dθ = (8/3)[−cos θ]₀^π = (8/3)(1 + 1) = 16/3

Thus the integral equals 16/3.

因此该积分等于 16/3。

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6. Example 3: Region between Two Circles | 例3：两个圆之间的区域

Evaluate ∬_R 1 dA (the area) of the annular region between the circles r = 1 and r = 3.

计算环状区域（介于 r = 1 与 r = 3 之间）的面积 ∬_R 1 dA。

This is simply the area of the annulus. In polar coordinates, the region is 0 ≤ θ ≤ 2π and 1 ≤ r ≤ 3:

这实际上就是圆环的面积。在极坐标中，该区域为 0 ≤ θ ≤ 2π，1 ≤ r ≤ 3：

A = ∫₀^2π ∫₁³ r dr dθ

Evaluate the inner integral:

计算内层积分：

∫₁³ r dr = [r²/2]₁³ = (9 − 1)/2 = 4

Then:

然后：

A = ∫₀^2π 4 dθ = 8π

This matches the known formula for the area of an annulus: π(3² − 1²) = 8π.

这与已知的圆环面积公式一致：π(3² − 1²) = 8π。

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7. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Students often make several predictable mistakes when using polar coordinates in double integrals. Here are the most important ones to avoid:

学生在二重积分中使用极坐标时，常犯一些可预见的错误。以下是需要避免的最重要几点：

• Forgetting the Jacobian factor r — Always multiply the integrand by r before integrating.
• Incorrect θ limits — For a full circle, θ goes from 0 to 2π. For the upper half, θ goes from 0 to π. Be careful with quadrants.
• Wrong r limits — The radial limits must be expressed as functions of θ if the boundary is not a circle centred at the origin.
• Mixing orders incorrectly — The inner integral is with respect to r, not θ, unless the region is better described differently.

• 忘记雅可比因子r——在积分之前务必将被积函数乘以r。
• θ界限错误——完整圆时，θ从0到2π；上半圆时，θ从0到π。注意象限。
• r界限错误——如果边界不是以原点为中心的圆，径向界限必须表示为θ的函数。
• 积分顺序错误——内层积分对r，而不是对θ，除非区域更适合其他描述方式。

In IB exams, always show the substitution step explicitly: write x = r cos θ, y = r sin θ, and dA = r dr dθ. This earns method marks even if the final arithmetic is wrong.

在IB考试中，务必明确写出替换步骤：x = r cos θ，y = r sin θ，dA = r dr dθ。即使最终计算有误，也能获得方法分。

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8. When the Centre Is Not at the Origin | 圆心不在原点的情况

If the circle is centred at (a, 0) or (0, a), the radial limits become more complicated. For example, the circle (x − a)² + y² = a², which passes through the origin, can be expressed in polar coordinates.

如果圆心在 (a, 0) 或 (0, a)，径向界限会变得更加复杂。例如，经过原点的圆 (x − a)² + y² = a² 可以用极坐标表达。

Expand the equation:

展开方程：

x² − 2ax + a² + y² = a² ⇒ x² + y² = 2ax

Using x² + y² = r² and x = r cos θ, we obtain:

利用 x² + y² = r² 和 x = r cos θ，我们得到：

r² = 2a r cos θ ⇒ r = 2a cos θ

This gives the radial upper limit for θ in [−π/2, π/2]. Such questions are common in IB Paper 3 style problems.

这就是 θ 在 [−π/2, π/2] 区间内的径向上限。这类问题常见于IB Paper 3风格的题目中。

Similarly, the circle x² + (y − a)² = a² has polar equation r = 2a sin θ.

类似地，圆 x² + (y − a)² = a² 的极坐标方程为 r = 2a sin θ。

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9. Transforming Integrals with Gaussian-Type Functions | 含高斯型函数的积分变换

A classic application of polar coordinates is evaluating the Gaussian integral:

极坐标的一个经典应用是计算高斯积分：

∫_−∞^∞ ∫_−∞^∞ e^{−(x² + y²)} dx dy

Although not always explicitly tested in IB AA HL, this technique demonstrates the power of polar coordinates. Convert to polar:

虽然IB AA HL不一定直接考查，但这一技巧展示了极坐标的强大之处。转换为极坐标：

∫₀^2π ∫₀^∞ e^−r² r dr dθ

The inner integral evaluates as follows:

内层积分计算如下：

∫₀^∞ r e^−r² dr = 1/2

using the substitution u = r². Then the outer integral gives 2π · (1/2) = π. This proves that the area under the bivariate normal curve is π.

使用代换 u = r²，可得内层积分为 1/2。然后外层积分为 2π · (1/2) = π。这证明了二元正态曲线下的面积为 π。

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10. Practice Problems | 练习题目

Each problem below is designed to reinforce one key aspect of polar coordinate transformations.

以下每道题目都旨在强化极坐标变换的一个关键方面。

Problem 1	Evaluate ∬R (x² + y²) dA where R is the disk x² + y² ≤ 9.
问题1	计算 ∬R (x² + y²) dA，其中R为圆盘 x² + y² ≤ 9。
Problem 2	Evaluate ∬R x dA where R is the right half of the circle x² + y² = 4.
问题2	计算 ∬R x dA，其中R为圆 x² + y² = 4 的右半部分。
Problem 3	Convert the integral ∫0a ∫0√(a²−x²) (x² + y²) dy dx to polar coordinates and evaluate.
问题3	将积分 ∫0a ∫0√(a²−x²) (x² + y²) dy dx 转换为极坐标并求值。

Answers: (1) 81π/2; (2) 16/3; (3) πa⁴/8.

答案：(1) 81π/2；(2) 16/3；(3) πa⁴/8。

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11. Integration Order and Limits for IB Exams | IB考试中的积分顺序与界限

In IB exams, you may be given an integral in Cartesian form and asked to convert it to polar. The key step is to sketch the region of integration carefully. For example, the region of the integral in Problem 3 above is the quarter circle in the first quadrant.

在IB考试中，你可能会被要求将直角坐标形式的积分转换为极坐标。关键步骤是仔细画出积分区域。例如，上面问题3中的积分区域是第一象限的四分之一圆。

The radial limits for a region bounded by a circle centred at the origin are usually constants. For regions bounded by lines or other curves, the radial limits become functions of θ.

以原点为中心的圆所围成的区域，其径向界限通常是常数。对于由直线或其他曲线围成的区域，径向界限变为θ的函数。

Always remember: the angular variable θ must have constant limits. If one of the boundaries is a ray from the origin, that ray defines one of the angular limits.

始终记住：角度变量θ必须具有常数界限。如果某条边界是从原点出发的射线，那么这条射线就定义了其中一个角度界限。

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12. Connection to the Jacobian and Change of Variables | 与雅可比行列式及变量替换的联系

For IB HL students seeking deeper understanding, the polar transformation is a special case of the general change-of-variables formula:

对于希望加深理解的IB HL学生，极坐标变换是通用变量替换公式的一个特例：

∬_R f(x, y) dx dy = ∬_S f(x(u,v), y(u,v)) |∂(x,y)/∂(u,v)| du dv

For polar coordinates, we set u = r and v = θ. The Jacobian determinant is:

对于极坐标，我们令 u = r，v = θ。雅可比行列式为：

|∂(x,y)/∂(r,θ)| = |cos θ −r sin θ; sin θ r cos θ| = r(cos²θ + sin²θ) = r

This is why the absolute value of the Jacobian is r, not r² or any other expression. In the region r ≥ 0, we simply write r.

这就是为什么雅可比行列式的绝对值为 r，而不是r²或其他表达式。在 r ≥ 0 的区域中，我们直接写 r。

Understanding this origin helps students remember the factor r rather than memorising it mechanically.

理解这一来源有助于学生记住因子 r，而不是机械地记忆。

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