Double Integration | 二重积分

📚 Double Integration | 二重积分

Double integration extends the idea of a definite integral to functions of two variables. It allows us to accumulate a quantity over a two‑dimensional region, whether that quantity represents volume, mass, charge, or probability. In the IB Mathematics: Analysis and Approaches HL course, double integrals appear as a natural progression from single‑variable calculus and are essential for solving problems involving areas, volumes, and moments.

二重积分将定积分的概念推广到了二元函数。它让我们能够在二维区域上累积某个量，这个量可以是体积、质量、电荷或概率。在IB数学分析与方法（HL）课程中，二重积分是单变量微积分的自然延伸，对于解决面积、体积和矩等问题至关重要。

1. From Single to Double Integrals | 从单积分到二重积分

A single definite integral ∫_a^b f(x) dx accumulates the signed area under the curve y = f(x) over an interval. When we move to a function of two variables, f(x, y), we need to accumulate over a region R in the xy‑plane. This leads to the double integral ∬_R f(x, y) dA, where dA represents an infinitesimal area element.

单变量定积分 ∫_a^b f(x) dx 计算的是曲线 y = f(x) 下方在区间上的带号面积。当处理二元函数 f(x, y) 时，我们需要在 xy 平面的区域 R 上进行累积，这就引出了二重积分 ∬_R f(x, y) dA，其中 dA 表示无穷小的面积元素。

In single‑variable calculus the region is an interval; in two variables the region can be a rectangle, a disk, or any region bounded by curves. The double integral can be interpreted as the volume under the surface z = f(x, y) and above the region R when f(x, y) ≥ 0.

在单变量微积分中，积分区域是一个区间；而在二元情况下，区域可以是矩形、圆盘或由曲线围成的任意形状。当 f(x, y) ≥ 0 时，二重积分可以解释为曲面 z = f(x, y) 下方、区域 R 上方的体积。

2. Notation and Fubini’s Theorem | 符号与富比尼定理

The double integral ∬_R f(x, y) dA is computed by writing it as an iterated integral, thanks to Fubini’s Theorem. If the region R is simple enough, the double integral splits into two single integrals. The notation is ∬_R f(x, y) dA = ∫_a^b ∫_g₁(x)^g₂(x) f(x, y) dy dx (or with the order reversed).

根据富比尼定理，二重积分 ∬_R f(x, y) dA 可以通过将它写成累次积分来计算。如果区域 R 足够简单，二重积分会拆分成两个单积分，记作 ∬_R f(x, y) dA = ∫_a^b ∫_g₁(x)^g₂(x) f(x, y) dy dx（也可交换顺序）。

In the iterated integral, the inner integral is evaluated first with respect to one variable while treating the other as constant. The result is a function of the outer variable, which is then integrated. Fubini’s Theorem guarantees that for continuous functions, the order of integration can be swapped provided the limits are adjusted accordingly.

在累次积分中，先计算内层积分，将其中的一个变量视作常量，结果是外变量的一个函数，再对其进行积分。富比尼定理保证，对于连续函数，只要相应地调整积分限，积分次序可以交换。

3. Type I Regions | I 型区域

A region in the xy‑plane is called Type I if it can be described by a ≤ x ≤ b and g₁(x) ≤ y ≤ g₂(x), where the bounding curves are functions of x. For such a region, the double integral becomes ∫_a^b ∫_g₁(x)^g₂(x) f(x, y) dy dx.

若 xy 平面上的区域可以表示为 a ≤ x ≤ b 且 g₁(x) ≤ y ≤ g₂(x)，其中边界曲线是 x 的函数，则称其为 I 型区域。对这样的区域，二重积分化为 ∫_a^b ∫_g₁(x)^g₂(x) f(x, y) dy dx。

The inner integration with respect to y is performed first, treating x as constant. The limits g₁(x) and g₂(x) provide the lower and upper boundaries of the region for each fixed x. This is the most common setup when the region is vertically simple.

先对 y 计算内层积分，将 x 看作常量。积分限 g₁(x) 和 g₂(x) 给出了每个固定 x 处区域的下边界和上边界。当区域是垂直简单区域时，这是最常见的表达形式。

4. Type II Regions | II 型区域

A region is Type II if it can be described by c ≤ y ≤ d and h₁(y) ≤ x ≤ h₂(y), where the boundaries are functions of y. The double integral is then set up as ∫_c^d ∫_h₁(y)^h₂(y) f(x, y) dx dy.

若区域可以表示为 c ≤ y ≤ d 且 h₁(y) ≤ x ≤ h₂(y)，其中边界是 y 的函数，则称其为 II 型区域。此时二重积分设为 ∫_c^d ∫_h₁(y)^h₂(y) f(x, y) dx dy。

Many regions can be described in both ways, and often one order of integration leads to simpler calculations. Recognising whether a region is Type I, Type II, or both is a critical skill for evaluating double integrals efficiently.

许多区域可同时用两种方式描述，通常某一种积分次序会使计算更简单。识别区域是 I 型、II 型还是两者均可，是高效计算二重积分的关键技能。

5. Changing the Order of Integration | 交换积分次序

Sometimes an iterated integral looks intractable in one order but becomes straightforward after swapping the order. To change the order, first sketch the region of integration from the original limits, then re‑express the region in the opposite type. The limits are updated accordingly.

有时累次积分在某一种次序下看似很难计算，但交换次序后就变得简单。要交换次序，首先根据原积分限画出积分区域，然后用另一种类型重新表示该区域，并相应地更新上下限。

For example, consider ∫₀¹ ∫_y¹ e^x² dx dy. The inner integral with respect to x cannot be expressed in elementary terms. By sketching the region 0 ≤ y ≤ 1, y ≤ x ≤ 1, we see it is the triangle under the line x = y. Reversing the order gives ∫₀¹ ∫₀^x e^x² dy dx = ∫₀¹ x e^x² dx = (½)(e – 1).

举个例子，∫₀¹ ∫_y¹ e^x² dx dy。内层对 x 的积分不能用初等函数表达。画出区域 0 ≤ y ≤ 1, y ≤ x ≤ 1 后，我们看到它是直线 x = y 下方的三角形。交换次序后得 ∫₀¹ ∫₀^x e^x² dy dx = ∫₀¹ x e^x² dx = (½)(e – 1)。

6. Double Integrals over Rectangles | 矩形区域上的二重积分

The simplest regions are rectangles: R = [a, b] × [c, d]. In this case, the double integral factors if the function is separable, i.e. f(x, y) = g(x)h(y). Then ∬_R g(x)h(y) dA = (∫_a^b g(x) dx)(∫_c^d h(y) dy).

最简单的区域是矩形：R = [a, b] × [c, d]。此时，如果函数可分离，即 f(x, y) = g(x)h(y)，二重积分可分解为 ∬_R g(x)h(y) dA = (∫_a^b g(x) dx)(∫_c^d h(y) dy)。

For non‑separable functions the iterated integration is still straightforward because the limits are constants. You integrate with respect to y first (or x) and then x, with no variable limits to worry about. Rectangular regions often appear in applications where the domain is naturally described by constant bounds.

对于不可分离的函数，累次积分依然简单，因为积分限均为常数。先对 y 积分（或先对 x），再对 x 积分，无需处理变动的积分限。在领域自然具有常数边界的应用中，矩形区域十分常见。

7. Polar Coordinates for Double Integrals | 极坐标下的二重积分

When the region R has circular symmetry, it is often easier to use polar coordinates: x = r cos θ, y = r sin θ. The area element transforms as dA = r dr dθ. The double integral becomes ∬_R f(x, y) dA = ∫_α^β ∫_r₁(θ)^r₂(θ) f(r cos θ, r sin θ) r dr dθ.

当区域 R 具有圆对称性时，使用极坐标通常更为简便：x = r cos θ, y = r sin θ。面积元素变为 dA = r dr dθ。二重积分化为 ∬_R f(x, y) dA = ∫_α^β ∫_r₁(θ)^r₂(θ) f(r cos θ, r sin θ) r dr dθ。

The extra factor r comes from the Jacobian of the transformation. It is essential to include this r term; forgetting it is a common mistake. Polar coordinates are particularly useful for integrals involving expressions like √(x² + y²) or x² + y² over disks, sectors, or rings.

额外的 r 因子来自变换的雅可比行列式，必须包含这个 r 项，忘记乘以 r 是常见的错误。极坐标对于圆盘、扇形或圆环上包含 √(x² + y²) 或 x² + y² 等表达式的积分特别有用。

8. Area Via Double Integrals | 用二重积分求面积

If we take f(x, y) = 1, the double integral ∬_R 1 dA gives the area of the region R. This provides a powerful way to compute areas of irregular shapes directly from their boundaries without needing separate geometric formulas.

若取 f(x, y) = 1，二重积分 ∬_R 1 dA 给出区域 R 的面积。这为直接从边界计算不规则图形的面积提供了强大的工具，无需依赖单独的几何公式。

For instance, the area of a region bounded by y = x² and y = 2 – x can be found as Area = ∫₋₂¹ ∫_x²^2−x 1 dy dx. Evaluating the inner integral gives (2 – x – x²), and integrating from –2 to 1 yields 9/2.

例如，由 y = x² 和 y = 2 – x 围成的区域面积可写为 Area = ∫₋₂¹ ∫_x²^2−x 1 dy dx。计算内层积分得到 (2 – x – x²)，再从 –2 到 1 积分得 9/2。

9. Volume Under a Surface | 曲面下的体积

The most intuitive application is computing the volume of the solid that lies under a surface z = f(x, y) ≥ 0 and above a region R in the xy‑plane. The volume is V = ∬_R f(x, y) dA. This generalises the disc/washer methods from single‑variable calculus to any base shape.

最直观的应用是计算位于曲面 z = f(x, y) ≥ 0 下方、xy 平面上区域 R 上方立体的体积。体积 V = ∬_R f(x, y) dA。它将单变量微积分中的圆盘/垫圈法推广到了任意底面形状。

For example, the volume under the plane z = 6 – 2x – y over the rectangle [0, 1] × [0, 2] is ∫₀¹ ∫₀² (6 – 2x – y) dy dx. Integrating first with respect to y, then x, gives the volume as 9 cubic units.

例如，平面 z = 6 – 2x – y 在矩形 [0, 1] × [0, 2] 下方的体积为 ∫₀¹ ∫₀² (6 – 2x – y) dy dx。先对 y 积分，再对 x 积分，得到体积为 9 立方单位。

10. Mass and Centre of Mass | 质量与质心

Double integrals can compute the mass of a flat plate (lamina) when the density ρ(x, y) varies across the plate. The total mass is M = ∬_R ρ(x, y) dA. The moments about the axes are M_y = ∬_R x ρ(x, y) dA, M_x = ∬_R y ρ(x, y) dA.

当薄板（平面薄片）的密度 ρ(x, y) 随位置变化时，二重积分可计算其质量。总质量 M = ∬_R ρ(x, y) dA。对坐标轴的矩为 M_y = ∬_R x ρ(x, y) dA, M_x = ∬_R y ρ(x, y) dA。

The centre of mass (x̄, ȳ) is then given by x̄ = M_y/M and ȳ = M_x/M. This technique is directly examinable in IB HL, often combining density functions that are polynomials or simple exponential forms.

质心 (x̄, ȳ) 随后由 x̄ = M_y/M 和 ȳ = M_x/M 给出。这一技巧在IB HL考试中会直接考查，经常结合多项式或简单指数形式的密度函数。

11. Common Mistakes and Tips | 常见错误与技巧

Forgetting the Jacobian r in polar form: Always double‑check that your integral includes the extra r factor. Setting up limits incorrectly: Sketch the region and draw a typical vertical or horizontal arrow to confirm the bounds. Not simplifying the integrand before integrating: Where possible, use symmetry or split the region to avoid difficult antiderivatives.

忘记极坐标中的雅可比因子 r：务必检查积分式中是否包含了额外的 r 因子。积分限设置错误：画出区域，并画一条典型的垂直或水平箭头来验证上下限。积分前未化简被积函数：尽可能利用对称性或拆分区域，以避开复杂的反导数。

Error 错误	Correction 改正
Using dx dy limits from a Type I sketch for a Type II setup	Always redraw or reinterpret the region for the new order
Integrating in the wrong order when one order is impossible	Check if the integrand can be integrated elementarily in that order
Dropping absolute value when dA = r dr dθ	r is always non‑negative in polar coordinates; no absolute value needed

Practice with both Type I and Type II descriptions, and convert between Cartesian and polar coordinates fluently. Many IB questions test the ability to switch representation rather than heavy computation.

同时练习I型和II型区域的描述，并熟练进行直角坐标与极坐标的转换。很多IB考题考查的是表示形式的转换能力，而非繁重的计算。

12. IB Exam‑Style Example | IB 考试题型示例

Question: A triangular plate occupies the region bounded by the x‑axis, the line x = 1, and the line y = x. The density at any point (x, y) is given by ρ(x, y) = 1 + x + y. Find the total mass of the plate.

题目：一块三角形薄板占据由 x 轴、直线 x = 1 和直线 y = x 所围成的区域。任一点 (x, y) 处的密度为 ρ(x, y) = 1 + x + y。求薄板的总质量。

Set up the mass integral as M = ∬_R (1 + x + y) dA. The region is Type I: 0 ≤ x ≤ 1, 0 ≤ y ≤ x. Thus M = ∫₀¹ ∫₀^x (1 + x + y) dy dx. Integrate with respect to y: ∫₀^x (1 + x + y) dy = [(1 + x)y + ½ y²]₀^x = (1 + x)x + ½ x² = x + (3/2)x².

建立质量积分 M = ∬_R (1 + x + y) dA。该区域是I型：0 ≤ x ≤ 1, 0 ≤ y ≤ x。因此 M = ∫₀¹ ∫₀^x (1 + x + y) dy dx。对 y 积分得：[(1 + x)y + ½ y²]₀^x = (1 + x)x + ½ x² = x + (3/2)x²。

Now integrate with respect to x: ∫₀¹ (x + (3/2)x²) dx = [½ x² + (1/2)x³]₀¹ = ½ + ½ = 1. Therefore the total mass is 1 unit.

现在对 x 积分：∫₀¹ (x + (3/2)x²) dx = [½ x² + (1/2)x³]₀¹ = ½ + ½ = 1。因此总质量为 1 单位。

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Double Integration | 二重积分