Double Integrals: Basic Concepts and Computation | 二重积分的基本概念与计算

📚 Double Integrals: Basic Concepts and Computation | 二重积分的基本概念与计算

Welcome to TutorHao’s IB Math revision series. In this article we will explore double integrals—a powerful tool for computing volumes, areas, and averages in two dimensions. You will learn the meaning of a double integral, how to evaluate it using iterated integrals, and how to handle different regions and coordinate systems.

欢迎阅读 TutorHao 的 IB 数学复习系列。在这篇文章中，我们将探讨二重积分——一个用于计算二维区域的体积、面积和平均值的强大工具。你将学习二重积分的含义、如何用累次积分来计算它，以及如何处理不同积分区域和坐标系。

1. What Is a Double Integral? | 什么是二重积分？

For a function of two variables \( f(x,y) \) defined on a closed rectangle \( R = [a,b] \times [c,d] \), the double integral \( \iint_R f(x,y)\,dA \) is defined as the limit of Riemann sums. We partition \( R \) into small rectangles with area \( \Delta A = \Delta x \Delta y \). In each small rectangle we choose a sample point \( (x_i, y_j) \), and form the sum of \( f(x_i, y_j)\Delta A \). As the partition becomes finer, the limit (if it exists) gives the double integral.

对于定义在闭矩形 \( R = [a,b] \times [c,d] \) 上的二元函数 \( f(x,y) \)，二重积分 \( \iint_R f(x,y)\,dA \) 定义为黎曼和的极限。我们把 \( R \) 分割成许多小矩形，每个小矩形面积为 \( \Delta A = \Delta x \Delta y \)。在每个小矩形中取一个样本点 \( (x_i, y_j) \)，构造和式 \( \sum \sum f(x_i,y_j)\Delta A \)。当分割无限变细时，若极限存在，即为二重积分。

Geometrically, if \( f(x,y) \ge 0 \), the double integral represents the volume of the solid that lies below the surface \( z = f(x,y) \) and above the region \( R \). If \( f(x,y) \) takes negative values, the integral is the signed volume, where parts below the \( xy \)-plane are subtracted.

从几何上看，若 \( f(x,y) \ge 0 \)，则二重积分表示位于曲面 \( z = f(x,y) \) 之下、区域 \( R \) 之上的立体体积。若 \( f(x,y) \) 可取负值，则该积分为带符号体积，即平面以下的部分被减去。

2. Properties of Double Integrals | 二重积分的性质

Double integrals share many algebraic properties with single-variable integrals. These properties are essential for simplifying computations and for solving exam problems.

二重积分与一元积分有许多相似的代数性质。这些性质对于简化计算和解决考试问题至关重要。

Linearity: \( \iint_R [c_1 f(x,y) + c_2 g(x,y)]\,dA = c_1 \iint_R f(x,y)\,dA + c_2 \iint_R g(x,y)\,dA \), where \( c_1, c_2 \) are constants.

线性性质：\( \iint_R [c_1 f(x,y) + c_2 g(x,y)]\,dA = c_1 \iint_R f(x,y)\,dA + c_2 \iint_R g(x,y)\,dA \)，其中 \( c_1, c_2 \) 为常数。
Additivity: If \( R \) is split into two non-overlapping regions \( R_1 \) and \( R_2 \), then \( \iint_R f\,dA = \iint_{R_1} f\,dA + \iint_{R_2} f\,dA \).

区域可加性：若 \( R \) 被分成两个不重叠的区域 \( R_1 \) 和 \( R_2 \)，则 \( \iint_R f\,dA = \iint_{R_1} f\,dA + \iint_{R_2} f\,dA \)。
Monotonicity: If \( f(x,y) \le g(x,y) \) throughout \( R \), then \( \iint_R f\,dA \le \iint_R g\,dA \).

单调性：若在 \( R \) 上处处有 \( f(x,y) \le g(x,y) \)，则 \( \iint_R f\,dA \le \iint_R g\,dA \)。
Constant integral: \( \iint_R k\,dA = k \cdot \text{Area}(R) \).

常数积分：\( \iint_R k\,dA = k \cdot \text{面积}(R) \)。

3. Iterated Integrals and Fubini’s Theorem | 累次积分与富比尼定理

To evaluate a double integral over a rectangle, we can compute two single-variable integrals in succession. This is called an iterated integral. The key result that allows us to do this is Fubini’s Theorem.

为了计算矩形区域上的二重积分，我们可以连续进行两次一元积分，称为累次积分。允许这样做的关键定理是富比尼定理。

If \( f(x,y) \) is continuous on \( R=[a,b]\times[c,d] \), then
\[ \iint_R f(x,y)\,dA = \int_a^b \left( \int_c^d f(x,y)\,dy \right) dx = \int_c^d \left( \int_a^b f(x,y)\,dx \right) dy. \]

In practice we often write the iterated integral without parentheses: \( \int_a^b \int_c^d f(x,y)\,dy\,dx \). The inner integral is evaluated first, holding the outer variable constant. Then the result is integrated with respect to the outer variable.

在实际书写中，我们通常省略括号，直接写为 \( \int_a^b \int_c^d f(x,y)\,dy\,dx \)。先计算内层积分，此时把外层变量视为常数；然后对外层变量进行积分。

Fubini’s theorem tells us that for continuous functions the order of integration does not affect the result. This flexibility is very useful in IB problems, where one order may be much easier than the other.

富比尼定理告诉我们，对于连续函数，积分的顺序不影响结果。这种灵活性在 IB 考试中非常有用，因为往往某一种积分顺序会远比另一种简单。

4. Computing Over Rectangular Regions | 矩形区域上的计算

Let us work through a simple example over a rectangle. Evaluate \( \int_1^2 \int_0^3 (x^2 + y)\,dy\,dx \).

让我们通过一个矩形区域上的简单例子来练习。计算 \( \int_1^2 \int_0^3 (x^2 + y)\,dy\,dx \)。

First integrate with respect to \( y \), treating \( x \) as a constant:

首先对 \( y \) 积分，将 \( x \) 视为常数：

\[ \int_0^3 (x^2 + y)\,dy = \left[ x^2 y + \frac{y^2}{2} \right]_{y=0}^{y=3} = 3x^2 + \frac{9}{2}. \]

Now integrate the result with respect to \( x \):

然后对得到的结果关于 \( x \) 积分：

\[ \int_1^2 \left( 3x^2 + \frac{9}{2} \right) dx = \left[ x^3 + \frac{9}{2}x \right]_{1}^{2} = (8 + 9) – (1 + \frac{9}{2}) = \frac{21}{2}. \]

Thus the double integral equals \( \frac{21}{2} \). Notice that we could have integrated in the opposite order and obtained the same numerical answer.

因此，该二重积分的值为 \( \frac{21}{2} \)。注意，如果我们交换积分顺序，也会得到同样的数值结果。

5. General Regions: Type I and Type II | 一般区域：第一类与第二类

Not every region is a rectangle. A Type I region lies between two vertical lines and between two curves of the form \( y = g_1(x) \) and \( y = g_2(x) \).

并非所有区域都是矩形。第一类区域位于两条竖直线之间，并且介于两条形如 \( y = g_1(x) \) 与 \( y = g_2(x) \) 的曲线之间。

\[ \iint_R f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx. \]

A Type II region lies between two horizontal lines and between two curves of the form \( x = h_1(y) \) and \( x = h_2(y) \).

第二类区域位于两条水平线之间，并且介于两条形如 \( x = h_1(y) \) 与 \( x = h_2(y) \) 的曲线之间。

\[ \iint_R f(x,y)\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy. \]

The following table summarises the two types. It is crucial to decide which type describes the region before setting up the limits.

下表总结了这两类区域。在确定积分限之前，必须判断该区域属于哪一类。

Type I	Type II
\( y \) is integrated first	\( x \) is integrated first
\( x \) bounds: \( a \le x \le b \)	\( y \) bounds: \( c \le y \le d \)
\( y \) bounds: \( g_1(x) \le y \le g_2(x) \)	\( x \) bounds: \( h_1(y) \le x \le h_2(y) \)

6. Setting Up Limits of Integration | 确定积分限

For a general region, the most difficult step is finding the correct limits. Follow these steps:

对于一般区域，最困难的步骤是确定正确的积分限。请遵循以下步骤：

Sketch the region \( R \).
画出区域 \( R \) 的草图。
Find all intersection points of the boundary curves.
求出边界曲线的所有交点。
Choose an order of integration (decide whether to treat \( x \) or \( y \) as the outer variable).
选择积分顺序（决定把 \( x \) 还是 \( y \) 作为外层变量）。
Slice the region according to the chosen order and write the appropriate outer and inner limits.
根据所选顺序切割区域，并写出相应的外层和内层积分限。

Example: Compute \( \iint_R xy\,dA \), where \( R \) is the region bounded by \( y = x^2 \) and \( y = 2x \).

例：计算 \( \iint_R xy\,dA \)，其中 \( R \) 是由 \( y = x^2 \) 和 \( y = 2x \) 围成的区域。

First find intersections: \( x^2 = 2x \Rightarrow x(x-2)=0 \Rightarrow x=0,2 \). So the region is between \( x=0 \) and \( x=2 \). This is a Type I region (for every \( x \), \( x^2 \le y \le 2x \)). Thus:

首先求交点：\( x^2 = 2x \Rightarrow x(x-2)=0 \Rightarrow x=0,2 \)。因此区域位于 \( x=0 \) 和 \( x=2 \) 之间。这是一个第一类区域（对于每个 \( x \)，\( x^2 \le y \le 2x \)）。因此：

\[ \int_0^2 \int_{x^2}^{2x} xy\,dy\,dx. \]

Evaluating the inner integral gives \( \int_{x^2}^{2x} xy\,dy = x \left[ \frac{y^2}{2} \right]_{y=x^2}^{y=2x} = x(2x^2 – \frac{x^4}{2}) = 2x^3 – \frac{x^5}{2} \). Then \( \int_0^2 (2x^3 – \frac{x^5}{2})\,dx = \left[ \frac{x^4}{2} – \frac{x^6}{12} \right]_0^2 = 8 – \frac{64}{12} = \frac{8}{3} \).

计算内层积分得 \( \int_{x^2}^{2x} xy\,dy = x \left[ \frac{y^2}{2} \right]_{y=x^2}^{y=2x} = x(2x^2 – \frac{x^4}{2}) = 2x^3 – \frac{x^5}{2} \)。于是 \( \int_0^2 (2x^3 – \frac{x^5}{2})\,dx = \left[ \frac{x^4}{2} – \frac{x^6}{12} \right]_0^2 = 8 – \frac{64}{12} = \frac{8}{3} \)。

7. Reversing the Order of Integration | 交换积分顺序

Sometimes the given integration order leads to a complicated inner integral, while the reverse order is much simpler. Reversing order means rewriting the same region using the other type (Type I ↔ Type II).

有时给定的积分顺序会导致内层积分很复杂，而交换顺序后简单得多。交换积分顺序意味着用另一种类型（第一类 ↔ 第二类）重新描述同一个区域。

Consider the iterated integral \( \int_0^1 \int_{\sqrt{x}}^1 e^{x^3} dy\,dx \). The inner integral is easy, but let us see why reversing may be useful when the inner integral is impossible to evaluate directly.

考虑累次积分 \( \int_0^1 \int_{\sqrt{x}}^1 e^{x^3} dy\,dx \)。内层积分很容易，但让我们看看为什么当内层积分无法直接计算时，交换顺序会很有用。

Example: Evaluate \( \int_0^1 \int_{y}^{1} e^{x^2} dx\,dy \). The function \( e^{x^2} \) has no elementary antiderivative with respect to \( x \), so we reverse the order. The region is described by \( 0 \le y \le 1 \) and \( y \le x \le 1 \). As a Type I region, \( 0 \le x \le 1 \) and \( 0 \le y \le x \). Therefore:

例：计算 \( \int_0^1 \int_{y}^{1} e^{x^2} dx\,dy \)。函数 \( e^{x^2} \) 关于 \( x \) 没有初等原函数，因此我们交换积分顺序。该区域可以描述为 \( 0 \le y \le 1 \)，且 \( y \le x \le 1 \)。作为第一类区域，\( 0 \le x \le 1 \)，且 \( 0 \le y \le x \)。因此：

\[ \int_0^1 \int_{y}^{1} e^{x^2} dx\,dy = \int_0^1 \int_0^x e^{x^2} dy\,dx. \]

Now \( \int_0^x e^{x^2} dy = x e^{x^2} \). The outer integral is straightforward: \( \int_0^1 x e^{x^2} dx = \frac{1}{2}(e-1) \).

此时 \( \int_0^x e^{x^2} dy = x e^{x^2} \)。外层积分就很简单了：\( \int_0^1 x e^{x^2} dx = \frac{1}{2}(e-1) \)。

8. Change of Variables: Polar Coordinates | 变量替换：极坐标

For regions that have circular symmetry, polar coordinates simplify the double integral. Recall the transformation \( x = r\cos\theta \), \( y = r\sin\theta \), with \( dA = r\,dr\,d\theta \). The factor \( r \) is the Jacobian of the transformation.

对于具有圆形对称性的区域，极坐标可以简化二重积分。回想变换 \( x = r\cos\theta \)，\( y = r\sin\theta \)，并且有 \( dA = r\,dr\,d\theta \)。因子 \( r \) 是变换的雅可比行列式。

\[ \iint_R f(x,y)\,dA = \iint_S f(r\cos\theta, r\sin\theta)\, r\,dr\,d\theta, \]

where \( S \) is the region \( R \) described in polar coordinates.

其中 \( S \) 是用极坐标描述 \( R \) 的区域。

Example: Find the volume under the paraboloid \( z = 4 – x^2 – y^2 \) and above the \( xy \)-plane in the first quadrant of the unit disk. The region is: \( 0 \le r \le 1 \), \( 0 \le \theta \le \frac{\pi}{2} \). In polar coordinates \( x^2 + y^2 = r^2 \), so:

例：求抛物面 \( z = 4 – x^2 – y^2 \) 与 \( xy \) 平面之间、单位圆第一象限部分上方的体积。区域为 \( 0 \le r \le 1 \)，\( 0 \le \theta \le \frac{\pi}{2} \)。在极坐标中 \( x^2 + y^2 = r^2 \)，所以：

\[ V = \int_0^{\pi/2} \int_0^1 (4 – r^2)\, r\,dr\,d\theta. \]

Inner integral: \( \int_0^1 (4r – r^3)\,dr = [2r^2 – \frac{r^4}{4}]_0^1 = 2 – \frac{1}{4} = \frac{7}{4} \). Then \( V = \int_0^{\pi/2} \frac{7}{4} d\theta = \frac{7\pi}{8} \).

内层积分：\( \int_0^1 (4r – r^3)\,dr = [2r^2 – \frac{r^4}{4}]_0^1 = 2 – \frac{1}{4} = \frac{7}{4} \)。于是 \( V = \int_0^{\pi/2} \frac{7}{4} d\theta = \frac{7\pi}{8} \)。

9. Applications: Area, Volume and Average Value | 应用：面积、体积与平均值

The most common applications of double integrals are computing areas of plane regions, volumes of solids, and average values of functions. In all cases, the first step is to choose the correct integration region and order.

二重积分最常用的应用包括计算平面区域面积、立体体积以及函数的平均值。在所有情形中，第一步都是选择正确的积分区域和顺序。

Area: The area of a region \( R \) is given by \( \iint_R 1\,dA \). This is the same as setting \( f(x,y)=1 \) in the double integral.

面积：区域 \( R \) 的面积为 \( \iint_R 1\,dA \)，即令二重积分中的 \( f(x,y)=1 \)。

Volume: If \( f(x,y) \ge 0 \), the volume of the solid under the surface \( z=f(x,y) \) and above \( R \) is \( \iint_R f(x,y)\,dA \).

体积：若 \( f(x,y) \ge 0 \)，则曲面 \( z=f(x,y) \) 之下、\(

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Double Integrals: Basic Concepts and Computation | 二重积分的基本概念与计算