IB Mathematics: Iterated Integrals over Non-Rectangular Regions | IB数学：非矩形区域上的累次积分

📚 IB Mathematics: Iterated Integrals over Non-Rectangular Regions | IB数学：非矩形区域上的累次积分

In multivariable calculus, we often begin with double integrals over rectangles, where all limits are constant. But many interesting regions are bounded by curves rather than straight lines. To integrate over such non-rectangular regions, we need iterated integrals with variable limits. This article explains how to set up and evaluate double integrals over such regions, focusing on the two standard descriptions: Type I and Type II regions.

在多变量微积分中，我们通常从矩形区域上的二重积分开始，所有积分限都是常数。然而，许多有趣的区域是由曲线而非直线围成的。为了在非矩形区域上积分，我们需要使用带有变量积分限的累次积分。本文将讲解如何在非矩形区域上建立并计算二重积分，重点讨论两种标准描述：第一类（Type I）和第二类（Type II）区域。

1. From Rectangles to Non-Rectangular Regions | 从矩形到非矩形区域

For a rectangle R = [a,b] × [c,d], the double integral of f(x,y) is written as ∫₁? Let us write it as ∫_a^b∫_c^d f(x,y) dy dx. The inner and outer limits are fixed constants. When the region is not a rectangle, at least one pair of limits depends on the other variable. The main idea is to slice the region with vertical or horizontal lines, express the boundary curves as functions, and then integrate slice by slice.

对于矩形区域 R = [a,b] × [c,d]，函数 f(x,y) 的二重积分可以写成 ∫_a^b∫_c^d f(x,y) dy dx。内层和外层的积分限都是固定的常数。当区域不是矩形时，至少有一对积分限依赖于另一个变量。核心思想是用竖直或水平直线将区域切片，把边界曲线表示为函数，然后逐片积分。

2. Type I Regions: Vertical Slices | 第一类区域：竖直切片

A region D is called Type I, or vertically simple, if it can be described as D = { (x,y) : a ≤ x ≤ b, g₁(x) ≤ y ≤ g₂(x) }. For each fixed x, the vertical segment inside D runs from the lower curve y = g₁(x) to the upper curve y = g₂(x). The curves must satisfy g₁(x) ≤ g₂(x) for all x in [a,b].

区域 D 被称为第一类区域，也称为竖直简单区域，如果它可以表示为 D = { (x,y) : a ≤ x ≤ b, g₁(x) ≤ y ≤ g₂(x) }。对每一个固定的 x，D 内的竖直线段从下曲线 y = g₁(x) 延伸到上曲线 y = g₂(x)。这些曲线必须满足对一切 x ∈ [a,b] 都有 g₁(x) ≤ g₂(x)。

3. Type II Regions: Horizontal Slices | 第二类区域：水平切片

A region D is called Type II, or horizontally simple, if it can be described as D = { (x,y) : c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y) }. For each fixed y, the horizontal segment inside D runs from the left curve x = h₁(y) to the right curve x = h₂(y). The curves must satisfy h₁(y) ≤ h₂(y) for all y in [c,d].

区域 D 被称为第二类区域，也称为水平简单区域，如果它可以表示为 D = { (x,y) : c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y) }。对每一个固定的 y，D 内的水平线段从左曲线 x = h₁(y) 延伸到右曲线 x = h₂(y)。这些曲线必须满足对一切 y ∈ [c,d] 都有 h₁(y) ≤ h₂(y)。

4. Setting Up an Iterated Integral: Type I | 建立第一类区域的累次积分

For a Type I region, we integrate first with respect to y while holding x fixed. The inner limits are the boundary functions g₁(x) and g₂(x). After computing the inner integral, we integrate the remaining expression with respect to x from a to b.

对于第一类区域，我们先对 y 积分，同时把 x 视为常数。内层积分限是边界函数 g₁(x) 和 g₂(x)。计算出内层积分后，再对 x 从 a 到 b 积分。

∫∫_D f(x,y) dA = ∫_a^b ( ∫_g₁(x)^g₂(x) f(x,y) dy ) dx

It is useful to write dA = dy dx. The inner integral is evaluated first, and the variable x acts as a parameter. The final value is a real number, not a function.

通常写成 dA = dy dx。内层积分先被计算，变量 x 相当于参数。最终结果是一个实数，而不是函数。

5. Setting Up an Iterated Integral: Type II | 建立第二类区域的累次积分

For a Type II region, we integrate first with respect to x while holding y fixed. The inner limits are h₁(y) and h₂(y). After evaluating the inner integral, we integrate with respect to y from c to d.

对于第二类区域，我们先对 x 积分，同时把 y 视为常数。内层积分限是 h₁(y) 和 h₂(y)。计算内层积分后，再对 y 从 c 到 d 积分。

∫∫_D f(x,y) dA = ∫_c^d ( ∫_h₁(y)^h₂(y) f(x,y) dx ) dy

In this order, we write dA = dx dy. The inner variable is x, and the outer variable is y. The functions h₁ and h₂ are found by solving the boundary curves for x.

在这种顺序下，我们写成 dA = dx dy。内层变量是 x，外层变量是 y。函数 h₁ 和 h₂ 通过将边界曲线解出 x 而得到。

6. Example 1: Region between Two Curves | 例 1：两条曲线之间的区域

Let D be the finite region bounded by y = x² and y = x + 2. To find the intersection points, solve x² = x + 2, which gives x² – x – 2 = 0. Hence x = -1 or x = 2. On the interval [-1,2], the line y = x + 2 lies above the parabola y = x². Therefore D is a Type I region with a = -1, b = 2, g₁(x) = x², and g₂(x) = x + 2.

设 D 是由 y = x² 和 y = x + 2 围成的有限区域。为求交点，解 x² = x + 2，即 x² – x – 2 = 0。因此 x = -1 或 x = 2。在区间 [-1,2] 上，直线 y = x + 2 位于抛物线 y = x² 的上方。所以 D 是第一类区域，其中 a = -1，b = 2，g₁(x) = x²，g₂(x) = x + 2。

If f(x,y) = 1, the integral gives the area of D:

若 f(x,y) = 1，该积分给出区域 D 的面积：

Area(D) = ∫_-1² ∫_x²^x+2 dy dx = ∫_-1² (x + 2 – x²) dx = 9/2

To verify, evaluate the antiderivative x²/2 + 2x – x³/3 at x = 2 and x = -1. At x

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）