2 Examples Leading to Repeated Integrals with Constant Limits | 两个导出常限累次积分的例子

📚 2 Examples Leading to Repeated Integrals with Constant Limits | 两个导出常限累次积分的例子

1. Introduction: Why Constant Limits? | 引言：为何是常数积分限

In multivariable calculus, double integrals over rectangular regions naturally lead to repeated (iterated) integrals where all limits of integration are constants. This simplifies computation and allows the use of Fubini’s theorem without worrying about variable-dependent bounds. Here, we present two clear examples that demonstrate how a double integral can be expressed and evaluated using repeated integrals with constant limits.

在多变量微积分中，矩形区域上的二重积分会很自然地导出累次积分（迭代积分），此时所有的积分限都是常数。这大大简化了计算，并且可以应用富比尼定理而无需担心与变量有关的积分限。在这里，我们通过两个清晰的例子来展示如何将二重积分表示并计算为具有常数积分限的累次积分。

2. The Double Integral over a Rectangular Region | 矩形区域上的二重积分

Consider a continuous function f(x,y) defined on a closed rectangle R = [a, b] × [c, d]. The double integral ∬_R f(x,y) dA can be expressed as an iterated integral ∫_a^b (∫_c^d f(x,y) dy) dx, where both the inner and outer integrals have constant limits (a, b, c, d are constants). This is a repeated integral with constant limits.

考虑定义在闭矩形 R = [a, b] × [c, d] 上的连续函数 f(x,y)。二重积分 ∬_R f(x,y) dA 可以表示为累次积分 ∫_a^b (∫_c^d f(x,y) dy) dx，其中内、外积分限均为常数（a、b、c、d 为常数）。这就是一个具有常数积分限的累次积分。

3. Example 1: Volume under a Paraboloid | 例1：抛物面下的体积

Find the volume under the surface z = x² + y² and above the rectangle R = [0, 1] × [0, 2] in the xy-plane. The desired volume V is given by the double integral V = ∬_R (x² + y²) dA. Since R is a rectangle, we can write this as a repeated integral with constant limits: V = ∫₀¹ ∫₀² (x² + y²) dy dx.

求曲面 z = x² + y² 在 xy 平面上的矩形区域 R = [0, 1] × [0, 2] 下方的体积。所求体积 V 由二重积分 V = ∬_R (x² + y²) dA 给出。由于 R 是矩形，我们可以将其写成具有常数积分限的累次积分：V = ∫₀¹ ∫₀² (x² + y²) dy dx。

4. Example 1: Evaluating the Inner Integral | 例1：计算内层积分

Hold x constant and integrate with respect to y: ∫₀² (x² + y²) dy = [x²y + y³/3]_y=0^y=2 = (x²·2 + 8/3) − 0 = 2x² + 8/3.

将 x 视为常数，对 y 进行积分：∫₀² (x² + y²) dy = [x²y + y³/3]_y=0^y=2 = (x²·2 + 8/3) − 0 = 2x² + 8/3。

5. Example 1: Completing the Outer Integral | 例1：完成外层积分

Now integrate the result with respect to x from 0 to 1: V = ∫₀¹ (2x² + 8/3) dx = [2x³/3 + 8x/3]₀¹ = (2/3 + 8/3) − 0 = 10/3. Hence, the volume is 10/3 cubic units.

现在将上一步的结果对 x 从 0 到 1 积分：V = ∫₀¹ (2x² + 8/3) dx = [2x³/3 + 8x/3]₀¹ = (2/3 + 8/3) − 0 = 10/3。因此，体积为 10/3 立方单位。

6. Example 2: A Separable Function on a Rectangle | 例2：矩形上的可分离函数

Consider the density function ρ(x,y) = eˣ cos y on the rectangular plate R = [0, ln 2] × [0, π/2]. The total mass M = ∬_R ρ(x,y) dA. Because the region is a rectangle, we can again set up a repeated integral with constant limits: M = ∫₀^{ln 2} ∫₀^π/2 eˣ cos y dy dx.

考虑密度函数 ρ(x,y) = eˣ cos y 定义在矩形平板 R = [0, ln 2] × [0, π/2]

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