Double Integrals of Separable Functions | 可分离变量的二重积分

📚 Double Integrals of Separable Functions | 可分离变量的二重积分

In IB Mathematics Higher Level, double integrals extend the idea of integration to functions of two variables. When the integrand can be written as a product of a function of x and a function of y, and the region of integration is a rectangle, the double integral simplifies dramatically into the product of two single integrals. This powerful technique reduces computational effort and deepens conceptual understanding.

在IB数学高级课程中，二重积分将积分的概念扩展到二元函数。当被积函数可以写成x的函数与y的函数的乘积，并且积分区域是矩形时，二重积分可以大大简化为两个单积分的乘积。这一强大技巧不仅减少了计算量，还加深了概念理解。

1. Review of Double Integrals | 二重积分回顾

A double integral over a region R in the xy-plane is written as ∬_R f(x,y) dA. It represents the signed volume under the surface z = f(x,y) above R. For a rectangular region R = [a,b] × [c,d], we compute it as an iterated integral:

二重积分在xy平面上的区域R上写作 ∬_R f(x,y) dA。它表示曲面 z = f(x,y) 在R上方所围成的有向体积。对于矩形区域 R = [a,b] × [c,d]，我们将其计算为累次积分：

∬_R f(x,y) dA = ∫_a^b ∫_c^d f(x,y) dy dx = ∫_c^d ∫_a^b f(x,y) dx dy

The order of integration can often be changed by Fubini’s theorem, provided the function is continuous on the region.

根据富比尼定理，如果函数在区域上连续，则积分的顺序通常可以交换。

2. Definition of Separable Functions | 可分离函数的定义

A function f(x,y) is called separable if it can be expressed as f(x,y) = g(x) · h(y), where g depends only on x and h depends only on y. For example, f(x,y) = x² y³ is separable because g(x) = x² and h(y) = y³. However, f(x,y) = x² + y³ is not separable in this multiplicative sense.

如果一个函数 f(x,y) 可以表示为 f(x,y) = g(x) · h(y)，其中 g 只依赖于 x，h 只依赖于 y，则称该函数是可分离的。例如，f(x,y) = x² y³ 是可分离的，因为 g(x) = x²，h(y) = y³。而 f(x,y) = x² + y³ 在这种乘法意义下是不可分离的。

Recognising separability is the first step to simplifying double integrals. Look for a product of powers, exponentials, trigonometric functions, or other expressions that factor neatly.

识别可分离性是简化二重积分的第一步。寻找幂、指数、三角函数或其他可以整洁分解的表达式之积。

3. The Product Formula on Rectangular Regions | 矩形区域上的乘积公式

If R is a rectangle [a,b] × [c,d] and f(x,y) = g(x) h(y), then the double integral splits into a product of two ordinary integrals:

如果 R 是矩形 [a,b] × [c,d]，且 f(x,y) = g(x) h(y)，则二重积分可以拆分为两个普通积分的乘积：

∬_R g(x) h(y) dA = (∫_a^b g(x) dx) · (∫_c^d h(y) dy)

This holds because the inner integral treats h(y) as a constant when integrating with respect to x, and the outer integral then integrates with respect to y.

这一公式成立，因为内层积分对x积分时，将 h(y) 视为常数，然后外层积分再对y积分。

4. Proof of the Product Formula | 乘积公式的证明

Starting from the iterated integral:

从累次积分出发：

∬_R g(x)h(y) dA = ∫_a^b [ ∫_c^d g(x)h(y) dy ] dx

For fixed x, g(x) is a constant with respect to y, so the inner integral becomes g(x) ∫_c^d h(y) dy. Then:

对于固定的x，g(x) 关于y是常数，因此内层积分变为 g(x) ∫_c^d h(y) dy。于是：

∫_a^b g(x) [∫_c^d h(y) dy] dx = (∫_a^b g(x) dx) · (∫_c^d h(y) dy)

because ∫_c^d h(y) dy is a constant number independent of x. This completes the proof.

因为 ∫_c^d h(y) dy 是不依赖于x的常数。证明完毕。

5. Example 1: Simple Separable Function | 例1：简单的可分离函数

Compute ∬_R x² y³ dA where R = [1,2] × [0,1].

计算 ∬_R x² y³ dA，其中 R = [1,2] × [0,1]。

Since x² y³ is separable, apply the product formula:

因为 x² y³ 是可分离的，应用乘积公式：

∬_R x² y³ dA = (∫_1^2 x² dx) · (∫_0^1 y³ dy)

Compute each integral:

分别计算每个积分：

∫_1^2 x² dx = [x³/3]_1^2 = 8/3 − 1/3 = 7/3
∫_0^1 y³ dy = [y⁴/4]_0^1 = 1/4

Therefore the double integral equals (7/3) × (1/4) = 7/12.

因此二重积分等于 (7/3) × (1/4) = 7/12。

6. Example 2: Exponential and Trigonometric Factorisation | 例2：指数与三角函数的因式分解

Evaluate ∬_R e^{2x} cos(3y) dA over R = [0, ln 2] × [0, π/2].

计算 ∬_R e^{2x} cos(3y) dA，其中 R = [0, ln 2] × [0, π/2]。

The integrand is already a product: g(x) = e^{2x} and h(y) = cos 3y. Thus:

被积函数已经是乘积形式：g(x) = e^{2x}，h(y) = cos 3y。因此：

∬_R e^{2x} cos(3y) dA = (∫_0^{ln 2} e^{2x} dx) · (∫_0^{π/2} cos(3y) dy)

First integral: ∫_0^{ln 2} e^{2x} dx = [e^{2x}/2]_0^{ln 2} = (e^{2 ln 2} − 1)/2 = (4 − 1)/2 = 3/2.

第一个积分：∫_0^{ln 2} e^{2x} dx = [e^{2x}/2]_0^{ln 2} = (e^{2 ln 2} − 1)/2 = (4 − 1)/2 = 3/2。

Second integral: ∫_0^{π/2} cos(3y) dy = [sin(3y)/3]_0^{π/2} = (sin(3π/2) − sin 0)/3 = (−1 − 0)/3 = −1/3.

第二个积分：∫_0^{π/2} cos(3y) dy = [sin(3y)/3]_0^{π/2} = (sin(3π/2) − sin 0)/3 = (−1 − 0)/3 = −1/3。

Multiplying gives (3/2) × (−1/3) = −1/2. The negative value indicates that the surface lies below the xy-plane over part of the region.

相乘得到 (3/2) × (−1/3) = −1/2。负值表示曲面在区域的一部分位于xy平面下方。

7. Non-Rectangular Regions | 非矩形区域

If the region R is not a rectangle, the product formula cannot be applied directly, even if f(x,y) is separable. For example, consider the triangle bounded by y = 0, x = 1, and y = x, with f(x,y) = x y. The limits for y depend on x:

如果区域R不是矩形，即使 f(x,y) 是可分离的，乘积公式也不能直接应用。例如，考虑由 y = 0，x = 1 和 y = x 围成的三角形，取 f(x,y) = x y。此时y的积分限依赖于x：

∬_R xy dA = ∫_0^1 ∫_0^x xy dy dx

We must integrate with respect to y first:

我们必须先对y积分：

∫_0^x xy dy = x · [y²/2]_0^x = x · (x²/2) = x³/2

Then ∫_0^1 x³/2 dx = [x⁴/8]_0^1 = 1/8. The product formula would incorrectly give (∫_0^1 x dx)(∫_0^1 y dy) = (1/2)(1/2) = 1/4, which is wrong.

然后 ∫_0^1 x³/2 dx = [x⁴/8]_0^1 = 1/8。如果错误地使用乘积公式，会得到 (∫_0^1 x dx)(∫_0^1 y dy) = (1/2)(1/2) = 1/4，这是错误的。

8. When the Region Can Be Decomposed | 区域可以分解的情况

Sometimes a non-rectangular region can be split into rectangles or into parts where the limits separate. For example, if R consists of two disjoint rectangles, the integral over R is the sum of integrals over each rectangle. The product formula applies to each rectangle separately, provided the integrand is separable.

有时非矩形区域可以分解为若干个矩形，或分解为积分限可分离的部分。例如，若R由两个不相交的矩形组成，则R上的积分等于每个矩形上积分之和。只要被积函数可分离，乘积公式就可以分别应用于每个矩形。

Another situation is when the region is a “cross product” of intervals that have been translated, such as [a,b] × [c,d] but with some parts removed. In IB, you are normally expected to recognise when simple rectangular separation works and when you must set up the iterated integral with variable limits.

另一种情况是区域是由区间平移形成的“笛卡尔积”，例如 [a,b] × [c,d] 但去掉某些部分。在IB中，通常期望你识别何时简单的矩形分解有效，何时必须建立带有变量积分限的累次积分。

9. Common Mistakes | 常见错误

Applying the product formula to non-rectangular regions: Always check that the limits of integration are constants. If the inner limits depend on the outer variable, you must integrate normally.
Forgetting to separate powers: For example, (x + y)² is not x² + y²; it expands to x² + 2xy + y². Only the xy term is separable. The x² and y² terms must be integrated separately or combined carefully.
Ignoring constant factors: If f(x,y) = 4 x² y³, then g(x) = 4 x² and h(y) = y³, or g(x) = x² and h(y) = 4 y³. Just ensure the constant appears exactly once.

将乘积公式应用于非矩形区域： 务必检查积分限是否为常数。如果内层积分限依赖于外层变量，则必须正常积分。

忘记正确展开幂： 例如，(x + y)² 不是 x² + y²，而应展开为 x² + 2xy + y²。只有 xy 项是可分离的。x² 和 y² 项必须单独积分或仔细合并。

忽略常数因子： 如果 f(x,y) = 4 x² y³，则可以将 g(x) = 4 x²，h(y) = y³，或 g(x) = x²，h(y) = 4 y³。只需确保常数恰好出现一次。

10. Applications in Probability | 在概率中的应用

Separable double integrals occur naturally in probability when two random variables are independent. If X and Y have joint probability density function f(x,y) = g(x) h(y) over a rectangular domain, the double integral of the density over the entire domain equals 1, and the product formula confirms that the total probability factors as (∫ g(x) dx)(∫ h(y) dy) = 1 × 1 = 1.

可分离二重积分在概率论中自然出现，如两个随机变量相互独立时。若X和Y的联合概率密度函数在矩形区域上为 f(x,y) = g(x) h(y)，则密度在整个区域上的二重积分等于1，乘积公式确认总概率可以分解为 (∫ g(x) dx)(∫ h(y) dy) = 1 × 1 = 1。

This is why independent variables can be analysed separately. The expectation E[XY] also becomes E[X]·E[Y] when the expectation integrals separate.

这就是独立变量可以分别分析的原因。当期望积分可分离时，E[XY] 也变为 E[X]·E[Y]。

11. Practice Questions | 练习问题

1. Evaluate ∬_R (x² + y²) dA over R = [0,2] × [1,3]. Hint: split into two integrals.

1. 计算 ∬_R (x² + y²) dA，其中 R = [0,2] × [1,3]。提示：拆分为两个积分。

2. Evaluate ∬_R x e^{x²} sin y dA over R = [0,1] × [0,π].

2. 计算 ∬_R x e^{x²} sin y dA，其中 R = [0,1] × [0,π]。

3. Determine whether ∬_R (x + y) dA over the triangle 0 ≤ y ≤ x ≤ 1 can be evaluated using the product formula. Compute it correctly.

3. 判断在三角形区域 0 ≤ y ≤ x ≤ 1 上的 ∬_R (x + y) dA 能否使用乘积公式，并正确计算。

Answers: 1. 56/3. 2. (e−1)/2 · 2 = e−1. 3. No; using iterated integral gives 1/3.

答案：1. 56/3。2. (e−1)/2 × 2 = e−1。3. 不能；使用累次积分得到1/3。

12. Summary | 总结

The separable double integral is one of the most efficient tools in multivariable calculus. When the integrand factors into g(x) h(y) and the region is rectangular, the double integral becomes the product of two single integrals. This technique is frequently tested in IB Paper 3 (HL) and is also essential for understanding independence in probability.

可分离二重积分是多变量微积分中最有效的工具之一。当被积函数分解为 g(x) h(y) 且区域为矩形时，二重积分变为两个单积分的乘积。这一技巧在IB HL Paper 3中经常考查，也是理解概率中独立性的关键。

Always verify the region is rectangular before using the product formula. For non-rectangular regions, set up the iterated integral with appropriate variable limits. Mastery of both approaches ensures you can handle any double integral question with confidence.

在使用乘积公式前，务必确认区域是矩形的。对于非矩形区域，应使用带有相应变量极限的累次积分。掌握这两种方法，你就能自信地处理任何二重积分问题。

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Double Integrals of Separable Functions | 可分离变量的二重积分