第11章多元函数的积分11.1 二重积分11.1.1 二重积分的概念11.1.2 二重积分的计算11.2 多变量积分的换元法11.3 重积分的应用11.4 三重积分

第11章多元函数的积分

Tip
在单变量积分中，我们通过定积分研究了一个变量在区间上的累积问题，例如曲线下的面积、物体的位移等。然而，现实世界中的许多问题涉及多个变量的相互作用，例如：三维空间中不均匀物体的质量（质量密度随位置变化），平面区域上的温度分布平均值等。这些问题要求我们将积分的概念从一维推广到高维空间。重积分（Multiple Integrals）正是解决这类问题的核心工具。本章将系统学习二重积分（Double Integrals）与三重积分（Triple Integrals）的定义、计算方法及其应用。

11.1 二重积分

Tip
跟单变量积分一样, 二重积分的本质仍然是“分割-近似-求和”的极限过程.

11.1.1 二重积分的概念

Tip

回顾: 单变量积分 $y = f(x)$ $\left[ a, b\right]$ $\left [a, b \right]$ $N$ $\displaystyle \Delta x = \frac{b-a}{N}$ $N$ $x_i$ $x_i = a + i\Delta x$ $\Delta x$ $f(x_i)$ $i$ $\Delta x \cdot f(x_i)$ $S_N$ 的近似值：

S_{N} = \sum_{i = 1}^{N} Δ x \cdot f (x_{i})

$N \to \infty$ $\Delta x \to 0$ $S_N$ $S$ ：

S = lim_{N \to \infty} S_{N} = lim_{Δ x \to 0} S_{N}

$S$ 可表示为定积分：

S = \int_{a}^{b} f (x) d x

Important

多变量积分

$D$ $z = f(x,y)$ $D$ $N$ $\Delta A$ $(x_i, y_i)$ $\Delta A$ $f(x_i, y_i)$ $\Delta A \cdot f(x_i, y_i)$ .将所有微元的贡献叠加，得到近似总体积：

V_{N} = \sum_{i = 1}^{N} Δ A \cdot f (x_{i}, y_{i})

$N \to \infty$ $\Delta A \to 0$ ）时，近似值趋于精确值：

V = lim_{N \to \infty} V_{N} = lim_{Δ A \to 0} V_{N}

这一极限过程即为二重积分的定义，可表示为：

V = \iint_{D} f (x, y) d A

$D$ $f(x,y)$ $\mathrm{d}A$ $\mathrm{d}x\mathrm{d}y$ ）.

如何计算？ [图略，待补充]

M = \sum_{i = 1}^{N} ρ (x_{i}, y_{i}) d A_{i} = \iint_{D} ρ (x, y) d A

11.1.2 二重积分的计算

Note

$f(x,y)=1-x^2-y^2,D=\{(x,y)|0\leq x\leq 1,0\leq y\leq 1\}$ $\displaystyle \iint_D f(x,y) \mathrm{d}A$ .

解：上式可展开为

\begin{aligned} \iint_{D} f (x, y) d A & = \iint_{D} f (x, y) d x d y \\ = \int_{0}^{1} \int_{0}^{1} (1 - x^{2} - y^{2}) d x d y \\ = \int_{0}^{1} [\int_{0}^{1} (1 - x^{2} - y^{2}) d x] d y \end{aligned}

Inner:

\int_{0}^{1} (1 - x^{2} - y^{2}) d x = {[x - \frac{1}{3} x^{3} - y^{2} x]}_{0}^{1} = \frac{2}{3} - y^{2}

Outer:

\int_{0}^{1} (\frac{2}{3} - y^{2}) d y = {[\frac{2}{3} y - \frac{1}{3} y^{3}]}_{0}^{1} = \frac{1}{3}

同理

\begin{aligned} \int_{0}^{1} [\int_{0}^{1} (1 - x^{2} - y^{2}) d x] d y & = \int_{0}^{1} [\int_{0}^{1} (1 - x^{2} - y^{2}) d y] d x \\ = \int_{0}^{1} ({[y - \frac{1}{3} y^{3} - x^{2} y]}_{0}^{1}) d x \\ = \int_{0}^{1} (\frac{2}{3} - x^{2}) d x \\ = {[\frac{2}{3} x - \frac{1}{3} x^{3}]}_{0}^{1} \\ = \frac{1}{3} \end{aligned}

$f(x,y)=1-x^2-y^2,D=\{(x,y)|x^2+y^2\leq 1,x> 0,y > 0 \}$ $\displaystyle \iint_D f(x,y) \mathrm{d}A$ .

解：由图 [图略，待补充]

\iint_{D} f (x, y) d A = \int_{0}^{1} [\int_{0}^{\sqrt{1 - x^{2}}} (1 - x^{2} - y^{2}) d y] d x

Inner:

\begin{aligned} \int_{0}^{\sqrt{1 - x^{2}}} (1 - x^{2} - y^{2}) d y & = {[y - x^{2} y - \frac{1}{3} y^{3}]}_{0}^{\sqrt{1 - x^{2}}} \\ = \sqrt{1 - x^{2}} - x^{2} \sqrt{1 - x^{2}} - \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} \\ = \sqrt{1 - x^{2}} (1 - x^{2} - \frac{1}{3} (1 - x^{2})) \\ = \sqrt{1 - x^{2}} (\frac{2}{3} - \frac{2}{3} x^{2}) \\ = \frac{2}{3} (1 - x^{2})^{\frac{3}{2}} \end{aligned}

$\displaystyle \int_0^1\frac{2}{3}(1-x^2)^{\frac{3}{2}}\mathrm{d}x$ $x=\sin\theta,\mathrm{d}x=\cos\theta\mathrm{d}\theta$

\begin{aligned} \int_{0}^{1} \frac{2}{3} (1 - x^{2})^{\frac{3}{2}} d x & = \int_{0}^{\frac{π}{2}} \frac{2}{3} \cos^{4} θ d θ \\ = \frac{3}{2} \int_{0}^{\frac{π}{2}} {(\frac{1 + \cos 2 θ}{2})}^{2} d θ \\ = \frac{3}{2} \int_{0}^{\frac{π}{2}} (\frac{1}{4} + \frac{1}{2} \cos 2 θ + \frac{1}{4} \cos^{2} 2 θ) d θ \\ = \frac{3}{2} ({[\frac{θ}{4}]}_{0}^{\frac{π}{2}} + {[\frac{\sin 2 θ}{4}]}_{0}^{\frac{π}{2}} + \frac{1}{4} \int_{0}^{\frac{π}{2}} {(\frac{1 + \cos 4 θ}{2})}^{2} d θ) \\ = \frac{3}{2} (\frac{π}{8} + \frac{π}{16} + \frac{1}{8} \cdot \frac{1}{4} {[\sin 4 θ]}_{0}^{\frac{π}{2}}) \\ = \frac{π}{8} \end{aligned}

Note

重积分可交换顺序？一般可以，注意上下限

$\displaystyle \iint_{D} xy \mathrm{d}x\mathrm{d}y$ $D$ $y = 1$ $x = 2$ $y = x$ 围成.

解法一：由图 [图略，待补充]

\begin{aligned} \iint_{D} x y d x d y & = \int_{1}^{2} [\int_{1}^{x} x y d y] d x \\ = \int_{1}^{2} ({[\frac{1}{2} x y^{2}]}_{1}^{x}) d x \\ = \int_{1}^{2} (\frac{x^{3}}{2} - \frac{x}{2}) d x \\ = {[\frac{x^{4}}{8} - \frac{x^{2}}{4}]}_{1}^{2} \\ = 1 + \frac{1}{8} \\ = \frac{9}{8} \end{aligned}

解法二：由图 [图略，待补充]

\begin{aligned} \iint_{D} x y d x d y & = \int_{1}^{2} [\int_{y}^{2} x y d x] d y \\ = \int_{1}^{2} {[\frac{1}{2} x^{2} y]}_{y}^{2} d y \\ = \int_{1}^{2} (2 y - \frac{y^{3}}{2}) d y \\ = {[y^{2} - \frac{y^{4}}{8}]}_{1}^{2} \\ = 2 - (1 - \frac{1}{8}) \\ = \frac{9}{8} \end{aligned}

$\displaystyle \iint_{D} y\sqrt{1+x^2 - y^2} \mathrm{d}\sigma$ $D$ $y = x$ $x = -1$ $y = 1$ 围成的闭区域.

解法一：由图 [图略，待补充]

\begin{aligned} \iint_{D} y \sqrt{1 + x^{2} - y^{2}} d x d y & = \int_{- 1}^{1} d x \int_{x}^{1} y \sqrt{1 + x^{2} - y^{2}} d y \\ = \int_{- 1}^{1} [{[- \frac{1}{3} (1 + x^{2} - y^{2})^{\frac{3}{2}}]}_{x}^{1}] d x \\ = \int_{- 1}^{1} \frac{1}{3} (1 - | x |^{3}) d x \\ = \frac{2}{3} \int_{0}^{1} (1 - x^{3}) d x \\ = \frac{1}{2} \end{aligned}

解法二：由图 [图略，待补充]

\iint_{D} y \sqrt{1 + x^{2} - y^{2}} d σ = \int_{- 1}^{1} y [\int_{- 1}^{y} \sqrt{1 + x^{2} - y^{2}} d x] d y

Note

极坐标下的积分

$f(x,y)=1-x^2-y^2,D= \{(x,y)|x^2+y^2\leq 1,x\geq 0,y\geq 0 \}$ $\displaystyle \iint_D f(x,y) \mathrm{d}A$ .

解:由图 [图略，待补充] 由于

d A = d x d y = r d θ d r

\begin{aligned} \iint_{D} (1 - x^{2} - y^{2}) d x d y & = \iint_{D} (1 - x^{2} - y^{2}) r d r d θ \\ = \iint_{D} (1 - r^{2}) r d r d θ \\ = \int_{0}^{\frac{π}{2}} [\int_{0}^{1} (1 - r^{2}) r d r] d θ \\ = \frac{π}{2} {[\frac{1}{2} r^{2} - \frac{1}{4} r^{4}]}_{0}^{1} \\ = \frac{π}{8} \end{aligned}

11.2 多变量积分的换元法

Important

变量替换与雅可比行列式

$\begin{cases} u = u(x,y) \\ v = v(x,y) \end{cases}$ ，其全微分形式可表示为：

{\begin{cases} d u = \frac{\partial u}{\partial x} d x + \frac{\partial u}{\partial y} d y \\ d v = \frac{\partial v}{\partial x} d x + \frac{\partial v}{\partial y} d y \end{cases}

该微分系统可写成矩阵形式：

[\begin{matrix} d u \\ d v \end{matrix}] = [\begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{matrix}] [\begin{matrix} d x \\ d y \end{matrix}]

[图略，待补充]

\begin{matrix} d u d v = J = \frac{\partial (u, v)}{\partial (x, y)} = | \begin{matrix} \frac{\partial u}{\partial x} d x & \frac{\partial u}{\partial y} d y \\ \frac{\partial v}{\partial x} d x & \frac{\partial v}{\partial y} d y \end{matrix} | = | \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x} | d x d y \end{matrix}

Important

雅可比行列式与坐标变换的深层关系

$(x,y) \leftrightarrow (u,v)$ 中，雅可比行列式不仅描述面积元的缩放比例，更揭示了双向变换的对称性。由基本关系：

d u d v = | J | d x d y = | \frac{\partial (u, v)}{\partial (x, y)} | d x d y

通过反函数定理，可得逆变换关系：

d x d y = | \frac{\partial (x, y)}{\partial (u, v)} | d u d v

这建立了双向变换的精确对应：雅可比行列式互为倒数，即：

| \frac{\partial (x, y)}{\partial (u, v)} | = {| \frac{\partial (u, v)}{\partial (x, y)} |}^{- 1}

经典示例：极坐标变换
$\begin{cases} x = \rho \cos\theta \\ y = \rho \sin\theta \end{cases}$ ，其雅可比矩阵为：

\begin{matrix} J (ρ, θ) = \frac{\partial (x, y)}{\partial (ρ, θ)} = | \begin{matrix} \frac{\partial x}{\partial ρ} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial ρ} & \frac{\partial y}{\partial θ} \end{matrix} | = | \begin{matrix} \cos θ & - ρ \sin θ \\ \sin θ & ρ \cos θ \end{matrix} | = ρ \end{matrix}

因此面积元变换为：

d x d y = | ρ | d ρ d θ = ρ d ρ d θ

矩阵互逆性的几何诠释
变换矩阵的互逆性直接体现在坐标系转换中：

[\begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix}] 与 [\begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{matrix}]

$J_{uv} = 5$ $\displaystyle \frac{1}{5}$ ，此时：

d x d y = \frac{1}{5} d u d v

这种精确的倒数关系确保了积分在不同坐标系下计算结果的一致性，是多重积分变量替换定理的核心数学基础。

Note

$\displaystyle \int_{-\infty}^{+\infty}e^{-x^2}\mathrm{d}x$ .

解: $\displaystyle I=\int_{-\infty}^{+\infty}e^{-x^2}\mathrm{d}x$ 则

I^{2} = \int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} e^{- (x^{2} + y^{2})} d x d y = \int_{- \infty}^{+ \infty} e^{- y^{2}} d y \int_{- \infty}^{+ \infty} e^{- x^{2}} d x

$x=r\cos\theta,y=r\sin\theta$ ，则

\begin{aligned} I^{2} & = \int_{0}^{2 π} \int_{0}^{+ \infty} e^{- r^{2}} r d r d θ \\ = \int_{0}^{2 π} d θ (\frac{1}{2} \int_{0}^{+ \infty} e^{- r^{2}} d r^{2}) \\ = \frac{1}{2} \int_{0}^{2 π} {[(- e^{- r^{2}})]}_{0}^{+ \infty} d θ \\ = π \end{aligned}

故

\int_{- \infty}^{+ \infty} e^{- x^{2}} d x = I = \sqrt{π}

11.3 重积分的应用

Important

质量

$D$ $\rho(x,y)$ ，则总质量可通过二重积分计算：

M = \iint_{D} ρ (x, y) d σ

$\mathrm{d}\sigma$ $\mathrm{d}x\mathrm{d}y$ $\rho$ $M = \rho \cdot A$ $\displaystyle A = \iint_D \mathrm{d}\sigma$ $D$ 的面积.

质心

$(\bar{x}, \bar{y})$ 定义为：

\bar{x} = \frac{\sum_{i = 1}^{n} m_{i} x_{i}}{\sum_{i = 1}^{n} m_{i}}, \bar{y} = \frac{\sum_{i = 1}^{n} m_{i} y_{i}}{\sum_{i = 1}^{n} m_{i}}

$\mu(x,y)$ 时：

\bar{x} = \frac{\iint_{D} x μ (x, y) d σ}{\iint_{D} μ (x, y) d σ}, \bar{y} = \frac{\iint_{D} y μ (x, y) d σ}{\iint_{D} μ (x, y) d σ}

Note

$\displaystyle I=\iint_Dr^2(\rho,\theta)\rho \mathrm{d}\rho \mathrm{d}\theta$ .

[图略，待补充] 解:

\begin{aligned} I & = \iint_{D} r^{2} (ρ, θ) ρ d ρ d θ \\ = \int_{- \frac{π}{2}}^{\frac{π}{2}} \int_{0}^{2 \cos θ} ρ^{3} d ρ d θ \\ = \int_{- \frac{π}{2}}^{\frac{π}{2}} {[\frac{1}{4} ρ^{4}]}_{0}^{2 \cos θ} d θ \\ = \int_{0}^{\frac{π}{2}} 4 \cos^{4} θ d θ \\ = \frac{3}{2} π \end{aligned}

$\displaystyle I=\iint_Dr^2(\rho,\theta)\rho \mathrm{d}\rho \mathrm{d}\theta$ .

[图略，待补充] 解法一：

\begin{aligned} I & = \iint_{D} r^{2} (ρ, θ) ρ d ρ d θ \\ = \int_{- \frac{π}{2}}^{\frac{π}{2}} \int_{0}^{2 \cos θ} (1 + ρ^{2} - 2 ρ \cos θ) ρ d ρ d θ \\ = \frac{π}{2} \end{aligned}

解法二:

I = \int_{0}^{2 π} \int_{0}^{1} ρ^{2} \cdot ρ d ρ d θ = \frac{π}{2}

Warning

转动惯量与质心的关系(不要求)

$D$ $\mu(x,y)$ $(x_0, y_0)$ 的转动惯量为：

I (x_{0}, y_{0}) = \iint_{D} μ (x, y) [(x - x_{0})^{2} + (y - y_{0})^{2}] d x d y

$\mu(x,y)\mathrm{d}x\mathrm{d}y$ 到参考点距离平方的加权积分.特别地，当参考点为原点时，转动惯量简化为：

I = \iint_{D} μ (x, y) r^{2} (x, y) d x d y (r^{2} = x^{2} + y^{2})

质心的极值特性
$I(x_0, y_0)$ $x_0, y_0$ 求偏导并令其为零：

\begin{matrix} \frac{\partial I}{\partial x_{0}} = \iint_{D} 2 μ (x, y) (x_{0} - x) d x d y = 0 \\ \frac{\partial I}{\partial y_{0}} = \iint_{D} 2 μ (x, y) (y_{0} - y) d x d y = 0 \end{matrix}

由此可得方程组：

\begin{matrix} x_{0} \iint_{D} μ d x d y = \iint_{D} μ x d x d y \\ y_{0} \iint_{D} μ d x d y = \iint_{D} μ y d x d y \end{matrix}

最终解得质心坐标为：

x_{0} = \frac{\iint_{D} μ x d x d y}{\iint_{D} μ d x d y}, y_{0} = \frac{\iint_{D} μ y d x d y}{\iint_{D} μ d x d y}

这与离散质点系的质心公式一致：

\begin{matrix} {\begin{cases} x_{0} = \frac{\sum m_{i} x_{i}}{\sum m_{i}} & （质量加权平均） \\ y_{0} = \frac{\sum m_{i} y_{i}}{\sum m_{i}} & （质量加权平均） \end{cases} \end{matrix}

Warning

平行轴定理

平行轴定理揭示了物体绕不同轴转动惯量之间的关系：绕任意轴的转动惯量等于绕质心轴的转动惯量加上总质量与该轴到质心距离平方的乘积。其数学推导可通过二重积分展开：

$(x_c, y_c)$ $(x_0, y_0)$ 的转动惯量为：

\begin{aligned} I (x_{0}, y_{0}) & = \iint_{D} μ [(x - x_{0})^{2} + (y - y_{0})^{2}] d x d y \\ = \iint_{D} μ [(x - x_{c} + x_{c} - x_{0})^{2} + (y - y_{c} + y_{c} - y_{0})^{2}] d x d y \end{aligned}

展开平方项并重组积分：

\begin{aligned} = \underset{I_{c} (绕质心的转动惯量)}{\underset{⏟}{\iint_{D} μ [(x - x_{c})^{2} + (y - y_{c})^{2}] d x d y}} \\ + \underset{M r^{2} (总质量 \times 距离平方)}{\underset{⏟}{\iint_{D} μ [(x_{c} - x_{0})^{2} + (y_{c} - y_{0})^{2}] d x d y}} \\ + 2 (x_{c} - x_{0}) \underset{0 (质心定义)}{\underset{⏟}{\iint_{D} μ (x - x_{c}) d x d y}} \\ + 2 (y_{c} - y_{0}) \underset{0 (质心定义)}{\underset{⏟}{\iint_{D} μ (y - y_{c}) d x d y}} \end{aligned}

$\displaystyle x_c = \frac{1}{M}\iint_D \mu x \mathrm{d}x\mathrm{d}y$ $\displaystyle \iint_D \mu(x - x_c)\mathrm{d}x\mathrm{d}y$ $\displaystyle \iint_D \mu(y - y_c)\mathrm{d}x\mathrm{d}y$ 必然为零。最终得到简洁形式：

I (x_{0}, y_{0}) = I_{c} + M r^{2}

$r = \sqrt{(x_c - x_0)^2 + (y_c - y_0)^2}$ $\displaystyle M = \iint_D \mu \mathrm{d}x\mathrm{d}y$ 为总质量.

Note

$x^2 + y^2 + z^2 \leq 4a^2$ $x^2 + y^2 = 2ax$ $a > 0$ ) 截得的含在圆柱面内的立体体积.

[图略，待补充] 解:

\begin{aligned} V & = 4 \iint_{D} \sqrt{4 a^{2} - x^{2} - y^{2}} d x d y \\ = 4 \int_{0}^{\frac{π}{2}} \int_{0}^{2 a \cos θ} \sqrt{4 a^{2} - ρ^{2}} ρ d ρ d θ \\ = 4 \int_{0}^{\frac{π}{2}} \frac{8}{3} a^{3} (1 - \sin^{3} θ) d θ \\ = \frac{16}{3} a^{3} π - \frac{8}{3} \cdot \frac{8}{3} a^{3} \\ = \frac{32}{3} a^{3} (\frac{π}{2} - \frac{2}{3}) \end{aligned}

Note
换元法

(x, y) \to (u, v)

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $\displaystyle \iint_D\mathrm{d}x\mathrm{d}y$ .

解法一: [图略，待补充]

\begin{aligned} \iint_{D} d x d y & = 4 \int_{0}^{a} \int_{0}^{b \sqrt{1 - \frac{x^{2}}{a^{2}}}} d y d x \\ = 4 \int_{0}^{a} b \sqrt{1 - \frac{x^{2}}{a^{2}}} d y \end{aligned}

$y=a\sin\theta,\mathrm{d}y=a\cos\theta\mathrm{d}\theta$ ,则

原 式 = 4 \int_{0}^{\frac{π}{2}} b \cos θ a \cos θ d θ = 4 a b \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 θ}{2} d θ = π a b

解法二 $x=au,y=bv,-1\leq u \leq 1,-1\leq v \leq 1,u^2+v^2=1$ ，则

\begin{matrix} d x = a d u \\ d y = b d v \\ d x d y = a d u \cdot b d v \end{matrix}

\iint_{D} d x d y = \iint_{D} a b d u d v = π a b

Note

$\begin{cases}u=3x-2y\\v=x+y\end{cases}$

[图略，待补充] 解: 方法一：

\begin{matrix} | ω_{1} \times ω_{2} | = | ω_{1} | \cdot | ω_{2} | \cdot \sin θ = | \begin{matrix} 3 & - 2 \\ 1 & 1 \end{matrix} | = 5 \end{matrix}

方法二：

\begin{matrix} J = | \begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{matrix} | = | \begin{matrix} 3 & - 2 \\ 1 & 1 \end{matrix} | = 5 \end{matrix}

则

d u d v = \frac{\partial (u, v)}{\partial (x, y)} d x d y = 5 d x d y

\iint_{D} d x d y = \iint_{D} \frac{1}{5} d u d v

11.4 三重积分

Note

$\displaystyle \iiint_{\Omega}x\mathrm{d}x\mathrm{d}y\mathrm{d}z$ $\Omega$ $x + 2y + z = 1$ $x=0$ $y=0$ $z=0$ ) 围成.

[图略，待补充] 解法一:

\begin{aligned} ∭_{Ω} x d x d y d z & = \int_{0}^{1} d z \int_{0}^{\frac{1}{2} (1 - z)} d y \int_{0}^{1 - 2 y - z} x d x \\ = \int_{0}^{1} d z \int_{0}^{\frac{1}{2} (1 - z)} \frac{1}{2} (1 - 2 y - z)^{2} d y \\ = \frac{1}{2} \int_{0}^{1} {[- \frac{(1 - 2 y - z)^{3}}{6}]}_{0}^{\frac{1}{2} (1 - z)} d z \\ = \frac{1}{12} \int_{0}^{1} (1 - z)^{3} d z \\ = \frac{1}{12} {[- \frac{(1 - z)^{4}}{4}]}_{0}^{1} \\ = \frac{1}{48} . \end{aligned}

解法二:

\begin{aligned} ∭_{Ω} x d x d y d z & = \int_{0}^{1} d x \int_{0}^{\frac{1}{2} (1 - x)} d y \int_{0}^{1 - x - 2 y} x d z \\ = \int_{0}^{1} x d x \int_{0}^{\frac{1}{2} (1 - x)} (1 - x - 2 y) d y \\ = \int_{0}^{1} \frac{1}{4} x (1 - x)^{2} d x \\ = \frac{1}{48} \end{aligned}

$\displaystyle \iiint_{\Omega} z \mathrm{d}x\mathrm{d}y\mathrm{d}z$ $\Omega$ $z = x^2 + y^2$ $z = 4$ 围成的闭区域.

[图略，待补充] 解:令

{\begin{cases} x = ρ \cos θ \\ y = ρ \sin θ \\ z = z \end{cases}

则

d V = d x d y d z = ρ d ρ d θ d z

\begin{aligned} ∭_{Ω} z d x d y d z & = ∭_{Ω} z ρ d ρ d θ d z \\ = \int_{0}^{4} d z \int_{0}^{2 π} d θ \int_{0}^{\sqrt{z}} z ρ d ρ \\ = \int_{0}^{4} d z \int_{0}^{2 π} \frac{1}{2} z^{2} d θ \\ = \int_{0}^{4} z^{2} π d z \\ = \frac{1}{3} π {[z^{3}]}_{0}^{4} \\ = \frac{64}{3} π \end{aligned}

第11章 多元函数的积分

11.1 二重积分

11.1.1 二重积分的概念

11.1.2 二重积分的计算

11.2 多变量积分的换元法

11.3 重积分的应用

11.4 三重积分

第11章多元函数的积分