475 lines
23 KiB
TeX
475 lines
23 KiB
TeX
\chapter{多重积分}
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\section{矩形区域上的积分}
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\begin{definition}[二重积分]
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设$I$是$\realnum^2$中的闭矩形，$I = [a, b] \times [c, d]$。$f: I \to \realnum$。作$[a, b]$的分割
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\[\pi_x: a = x_0 < x_1 < \dots < x_n = b\]
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又作$[c, d]$的分割
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\[\pi_y: c = y_0 < y_1 < \dots < y_m = d\]
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两组平行线把$I$分割成$k = n \times m$个子矩形
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\[[x_{i - 1}, x_i] \times [y_{j - 1}, y_j], i = 1, 2, \dots, n, j = 1, 2, \dots, m\]
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这$k$个子矩形的全体组成$I$的一个分割$\pi = \pi_x \times \pi_y$，用一定的次序重排这$k$个子矩形，将它们编号为$I_1, I_2, \dots, I_k$，在每一个$I_i$中任取一点$\xi_i(i = 1, 2, \dots, k)$，作积分和（也称Riemann和）
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\[\sum_{i = 1}^k f(\xi_i) \sigma(I_i)\]
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记
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\[\norm{\pi} = \max\{\diam (I_1), \dots, \diam(I_k)\}\]
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这里$\diam(I_i)$是矩形$I_i$对角线的长度，称$\norm{\pi}$为分割$\pi$的宽度；$\sigma(I_i)$表示矩形$I_i$的面积。
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如果存在数$A$使得对任意给定的$\varepsilon > 0$，有$\delta > 0$，凡是$\norm{\pi} < \delta$时，不论值点$\xi_i$在子矩形$I_i$中如何选择，都有
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\[\abs{\sum_{i = 1}^k f(\xi_i) \sigma(I_i) - A} < \varepsilon\]
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则称函数$f$在矩形$I$上可积，并将$A$写作
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\[\iint \limits_I f(x, y) \dif x \dif y \qquad \text{或者} \qquad \int \limits_I f \dif \sigma\]
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称之为$f$在矩形$I$上的二重积分，或简称$f$在$I$上的积分。这里$f$称为被积函数，$I$称为积分区域。
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\end{definition}
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\begin{theorem}[可积函数的有界性]
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如果$f$在$I$上可积，那么$f$必在$I$上有界。
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\end{theorem}
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\begin{theorem}[可积函数线性性质]
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设$f, g \in R(D)$，$\alpha, \beta \in \realnum$，则$\alpha f + \beta g \in R(D)$，且
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\[\int \limits_D (\alpha f + \beta g) \dif \sigma = \alpha \int \limits_D f \dif \sigma + \beta \int \limits_D g \dif \sigma\eqper\]
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\end{theorem}
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\begin{theorem}
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若$f$和$g$在$I$上可积且$f \geq g$，那么
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\[\int \limits_I f \dif \sigma \geq \int \limits_I g \dif \sigma\eqper\]
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\end{theorem}
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\begin{theorem}
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设$D = D_1 \cup D_2$，且$D_1$与$D_2$没有公共内点，若$f(x, y) \in R(D)$，则$f(x, y) \in R(D_1), f(x, y) \in R(D_2)$，且
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\[\iint \limits_D f(x, y) \dif \sigma = \iint \limits_{D_1} f(x, y) \dif \sigma + \iint \limits_{D_2} f(x, y) \dif \sigma\eqper\]
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\end{theorem}
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\begin{theorem}
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若$f(x, y) \in R(D)$，则$\abs{f(x, y)} \in R(D)$，并且有
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\[\abs{\iint \limits_D f(x, y) \dif \sigma} \leq \iint \limits_D \abs{f(x, y)} \dif \sigma\eqper\]
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\end{theorem}
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\begin{theorem}
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若$f(x, y) \in R(D)$，则$f$有界，设$m \leq f(x, y) \leq M$，则有
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\[m \sigma(D) \leq \iint \limits_D f(x, y) \dif \sigma \leq M \sigma(D)\eqper\]
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\end{theorem}
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\begin{theorem}[积分中值定理]
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$D \subset \realnum^2$连通、有界闭，$\partial D$为零面积集，$f \in C(D)$，则存在$(\xi, \eta) \in D$，满足
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\[\iint \limits_D f(x, y) \dif \sigma = f(\xi, \eta) \sigma(D)\]
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其中
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\[f(\xi, \eta) = \frac{1}{\sigma(D)} \iint \limits_D f(x, y) \dif \sigma\]
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称为$f(x, y)$在$D$上的平均值。
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\end{theorem}
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\begin{theorem}
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$D \subset \realnum^2$为连通有界闭集，$\partial D$为零面积集，$g$在$D$上不变号，$f, g \in C(D)$。则存在$(\xi, \eta) \in D$，满足
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\[\iint \limits_D f(x, y) g(x, y) \dif \sigma = f(\xi, \eta) \iint \limits_D g(x, y) \dif \sigma\]
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\end{theorem}
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\begin{theorem}
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设$f \in R(D)$，$D$关于$OX$轴对称，则
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\begin{itemize}
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\item 若$f(x, y)$关于$y$为己函数，则$\displaystyle\iint \limits_D f(x, y) \dif \sigma = 0$;
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\item 若$f(x, y)$关于$y$为偶函数，记$D_1$为$D$位于$OX$轴上方的部分，则$\displaystyle\iint \limits_D f(x, y) \dif \sigma = 2 \iint \limits_{D_1} f(x, y) \dif \sigma$。
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\end{itemize}
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\end{theorem}
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\begin{theorem}[轮换不变性]
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若$D \subset \realnum^2$关于$x, y$是轮换对称的，即$(x, y) \in D \Leftrightarrow (y, x) \in D$，则
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\[\iint \limits_D f(x, y) \dif \sigma = \iint \limits_D f(y, x) \dif \sigma\eqper\]
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\end{theorem}
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\section{函数可积性}
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考虑$D = [a, b] \times [c, d]$，$f: D \to \realnum$有界。
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引入Riemann和记号：令$T$是$D$的一个矩形分割，$\xi = \{\xi_i\}$表示关于分割$T$的小矩形组中任取的一组点集，记
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\[S(T, \xi) = \sum_i f(\xi_i) \sigma (D_i)\eqper\]
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再引入Darboux和记号，定义上/下和
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\[\overline{S}(T) = \sum_i M_i \sigma (D_i), \underline{S}(T) = \sum_i m_i \sigma(D_i)\]
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其中$M_i = \sup f(D_i), m_i = \inf f(D_i), i = 1, 2, \dots, k$。
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\begin{lemma}
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对任意的$T$，
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\[\underline{S}(T) \leq S(T, \xi) \leq \overline{S}(T)\eqper\]
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\end{lemma}
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\begin{lemma}
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设$T^\prime$为分割$T$的加细，则
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\[\underline{S}(T) \leq \underline{S}(T^\prime) \leq \overline{S}(T^\prime) \leq \overline{S}(T)\eqper\]
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\end{lemma}
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\begin{lemma}
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设$T_1$和$T_2$是$D$的任意两个矩形分割，则
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\[\underline{S}(T_1) \leq \overline{S}(T_2)\eqper\]
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\end{lemma}
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这说明上下和都是有界的，因此都有上下确界。因此我们可以定义Darboux上下积分：
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\[\underline{\int \limits_{D}} f \dif \sigma = \sup \limits_T \underline{S}(T), \overline{\int \limits_D} f \dif \sigma = \inf \limits_T \overline{S}(T)\eqper\]
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\begin{lemma}
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设$T_1$和$T_2$是$D$的任意两个矩形分割，则
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\[\underline{S}(T_1) \leq \underline{\int \limits_D} f \dif \sigma \leq \overline{\int \limits_D} f \dif \sigma \leq \overline{S}(T_2)\eqper\]
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\end{lemma}
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\begin{theorem}[Darboux定理]
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等式
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\[\underline{\int \limits_D} f \dif \sigma = \overline{\int \limits_D} f \dif \sigma\]
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成立当且仅当对任意的$\varepsilon > 0$，都存在分割$T$使得$0 \leq \overline{S}(T) - \underline{S}(T) < \varepsilon$。
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\end{theorem}
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\begin{theorem}[Riemann定理]
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$f \in R(D)$的充分必要条件是：对任意$\varepsilon > 0$，存在$\delta > 0$，对于任意分割$T$，只要满足$\norm{T} < \delta$，就有$0 \leq \overline{S}(T) - \underline{S}(T) < \varepsilon$。
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\end{theorem}
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\begin{theorem}[Riemann-Darboux定理]
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$f \in R(D)$的充分必要条件是等式
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\[\underline{\int \limits_D} f \dif \sigma = \overline{\int \limits_D} f \dif \sigma\]
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成立，这时
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\[\int \limits_D f \dif \sigma = \underline{\int \limits_D} f \dif \sigma = \overline{\int \limits_D} f \dif \sigma\eqper\]
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\end{theorem}
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\section{Riemann可积函数类}
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\begin{theorem}
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$C(D) \subset R(D)$。
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\end{theorem}
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\begin{definition}[二维零测集]
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若$E \subset \realnum^2$满足：对任意的$\varepsilon > 0$，存在一列闭矩形$\{D_i\}_{i = 1}^\infty$，使得$E \subset \bigcup \limits_{i = 1}^\infty D_i$且$\sum_{i = 1}^\infty \sigma(D_i) < \varepsilon$，则称$E$为二维零测集。
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\end{definition}
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\begin{definition}[零面积集]
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若$E \subset \realnum^2$满足：对任意的$\varepsilon > 0$，存在有限个闭矩形$\{D_i\}_{i = 1}^m$，使得$E \subset \bigcup \limits_{i = 1}^m D_i$且$\sum_{i = 1}^m \sigma(D_i) < \varepsilon$，则称$E$为二维零测集。
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\end{definition}
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\begin{corollary}
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零面积集是二维零测集。
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\end{corollary}
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二维零测集的实例：
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\begin{itemize}
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\item 至多可数的点集；
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\item 二维零测集的子集；
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\item 至多可数个二维零测集的并集。
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\end{itemize}
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零面积集的实例：
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\begin{itemize}
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\item 有限点集；
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\item 零面积集的子集；
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\item 有限多个零面积集的并集。
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\end{itemize}
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推广：
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\begin{itemize}
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\item 平面上任意有限长直线段是零面积集；
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\item 平面上任意有限长折线是零面积集；
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\item 平面上任意有限长光滑曲线是二维零测集。
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\end{itemize}
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仿照一元函数的情况，定义
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\begin{definition}[间断点]
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\[D_{is} (f) = \{(x, y) \in D \mid \text{$f$在$(x, y)$点不连续}\}\]
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称为$f$的间断点集。
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\end{definition}
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\begin{theorem}[Lebesgue定理]
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设$D = [a, b] \times [c, d]$，$f: D \to \realnum$有界，则$f$在$D$上Riemann可积当且仅当$f$的间断点集是二维零测集。
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\end{theorem}
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\begin{corollary}
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若$f$有界且间断点集是零面积集，则$f$是Riemann可积的。
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\end{corollary}
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\begin{corollary}
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记$D_0 = \{(x, y) \in D \mid f(x, y) \neq 0\}$。设$f$有界且$D_0$是零面积集，则$f$可积，且
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\[\int \limits_D f \dif \sigma = 0 \eqper\]
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\end{corollary}
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\begin{corollary}
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设$f, g$都在$D$上有界且仅在零面积集上不想等。若$f \in R(D)$，则必有$g \in R(D)$，且
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\[\int \limits_D f \dif \sigma = \int \limits_D g \dif \sigma \eqper\]
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\end{corollary}
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\begin{corollary}
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\begin{enumerate}
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\item 若$f \in R(D)$，则$\abs{f} \in R(D)$；
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\item 若$f, g \in R(D)$，则$fg \in R(D)$；
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\item 若$f, g \in R(D)$且$\dfrac{1}{g}$有界，则$\dfrac{f}{g} \in R(D)$。
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\end{enumerate}
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\end{corollary}
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\section{矩形区域上积分的计算}
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令$D = [a, b] \times [c, d]$，$f \in C(D)$记$I = \dint \limits_D f \dif \sigma$。
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\begin{enumerate}
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\item 固定$x \in [a, b]$，作为$y$的函数$f(x, y) \in C[c, d]$。令
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\[u(x) = \int_c^d f(x, y) \dif y\]
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易证$u(x) \in C[a,b]$记
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\[A = \int_a^b u(x) \dif x\]
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\item 类似地，固定$y \in [c, d]$可以得到$v(y)$，记
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\[B = \int_c^d v(y) \dif y\]
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\end{enumerate}
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\begin{theorem}[Fubini定理]
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设$f \in C(D)$，则$I = A = B$：
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\[\int \limits_D f \dif \sigma = \int_a^b \dif x \int_c^d f(x, y) \dif y = \int_c^d \dif y \int_a^b f(x, y) \dif x\]
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其中规定记号
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\begin{align*}
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\int_a^b \dif x \int_c^d f(x, y) \dif y & = \int_a^b \left[\int_c^d f(x, y) \dif y \right]\dif x\\
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\int_c^d \dif y \int_a^b f(x, y) \dif x & = \int_c^d \left[\int_a^b f(x, y) \dif x\right] \dif y\eqper
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\end{align*}
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\end{theorem}
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\begin{corollary}
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设$f(x, y) = f_1(x) f_2(y)$，其中$f_1 \in C[a, b], f_2 \in C[c, d]$。令$D = [a, b] \times [c, d]$，则$f \in C(D)$，且
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\[\int \limits_D f \dif \sigma = \left[\int_a^b f_1(x) \dif x\right]\left[\int_c^d f_2(y) \dif y\right]\eqper\]
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\end{corollary}
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\section{有界集合上的二重积分}
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\begin{definition}
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设$D \subset \realnum^2$，函数$f: B \to \realnum$。令
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\[f_D(x, y) = \left\{\begin{aligned}
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& f(x, y), \qquad (x, y) \in D\\
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& 0, \qquad (x, y) \not \in D
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\end{aligned}\right.\]
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\end{definition}
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\begin{definition}
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任取有界的闭矩形$I \supset D$，如果函数$f_D$在$I$上可积，则说函数$f$在$B$上可积，并把数值$\dint \limits_I f_D \dif \sigma$称为函数$f$在$B$上的（二重）积分，记作
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\[\iint \limits_B f(x, y) \dif x \dif y \quad \text{或者} \quad \int \limits_B f \dif \sigma\]
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\end{definition}
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在矩形区域上的二重积分性质在一般的有界集合的二重积分上也成立。
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\begin{theorem}[区域可加性]
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设$D_1, D_2 \subset \realnum^2$都是有界区域，且$D_1 \cap D_2$是零面积集。则$f \in R(D_1 \cup D_2)$当且仅当$f \in R(D_1)$且$f \in R(D_2)$。这时有
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\[\int \limits_{D_1 \cup D_2} f \dif \sigma = \int \limits_{D_1} f \dif \sigma + \int \limits_{D_2} f \dif \sigma\eqper\]
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\end{theorem}
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\begin{theorem}[一般有界区域上连续函数的可积性]
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设$D \subset \realnum^2$是有界区域，$\partial D$是零面积集，则$C(D) \subset R(D)$。
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\end{theorem}
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\begin{definition}[一般有界区域的面积]
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设$D \subset \realnum^2$是有界区域，且$\partial D$是零面积集，则$1 \in C(D) \subset R(D)$，可以定义$D$的面积
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\[\sigma(D) = \int \limits_D 1 \dif \sigma\eqper\]
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\end{definition}
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\begin{proposition}[平面区域面积可加性]
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设$D_1, D_2 \subset \realnum^2$都是有界区域，边界都是零面积集，又设$D_1 \cap D_2$也是零面积集，则
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\[\sigma(D_1 \cup D_2) = \sigma(D_1) + \sigma(D_2) \eqper\]
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\end{proposition}
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\section{有界集合上积分的计算}
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\begin{theorem}
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设点集$D = \{(x, y) \mid y_1(x) \leq y \leq y_2(x), a \leq x \leq b\}$，其中$y_1, y_2 \in C[a, b]$且$y_1 \leq y_2$，则
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\[\int \limits_D f \dif \sigma = \int_a^b \dif x \int_{y_1(x)}^{y_2(x)} f(x, y) \dif y\eqper\]
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类似地，如果有$D = \{(x, y) \mid x_1(y) \leq x \leq x_2(y), c \leq y \leq d\}$，其中$x_1, x_2 \in C[c, d]$，且$x_1 \leq x_2$则
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\[\int \limits_D f \dif \sigma = \int_c^d \dif y \int_{x_1(y)}^{x_2(y)} f(x, y) \dif x\eqper\]
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\end{theorem}
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\section{二重积分换元}
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\begin{theorem}
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设有界闭区域$D \subset \realnum^2$，连续函数$f: D \to \realnum$。$\Omega$是$uOv$平面上的有界闭区域。映射
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\[\bvec{\varphi}: \begin{cases}
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x = x(u, v)\\
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y = y(u, v)
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\end{cases}\]
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是$\Omega$到$D$的一一对应的连续可微映射，满足
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\[\frac{\partial(x, y)}{\partial (u, v)} = \begin{vmatrix}
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\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v}\\[1em]
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\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}
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\end{vmatrix} \neq 0\]
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有换元公式
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\[\iint \limits_D f(x, y) \dif x \dif y = \iint \limits_\Omega f (x(u, v), y(u, v)) \abs{\frac{\partial (x, y)}{\partial (u, v)}} \dif u \dif v\eqper\]
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\end{theorem}
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下面介绍两种常用的换元：极坐标换元与正交变换。
|
||
|
||
首先介绍极坐标换元。令
|
||
\[\bvec{\varphi}: \begin{cases}
|
||
x = r \cos \theta\\
|
||
y = r \sin \theta
|
||
\end{cases}\]
|
||
这时
|
||
\[\frac{\partial (x, y)}{\partial (r, \theta)} = \begin{vmatrix}
|
||
\cos \theta & -r \sin \theta\\
|
||
\sin \theta & r \cos \theta
|
||
\end{vmatrix}
|
||
= r\]
|
||
如果当$(r, \theta) \in \Omega$时映射$\bvec{\varphi}$将$\Omega$一对一地变为$D$，那么
|
||
\[\iint \limits_D f(x, y) \dif x \dif y = \iint \limits_\Omega f(r\cos \theta, r \sin \theta) r \dif r \dif \theta\eqper\]
|
||
|
||
\begin{example}
|
||
求椭球面
|
||
\(\dfrac{x^2}{a^1} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1\)
|
||
包围的体积$V$。
|
||
\end{example}
|
||
|
||
\begin{proof}[解]
|
||
由对称性，$V$等于$z = 0$平面上部分体积的两倍。在$z = 0$平面上的部分可以看作曲顶柱体，顶部曲面方程
|
||
\[z = c \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}, D: \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\]
|
||
因此
|
||
\[V = 2 \int \limits_D z \dif \sigma\eqper\]
|
||
|
||
进一步引入广义极坐标变换$x = a r \cos \theta, y = br \sin \theta$。则上面区域$D$转化为矩形区域$\tilde{D}: 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$。计算Jacobian行列式得$\dif x \dif y = abr \dif r \dif \theta$。因此
|
||
\[V = 2abc \iint \limits_{\tilde{D}} \sqrt{1 - r^2} r \dif r \dif \theta = 4\pi abc \int_0^1 r \sqrt{1 - r^2} \dif r = \frac{4}{3} \pi abc\eqper \qedhere\]
|
||
\end{proof}
|
||
|
||
现在再介绍正交变换。令
|
||
\[\left\{\begin{aligned}
|
||
& u = \frac{ax + by}{\sqrt{a^2 + b^2}}\\
|
||
& v = \frac{ay - bx}{\sqrt{a^2 + b^2}}
|
||
\end{aligned}\right.\]
|
||
那么
|
||
\[\frac{\partial(u, v)}{\partial(x, y)} = \frac{1}{a^2 + b^2} \begin{vmatrix}
|
||
a & b\\
|
||
-b & a
|
||
\end{vmatrix} = 1\]
|
||
注意到正交变换有性质
|
||
\[u^2 + v^2 = \frac{(ax + by)^2 + (bx - ay)^2}{a^2 + b^2} = x^2 + y^2\]
|
||
这在变换积分域时通常能给出不少好处。
|
||
|
||
\begin{example}
|
||
设$f \in C(\realnum)$，$D = \{(x, y) \mid x^2 + y^2 \leq R^2\}$。将二重积分
|
||
\[\iint \limits_{x^2 + y^2 \leq R^2} f(ax + by + c) \dif x \dif y\]
|
||
约化为一重积分。
|
||
\end{example}
|
||
|
||
\begin{proof}[解]
|
||
引入正交变换
|
||
\[\left\{\begin{aligned}
|
||
& u = \frac{ax + by}{\sqrt{a^2 + b^2}}\\
|
||
& v = \frac{ay - bx}{\sqrt{a^2 + b^2}}
|
||
\end{aligned}\right.\]
|
||
因此
|
||
\begin{align*}
|
||
\text{原式} & = \iint \limits_{u^2 + v^2 \leq R^2} f\left(\sqrt{a^2 + b^2} u + c\right) \dif u \dif v\\
|
||
& = \int_{-R}^R \dif u \int_{-\sqrt{R^2 - u^2}}^{\sqrt{R^2 - u^2}} f \left(\sqrt{a^2 + b^2}u + c\right) \dif v\\
|
||
& = 2\int_{-R}^R \sqrt{R^2 - u^2} f\left(\sqrt{a^2 + b^2}u + c\right) \dif u\eqper \qedhere
|
||
\end{align*}
|
||
\end{proof}
|
||
|
||
\section{三重积分}
|
||
\begin{definition}[三重积分]
|
||
令$\Omega = [a, b] \times [c, d] \times [g, h]$是$\realnum^3$中的长方体。$f: \Omega \to \realnum$。将$\Omega$做有限的分割$T = \pi_x \times \pi_y \times \pi_z$，其中
|
||
\begin{align*}
|
||
& \pi_x: a = x_0 < x_1 < \dots < x_n = b\\
|
||
& \pi_y: c = y_0 < y_1 < \dots < y_n = d\\
|
||
& \pi_z: g = z_0 < z_1 < \dots < z_k = h
|
||
\end{align*}
|
||
将$\Omega$分割为$nmk$个小长方体。将他们重新编号为$\Omega_1, \Omega_2, \dots, \Omega_{nmk}$。在每个$\Omega_i$中任取一点$\xi_i$，作Riemann和
|
||
\[\sum_{i = 1}^{nmk} f(\xi_i) \mu(\Omega_i)\]
|
||
记
|
||
\[\norm{T} = \max\{\diam (\Omega_1), \dots, \diam (\Omega_{nmk})\}\]
|
||
这里$\diam (\Omega_i)$是长方体$\Omega_i$对角线的长度。称$\norm{T}$为分割$T$的直径；$\mu(\Omega_i)$表示长方体$\Omega_i$的体积。
|
||
|
||
如果存在数$A$使得对任意给定的$\varepsilon > 0$，有$\delta > 0$，凡是$\norm{T} < \delta$时，不论值点$\xi_i$在子长方体$\Omega_i$中如何选择，都有
|
||
\[\abs{\sum_{i = 1}^{nmk} f(\xi_i) \mu(\Omega_i) - A} < \varepsilon\]
|
||
则称$A$为$f$在$\Omega$上的积分，记作
|
||
\[\iiint \limits_\Omega f(x, y, z) \dif x \dif y \dif z \qquad \text{或者} \qquad \int \limits_\Omega f \dif \mu\eqper\]
|
||
\end{definition}
|
||
|
||
类似二维的情况，我们可以定义零测集与零体积集：
|
||
\begin{definition}[三维零测集]
|
||
若$E \subset \realnum^3$满足：对任意的$\varepsilon > 0$，存在一系列闭长方体$\{\Omega_i\}_{i = 1}^\infty$使得$E \subset \bigcup \limits_{i = 1}^\infty \Omega_i$且$\sum_{i = 1}^\infty \mu(\Omega_i) < \varepsilon$，则称$E$为三维零测集。
|
||
\end{definition}
|
||
|
||
\begin{definition}[零体积集]
|
||
若$E \subset \realnum^3$满足：对任意的$\varepsilon > 0$，存在有限的闭长方体$\{\Omega_i\}_{i = 1}^m$使得$E \subset \bigcup \limits_{i = 1}^m \Omega_i$且$\sum_{i = 1}^m \mu(\Omega_i) < \varepsilon$，则称$E$为零体积集。
|
||
\end{definition}
|
||
|
||
\begin{theorem}[Lebesgue定理]
|
||
对于长方体$\Omega$上的有界函数$f$，积分
|
||
\(\dint \limits_\Omega f \dif \mu\)
|
||
存在的充分必要条件是$f$在$\Omega$上的间断点集是一零测集。
|
||
\end{theorem}
|
||
|
||
\begin{theorem}[Fubini定理]
|
||
设$f \in C(\Omega)$，$\Omega = [a, b] \times [c, d] \times [g, h]$，则
|
||
\begin{align*}
|
||
\int \limits_\Omega f \dif \mu & = \int_a^b \dif x \int_c^d \dif y \int_g^h f(x, y, z) \dif z\\
|
||
& = \int_c^d \dif y \int_a^b \dif x \int_g^h f(x, y, z) \dif z\\
|
||
& = \int_g^h \dif z \int_a^b \dif x \int_c^d f(x, y, z) \dif y\\
|
||
& = \int_a^b \dif x \int_g^h \dif z \int_c^d f(x, y, z) \dif y\\
|
||
& = \int_c^d \dif y \int_g^h \dif z \int_a^b f(x, y, z) \dif x\\
|
||
& = \int_g^h \dif z \int_c^d \dif y \int_a^b f(x, y, z) \dif x\eqper1
|
||
\end{align*}
|
||
\end{theorem}
|
||
|
||
一般有界区域上的三重积分和及三重积分的性质与二重积分完全类似。
|
||
|
||
对一般有界区域上的三重积分，设$\Omega \subset \realnum^3$是有界区域，$f: \Omega \to \realnum$。引入延拓函数
|
||
\[f_\Omega (x, y, z) = \begin{cases}
|
||
f(x, y, z), \quad (x, y, z) \in \Omega\\
|
||
0, \quad (x, y, z) \not \in \Omega
|
||
\end{cases}\]
|
||
记$\Omega_M = [-M, M] \times [-M, M] \times [-M, M] \supset \Omega$。如果$f_\Omega \in R(\Omega_M)$，则称$f \in R(\Omega)$。定义$f$在$\Omega$上的积分值
|
||
\[\int_\Omega f \dif \mu = \int \limits_{\Omega_M} f_\Omega \dif \mu\]
|
||
再记$D_M = [-M, M] \times [-M, M]$，结合Fubini定理可得
|
||
\begin{align*}
|
||
\int \limits_\Omega f \dif \mu & = \iint \limits_{D_M} \dif x \dif y \int_{-M}^M f_\Omega(x, y, z) \dif z\\
|
||
& = \int_{-M}^M \dif z \iint \limits_{D_M} f_\Omega(x, y, z) \dif x \dif y\eqper
|
||
\end{align*}
|
||
|
||
下面我们考虑在一般区域上的三重积分的计算。
|
||
\begin{theorem}
|
||
设有界集$\Omega \subset \realnum^3$有体积，有界的函数$f: \Omega \to \realnum$连续。
|
||
|
||
\begin{enumerate}
|
||
\item 设$\Omega$在$xy$平面上的垂直投影为$D$，且当$(x, y) \in D$时，过这一点且垂直于$D$的直线与$V$交成一个区间$[z_1(x, y), z_2(x, y)]$，即$\Omega$可以表示为
|
||
\[\{(x, y, z) \mid z_1(x, y) \leq z \leq z_2(x, y), (x, y) \in D\}\]
|
||
那么
|
||
\[\int \limits_\Omega f \dif \mu = \iint \limits_D \dif x \dif y \int_{z_1(x, y)}^{z_2(x, y)} f(x, y, z) \dif z\text{；}\]
|
||
\item 设$\Omega$在$z$轴上的垂直投影为区间$J$，且当$z \in J$时，通过点$(0, 0, z)$同时垂直于$z$轴的平面与$\Omega$交成的图形在$xy$平面上的垂直投影是一个有面积的点集$D$，即$\Omega$可以表示为
|
||
\[\{(x, y, z) \mid (x, y) \in \bvec{D}(z), z \in J\}\]
|
||
那么
|
||
\[\int \limits_\Omega f \dif \mu = \int \limits_J \dif z \iint \limits_D f(x, y, z) \dif x \dif y\eqper\]
|
||
\end{enumerate}
|
||
\end{theorem}
|
||
|
||
最后我们仿照二元积分的换元公式，给出三重积分的换元公式。
|
||
\begin{theorem}
|
||
设函数$f: \Omega \to \realnum$连续，映射
|
||
\[\bvec{\varphi}: \begin{cases}
|
||
x = x(u, v, w)\\
|
||
y = y(u, v, w)\\
|
||
z = z(u, v, w)
|
||
\end{cases}, (u, v, w) \in \tilde{\Omega}\]
|
||
是$\tilde{\Omega}$到$\Omega$的一一对应，且$\bvec{\varphi} \in C^1 (\tilde{\Omega}, \realnum^3)$满足
|
||
\[\det J\bvec{\varphi} = \frac{\partial (x, y, z)}{\partial (u, v, w)} = \begin{vmatrix}
|
||
\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w}\\[1em]
|
||
\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w}\\[1em]
|
||
\dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w}\\
|
||
\end{vmatrix} \neq 0\]
|
||
则有
|
||
\[\int \limits_\Omega f \dif \mu = \int \limits_{\tilde{\Omega}} f \circ \bvec{\varphi} \abs{\det J \bvec{\varphi}} \dif \mu\]
|
||
也即
|
||
\[\iiint \limits_\Omega f(x, y, z) \dif x \dif y \dif z = \iiint \limits_{\tilde{\Omega}} f(x(u, v, w), y(u, v, w), z(u, v, w)) \abs{\frac{\partial (x, y, z)}{\partial (u, v, w)}} \dif u \dif v \dif w \eqper\]
|
||
\end{theorem}
|
||
|
||
下面给出这换元公式的几个常见的特例。
|
||
|
||
第一个特例是柱坐标换元。设映射
|
||
\[\bvec{\varphi}: \begin{cases}
|
||
x = r \cos \theta\\
|
||
y = r \sin \theta\\
|
||
z = z
|
||
\end{cases}\]
|
||
这时
|
||
\[\frac{\partial (x, y, z)}{\partial (r, \theta, z)} \begin{vmatrix}
|
||
\cos \theta & \sin \theta & 0\\
|
||
-r \sin \theta & r \cos \theta & 0\\
|
||
0 & 0 & 1
|
||
\end{vmatrix}= r\]
|
||
则
|
||
\[\iiint \limits_\Omega f \dif \mu = \iiint \limits_{\tilde{\Omega}} f(r \cos \theta, r \sin \theta, z) r \dif r \dif \theta \dif z\eqper\]
|
||
|
||
第二个特例是球坐标变换。设映射
|
||
\[\bvec{\varphi}: \begin{cases}
|
||
x = r \sin \theta \cos \varphi\\
|
||
y = r \sin \theta \sin \varphi\\
|
||
z = r \cos \theta
|
||
\end{cases}\]
|
||
这时
|
||
\[\frac{\partial (x, y, z)}{\partial (r, \theta, z)} = r^2 \sin \theta\]
|
||
则
|
||
\[\int \limits_\Omega f \dif \mu = \iiint \limits_{\tilde{\Omega}} f(r \sin \theta \cos \varphi, r \sin \theta \sin \varphi, r \cos \theta)\eqper\] |