二重积分

多变量函数的积分学

张瑞
中国科学技术大学数学科学学院

二重积分

二重积分概念和性质

仿照一维情形下求曲边梯形的面积，可以求解空间中以$Oxy$平面上有界矩形区域$R$为底，$z=f(x,y)$为曲顶的柱体的体积。

  1. 将$Oxy$平面矩形区域$R$分割成互不重叠的小矩阵$R_{ij}$, $i=1,\cdots,n$, $j=1,\cdots,m$。
  2. 在$R_{ij}$内任取一点$(x_{ij}^*, y^*_{ij})$， 则曲顶小柱体的近似体积为
    \[f(x_{ij}^*, y^*_{ij})\Delta R_{ij} \]
    其中$\Delta R_{ij}$为$R_{ij}$的面积
\begin{tikzpicture}[global scale=0.8] %draw the axes \draw[->] (0,0,0) -- (3,0,0) node[anchor=west]{$y$}; \draw[->] (0,0,0) -- (0,2,0) node[anchor=south]{$z$}; \draw[->] (0,0,0) -- (0,0,3) node[anchor=east]{$x$}; % draw domain in plane Oxy \draw[thick] (0,0,0) -- (0,0,2.5) -- (2.5,0,2.5) -- (2.5,0,0) -- cycle; \draw[dashed, red] (0,0,2.0) -- (2.5,0,2.0); \draw[dashed, red] (0,0,1.5) -- (2.5,0,1.5); \draw[dashed, red] (0,0,1.0) -- (2.5,0,1.0); \draw[dashed, red] (0,0,0.5) -- (2.5,0,0.5); \draw[dashed, blue] (2.0,0,0) -- (2.0,0,2.5); \draw[dashed, blue] (1.5,0,0) -- (1.5,0,2.5); \draw[dashed, blue] (1.0,0,0) -- (1.0,0,2.5); \draw[dashed, blue] (0.5,0,0) -- (0.5,0,2.5); % draw the surface % == define coordinate matrix \foreach \x/\z in {0/1.75,1/1.83, 2/1.9, 3/1.85, 4/1.78, 5/1.8}{ \coordinate (p0\x) at (\x*0.5, \z, 0); } \foreach \x/\z in {0/1.70,1/1.83, 2/1.9, 3/1.85, 4/1.78, 5/1.88}{ \coordinate (p1\x) at (\x*0.5, \z, 0.5); } \foreach \x/\z in {0/1.55,1/1.83, 2/1.69, 3/1.85, 4/1.78, 5/1.68}{ \coordinate (p2\x) at (\x*0.5, \z, 1.0); } \foreach \x/\z in {0/1.4, 1/1.43, 2/1.5, 3/1.5, 4/1.53, 5/1.62}{ \coordinate (p3\x) at (\x*0.5, \z, 1.5); } \foreach \x/\z in { 0/1.28 , 1/1.32, 2/1.35, 3/1.32, 4/1.3, 5/1.43}{ \coordinate (p4\x) at (\x*0.5, \z, 2.0); } \foreach \x/\z in {0/1.25, 1/1.28, 2/1.35, 3/1.4, 4/1.39, 5/1.3}{ \coordinate (p5\x) at (\x*0.5, \z, 2.5); } % == draw surface border \draw[thick] plot[smooth] coordinates { (p00) (p10) (p20) (p30) (p40) (p50)}; \draw[thick] plot[smooth] coordinates { (p05) (p15) (p25) (p35) (p45) (p55)}; \draw[thick] plot[smooth] coordinates { (p00) (p01) (p02) (p03) (p04) (p05)}; \draw[thick] plot[smooth] coordinates { (p50) (p51) (p52) (p53) (p54) (p55)}; % == draw curve in surface \draw[dashed, red] plot[smooth] coordinates { (p40) (p41) (p42) (p43) (p44) (p45)}; \draw[dashed, red] plot[smooth] coordinates { (p30) (p31) (p32) (p33) (p34) (p35)}; \draw[dashed, red] plot[smooth] coordinates { (p20) (p21) (p22) (p23) (p24) (p25)}; \draw[dashed, red] plot[smooth] coordinates { (p10) (p11) (p12) (p13) (p14) (p15)}; \draw[dashed, blue] plot[smooth] coordinates { (p01) (p11) (p21) (p31) (p41) (p51)}; \draw[dashed, blue] plot[smooth] coordinates { (p02) (p12) (p22) (p32) (p42) (p52)}; \draw[dashed, blue] plot[smooth] coordinates { (p03) (p13) (p23) (p33) (p43) (p53)}; \draw[dashed, blue] plot[smooth] coordinates { (p04) (p14) (p24) (p34) (p44) (p54)}; % 画一个积分元 \draw[thin] (p42) -- (1.0,0,2.0); \draw[thin] (p43) -- (1.5,0,2.0); \draw[thin] (p33) -- (1.5,0,1.5); \draw[thin] (p32) -- (1.0,0,1.5); \filldraw[fill=red!20, opacity=0.5] (p42)--(p43)--(p33)--(p32)--cycle; \filldraw[fill=yellow!20, opacity=0.5] (1,0,2)--(1.5,0,2)--(1.5,0,1.5)--(1,0,1.5)--cycle; \node at (0,0,2.0) [anchor=east] {$x_i$}; %\node at (2.0,0,0) [anchor=south] {$y_j$}; \node at (1.5,0,2.50) [anchor=north] {$y_j$}; \end{tikzpicture}
  1. 所有这些近似小柱体的体积和为
    \[\sigma=\sum_{i=1}^n\sum_{j=1}^m f(x^*_{ij},y^*_{ij})\Delta R_{ij} \]
  2. 当$R$越分越细时，$\sigma$的极限就是曲顶柱体的体积。

定义 1.
设$R=[a,b]\times[c,d]$，$f(x,y)$是$R$上的一个函数。作$[a,b]$和$[c,d]$的分割：

\[\begin{aligned} T_x: a=x_0<x_1<\cdots<x_n=b, \\ T_y: c=y_0<y_1<\cdots<y_m=d \end{aligned} \]

两组平行直线将$R$分割成$n\times m$个二维子区间

\[R_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j] \]

这样，就得到$R$的一个分割$T$，记为$T=T_x\times T_y$，

$\|T\|=\max\{\|T_x\|,\|T_y\|\}$为分割$T$的宽度。

  • 在$R_{ij}$中任取一点$M_{ij}(x^*_{ij}, y^*_{ij})$，用$M$表示这样一组取值，做Riemann和(或称为积分和)
    \[S(f,T,M)=\sum_{i,j}f(M_{ij})\Delta x_i \Delta y_j %\Delta R_{ij} \]
    其中$\Delta x_i=x_i-x_{i-1}$, $\Delta y_j=y_j-y_{j-1}$, $i=1,\cdots,n$, $j=1,\cdots,m$
  • 若存在实数$A$，使得$\forall \epsilon>0$, $\exists \delta>0$，当$\|T\|<\delta$时，满足
    \[|S(f,T,M)-A|<\epsilon , \forall M_{ij}\in R_{ij} \]
    则称二元函数$f(x,y)$在二维闭区间$R$上Riemann可积，

而

\[A=\lim_{\|T\|\to0}\sum_{i,j}f(M_{ij})\Delta x_i \Delta y_j %\Delta R_{ij} \]

称为$f(x,y)$在$R$上的二重积分，记为

\[\iint\limits_Rf(x,y)dxdy , \mbox{ or } \int_Rf \]

$f(x,y)$称为被积函数，$R$为积分区间，$f(x,y)dxdy$为被积表达式，$dxdy$为面积元。

定义 2.
$f(x,y)$是有界集$D$上的函数，取二维闭区间$R\supset D$，做$f(x,y)$的零延拓

\[f_D(x,y)=\begin{cases} f(x,y), & (x,y)\in D \\ 0 , & (x,y)\notin D \end{cases} \]

若$f_D(x,y)$在$R$上可积，则称f(x,y)在D上可积，

并称$\displaystyle\iint\limits_Rf_D(x,y)dxdy$为f(x,y)在D上的二重积分，记为

\[\iint\limits_Df(x,y)dxdy , \mbox{ or } \int_Df \]

这样定义的二重积分与闭区间$R$的选取无关。

  • 几何上看，$\displaystyle \iint_D f(x,y)dxdy$就是以$D$上的函数$f(x,y)$为顶的曲顶柱体的体积，它与一元定积分是一样的。
  • 物理上看，$f(x,y)$是面积为$D$的薄板的密度函数，则二重积分$\displaystyle \iint_D f(x,y)dxdy$就是薄板的质量。

定义 3.
设$D$是有界的平面点集，如果$D$上取值为$1$的常值函数可积， 则称$D$是有面积的，并称 $\displaystyle\iint\limits_D1dxdy$ 为$D$的面积。

定理 1.
有界平面点集$D$是有面积的$\Leftrightarrow$$\partial D$的面积为$0$。

特别地，由有限条分段光滑曲线围成的区域或闭域是有面积的

定理不做证明。

以后，总假定积分区域$D$是由有限条分段光滑曲线围成的区域。

定理 2.
$D$是由有限条分段光滑曲线围成的区域，$f(x,y)$是$D$上的函数。

  1. 若$f(x,y)$在$D$上可积，则$f(x,y)$在$D$上有界；
  2. 若有界函数$f(x,y)$的不连续点分布在$D$中的有限条光滑曲线上，则$f(x,y)$在$D$上可积；
  3. 若有界函数$f(x,y)$，$g(x,y)$是$D$上的函数，且$f(x,y)\neq g(x,y)$的点分布在有限条光滑曲线上，则$f(x,y)$和$g(x,y)$在$D$上有相同的可积性。当它们可积时，有
\[\int_D f=\int_D g \]

定理 3.
$D$是由有限条分段光滑曲线围成的区域，$f(x,y)$, $g(x,y)$是$D$上的可积函数。

  1. （线性）对任意常数$c_1$, $c_2$ 和$c_1f+c_2g$在$D$上可积，且
    \[\int_D(c_1f+c_2g)=c_1\int_Df+c_2\int_D g \]
  2. （乘积） $f(x,y)g(x,y)$在$D$上可积
  3. （保序性）若在$D$上$f(x,y)\geq g(x,y)$，则$\int_Df\geq\int_D g$
  4. （绝对可积性）$|f(x,y)|$在$D$上可积，且有
    \[|\int_D f|\leq \int_D|f| \]

定理 4.
$D_1$, $D_2$是由有限条分段光滑曲线围成的区域，且$D_1\cap D_2=\emptyset$。 函数$h(x,y)$在$D_1$, $D_2$上都可积。 则$h(x,y)$在$D_1\cup D_2$上可积，且

\[\int_{D_1\cup D_2}h=\int_{D_1}h+\int_{D_2}h \]

定理 5. (积分中值定理)
$f(x,y)$在闭域$\bar D$中连续，则存在$(x_0,y_0)\in D$, 使得

\[\int_{D}f=f(x_0,y_0)S_D \]

其中$S_D$是$D$的面积。

定理 6.
$D$是有面积的平面点集，$f(x,y)$为$D$上的函数， 那么$f(x,y)$在$D$上可积且积分等于$A$的充要条件是，

对$\forall \epsilon>0$, $\exists \delta>0$， 将$D$分割为有限个内部互不相交的有面积的小块 $D_1$,$D_2$,$\cdots$,$D_n$，记

\[\lambda_i=\sup|M_i-N_i|, (M_i,N_i\in D_i) \]

为$D_i$的直径， 只要分割的宽度$\displaystyle\lambda=\max_i\lambda_i$满足$\lambda<\delta$， 对$\forall (x_i,y_i)\in D_i$就有

\[\left|\sum_{i=1}^nf(x_i,y_j)\Delta D_i-A\right|<\epsilon \]
\begin{tikzpicture} %draw the axes \draw[->] (0,0,0) -- (3,0,0) node[anchor=west]{$y$}; \draw[->] (0,0,0) -- (0,2,0) node[anchor=west]{$z$}; \draw[->] (0,0,0) -- (0,0,3) node[anchor=west]{$x$}; % 画曲面和定义域 \draw[dashed] (0.2, -0.4)--(1.4, -0.85)--(2.3,-0.25); \draw[dashed] (0.2, -0.4) .. controls (1,0) .. (2.3,-0.25); \draw[thick] (0.2, 1.4).. controls(0.75, 1.35) ..(1.4, 0.75); \draw[thick] (2.3, 1.25).. controls(1.7, 1.2)..(1.4, 0.75); \draw[thick] (0.2, 1.4) .. controls (1,2) .. (2.3,1.25); \draw[dashed] (0.2,1.4)--(0.2,-0.4); \draw[dashed] (2.3, 1.25)--(2.3, -0.25); \draw[dashed] (1.4, 0.75)--(1.4, -0.85); % 画小单元 \coordinate (a1) at (0.7, -0.4); \coordinate (a2) at (0.9, -0.6); \coordinate (a3) at (1.2, -0.35); \coordinate (c1) at (1.0, -0.3); \draw[dotted, red] (a1)--(a2)--(a3).. controls (c1) .. (a1); \coordinate (b1) at (0.7, 1.4); \coordinate (b2) at (0.9, 1.3); \coordinate (b3) at (1.2, 1.35); \coordinate (d1) at (0.8, 1.4); \coordinate (d2) at (1.1, 1.37); \coordinate (d3) at (1.0, 1.6); \filldraw[thick, red, draw=black] (b1).. controls (d1) ..(b2).. controls (d2) ..(b3).. controls (d3) .. (b1); \draw[dashed] (a1)--(b1); \draw[dashed] (a2)--(b2); \draw[dashed] (a3)--(b3); \end{tikzpicture}

二重积分的累次积分法

函数$f(x,y)$在二维闭区间$R=[a,b]\times[c,d]$上可积。 把二重积分$\displaystyle \int_R f$看做是以$R$为底、 $z=f(x,y)$为顶的曲顶柱体的体积，则

  • 这个体积也可以用截面积的积分来计算。
  • 若用垂直于$x$轴的平面截柱体得到的截面积为$A(x)$， 则柱体的体积为
    \[V= \int_a^b A(x)dx \]
\begin{tikzpicture}[global scale=0.8] %draw the axes \draw[->] (0,0,0) -- (3,0,0) node[anchor=west]{$y$}; \draw[->] (0,0,0) -- (0,2,0) node[anchor=south]{$z$}; \draw[->] (0,0,0) -- (0,0,3.5);% node[anchor=east]{$x$}; % draw domain in plane Oxy \draw[thick] (0,0,0) -- (0,0,2.5) -- (2.5,0,2.5) -- (2.5,0,0) -- cycle; \draw[dashed, red] (0,0,1.5) -- (2.5,0,1.5); % draw the surface % == define coordinate matrix \foreach \x/\z in {0/1.75,1/1.83, 2/1.9, 3/1.85, 4/1.78, 5/1.8}{ \coordinate (p0\x) at (\x*0.5, \z, 0); } \foreach \x/\z in {0/1.70,1/1.83, 2/1.9, 3/1.85, 4/1.78, 5/1.88}{ \coordinate (p1\x) at (\x*0.5, \z, 0.5); } \foreach \x/\z in {0/1.55,1/1.83, 2/1.69, 3/1.85, 4/1.78, 5/1.68}{ \coordinate (p2\x) at (\x*0.5, \z, 1.0); } \foreach \x/\z in {0/1.4, 1/1.43, 2/1.5, 3/1.5, 4/1.53, 5/1.62}{ \coordinate (p3\x) at (\x*0.5, \z, 1.5); } \foreach \x/\z in { 0/1.28 , 1/1.32, 2/1.35, 3/1.32, 4/1.3, 5/1.43}{ \coordinate (p4\x) at (\x*0.5, \z, 2.0); } \foreach \x/\z in {0/1.25, 1/1.28, 2/1.35, 3/1.4, 4/1.39, 5/1.3}{ \coordinate (p5\x) at (\x*0.5, \z, 2.5); } % == draw surface border \draw[thick] plot[smooth] coordinates { (p00) (p10) (p20) (p30) (p40) (p50)}; \draw[thick] plot[smooth] coordinates { (p05) (p15) (p25) (p35) (p45) (p55)}; \draw[thick] plot[smooth] coordinates { (p00) (p01) (p02) (p03) (p04) (p05)}; \draw[thick] plot[smooth] coordinates { (p50) (p51) (p52) (p53) (p54) (p55)}; % == draw curve in surface \draw[dashed, red] plot[smooth] coordinates { (p30) (p31) (p32) (p33) (p34) (p35)}; \draw[thick] (0,0,0) -- (p00) (0,0,2.5) -- (p50) (2.5,0,2.5) -- (p55) (2.5,0,0) -- (p05); \draw[thin] (0,0,1.5)--(p30); \draw[thin] (2.5,0,1.5)--(p35); \fill[fill=red!20, opacity=0.5] (0,0,1.5)-- plot[smooth] coordinates { (p30) (p31) (p32) (p33) (p34) (p35)} -- (2.5,0,1.5) -- node[above, opacity=1] {$A(x)$} cycle; \node at (0,0,1.5) [anchor=north] {$x$}; \node at (0,0,2.5) [anchor=north] {$b$}; \end{tikzpicture}

可以得到

\[A(x)=\int_c^df(x,y)dy \]

这样，柱体的体积用截面积的积分表示为

\[\begin{aligned} V=&\iint\limits_Rf(x,y)dxdy=\int_a^bA(x)dx \\ =&\int_a^b\left[\int_c^df(x,y)dy\right]dx \end{aligned} \]

定理 7. (Fubini定理)
函数$f(x,y)$在二维闭区间$R=[a,b]\times[c,d]$上可积。

  1. 如果对每个$y\in[c,d]$，$f(x,y)$作为$x$的函数在$[a,b]$上可积，记 $\phi(y)=\int_a^bf(x,y)dx$ 则$\phi(y)$在$[c,d]$上可积，且有
    \[\int_c^d\phi(y)dy=\int_c^d\left[\int_a^bf(x,y)dx\right]dy=\iint\limits_Rf(x,y)dxdy \]
  2. 如果对每个$x\in[a,b]$，$f(x,y)$作为$y$的函数在$[c,d]$上可积，记 $\psi(x)=\int_c^df(x,y)dy$ 则$\psi(x)$在$[a,b]$上可积，且有
    \[\int_a^b\psi(x)dx=\int_a^b\left[\int_c^df(x,y)dy\right]dx=\iint\limits_Rf(x,y)dxdy \]

证明： 由函数$f(x,y)$在$R$上可积，则 $\color{red}\forall\epsilon>0$，由

\[A=\lim_{|T|\to0}\sum_{i,j=1}^{n,m}f(M_{ij})\Delta x_i\Delta y_j \]

存在$\color{red}\delta>0$，当$|T|<\delta$时，对任意$(\xi_i,\eta_j)\in R_{ij}$成立

\[{\color{blue}(*)} \quad A-\frac{\epsilon}2<\sum_{j=1}^m\Delta y_j\left[\sum_{i=1}^nf(\xi_i,\eta_j)\Delta x_i\right]<A+\frac{\epsilon}2 \]

注意到，对于$\eta_j\in[c,d]$， $\displaystyle\sum_{i=1}^nf(\xi_i,\eta_j)\Delta x_i$是 $f(x,\eta_j)$在$[a,b]$上的Riemann和。

若对每个y，f(x,y)作为x的函数在[a,b]上可积， 记积分值为$\phi(y)=\int_a^bf(x,y)dx$，则

\[\lim_{|T_x|\to0}\sum_{i=1}^nf(\xi_i,\eta_j)\Delta x_i=\int_a^b f(x,\eta_j)dx=\phi(\eta_j) \]

对${\color{blue}(*)}$式取极限$|T_x|\to0$，则有

\[{\color{red}A-\epsilon<}A-\frac{\epsilon}2 \leq {\color{red}\sum_{j=1}^m\phi(\eta_j)\Delta y_j} \leq A+\frac{\epsilon}2{\color{red}<A+\epsilon} \]

对$\color{red}|T_y|<\delta$，$\color{red}\forall \eta_j\in D_{ij}$成立。

因此，$\phi(y)$在$[c,d]$上可积，且积分为$A$。

定理表明，二维区域上函数的二重积分，可化为先对一个变量的积分，再对另一个变量的积分。 这种积分过程称为累次积分。

几何上看，面包的体积，可以分为切片面包的体积来表示。

物理上看，薄板的质量，可以分为一些细长条来计算。

例 1. (例7.1.1) 计算二重积分

\[\iint\limits_D\frac1{(x+y)^2}dxdy , D=[3,4]\times[1,2] \]
\[\begin{aligned} \iint_D \frac1{(x+y)^2}dxdy =&\int_3^4\left[\int_1^2\frac1{(x+y)^2}dy\right]dx \\ \end{aligned} \]
  • $R=[a,b]\times[c,d]$，则
    \[\begin{aligned} \iint\limits_RX(x)Y(y)dxdy =&\int_c^d\left( \int_a^bX(x) Y(y) dx \right) dy \\ =&\int_c^d Y(y) \left( \int_a^bX(x) dx \right) dy \\ =&\int_c^d Y(y)dy \int_a^b X(x)dx \end{aligned} \]
  • 若$F''_{xy}(x,y)=f(x,y)$，则记$F'_y(x,y)=g(x,y)$，则
    \[g'_x(x,y)=F''_{xy}(x,y)=f(x,y) \]
    从而 $\begin{cases}\displaystyle\int_a^b f(x,y)dx = g(b,y)-g(a,y), \\\displaystyle\int_c^d g(x,y)dx = F(x,d)-F(x,c),\end{cases}$ 有
    \[\begin{aligned} \int_a^b\int_c^df(x,y)dxdy =&\int_c^d \left(\int_a^b f(x,y) dx\right) dy \\ =&\int_c^d \left( g(b,y)-g(a,y)\right) dy \\ =&F(b,d)-F(b,c)-F(a,d)+F(a,c) \end{aligned} \]

例 2. $f(x)$为闭区间$[a,b]$上的连续函数，则有

\[\left(\int_a^bf(x)dx\right)^2\leq(b-a)\int_a^bf^2(x)dx \]

等号成立当且仅当$f(x)$为常数

积分区域是曲边的情形

I型区域$D$是由曲线$y=\phi_1(x)$, $y=\phi_2(x)$和直线$x=a$, $x=b$围成 的区域，即

\[D=\{(x,y)|\phi_1(x)\leq y\leq\phi_2(x), a\leq x\leq b\} \]
%I型区域 \begin{tikzpicture}[scale=1.5] \draw[->] (-0.50,0) -- (3,0) node[anchor=west]{$x$}; \draw[->] (0,-0.50) -- (0,2) node[anchor=south]{$y$}; \node at (0,0) [anchor=north east] {$O$}; %\filldraw[fill=black!20, opacity=0.5] (0.5, 0.2)--(2.65, 0.2)--(2.65, 1.9)--(0.5, 1.9)--cycle; %\node at ( 0, 0.2) [anchor=east] {$c$}; %\node at ( 0, 1.9) [anchor=east] {$d$}; %\draw[dashed] (0,0.2)--(0.5,0.2); %\draw[dashed] (0,1.9)--(0.5,1.9); \coordinate (a1) at (0.5, 1.3); \coordinate (a2) at (0.5, 0.7); \coordinate (a3) at (0.5, 0.0); \coordinate (b1) at (2.65, 1.3); \coordinate (b2) at (2.65, 0.7); \coordinate (b3) at (2.65, 0.0); \draw[dashed] (a3) -- (a1); \draw[dashed] (b3) -- (b1); \draw[fill=black!20] (a1) .. controls (1.0, 2.0) .. (b1) -- (b2) .. controls (1.0, 0.5) .. (a2) -- cycle; \draw (a1) .. controls (1.0, 2.0) .. (b1); \draw (a2) .. controls (1.0, 0.5) .. (b2); \node at (0.5, 0) [anchor=north] {$a$}; \node at (2.65, 0) [anchor=north] {$b$}; \node at (1.8, 1.0) {$D$}; \node at (1.50, 0.3) {$y=\phi_1(x)$}; \node at (2.0, 2.1) [anchor=north] {$y=\phi_2(x)$}; \end{tikzpicture}

定理 8.
I型区域$D$

\[D=\{(x,y)|\phi_1(x)\leq y\leq\phi_2(x), a\leq x\leq b\} \]

其中$\phi_1(x)$, $\phi_2(x)$为连续函数。$f(x,y)$在$D$上可积， 且对于$\forall x\in[a,b]$，积分 $\displaystyle\int_{\phi_1(x)}^{\phi_2(x)}f(x,y)dy$ 存在，则

\[\iint\limits_Df(x,y)dxdy=\int_a^b\left[\int_{\phi_1(x)}^{\phi_2(x)}f(x,y)dy\right]dx \]

证明：

\begin{tikzpicture}[scale=1.5] \draw[->] (-0.50,0) -- (3,0) node[anchor=west]{$x$}; \draw[->] (0,-0.50) -- (0,2) node[anchor=south]{$y$}; \node at (0,0) [anchor=north east] {$O$}; \filldraw[fill=black!20, opacity=0.5] (0.5, 0.2)--(2.65, 0.2)--(2.65, 1.9)--(0.5, 1.9)--cycle; \node at ( 0, 0.2) [anchor=east] {$c$}; \node at ( 0, 1.9) [anchor=east] {$d$}; \draw[dashed] (0,0.2)--(0.5,0.2); \draw[dashed] (0,1.9)--(0.5,1.9); \coordinate (a1) at (0.5, 1.3); \coordinate (a2) at (0.5, 0.7); \coordinate (a3) at (0.5, 0.0); \coordinate (b1) at (2.65, 1.3); \coordinate (b2) at (2.65, 0.7); \coordinate (b3) at (2.65, 0.0); \draw[dashed] (a3) -- (a1); \draw[dashed] (b3) -- (b1); \draw (a1) .. controls (1.0, 2.0) .. (b1); \draw (a2) .. controls (1.0, 0.5) .. (b2); \node at (0.5, 0) [anchor=north] {$a$}; \node at (2.65, 0) [anchor=north] {$b$}; \node at (1.8, 1.0) {$D$}; \node at (1.50, 0.3) {$y=\phi_1(x)$}; \node at (2.0, 2.3) [anchor=north] {$y=\phi_2(x)$}; \end{tikzpicture}

定理 9.
II型区域

\[D=\{(x,y)|\psi_1(y)\leq x\leq\psi_2(y), c\leq y\leq d\} \]

其中$\psi_1(y)$, $\psi_2(y)$为连续函数。$f(x,y)$在$D$上可积， 且对于$\forall y\in[c,d]$，积分 $\displaystyle\int_{\psi_1(y)}^{\psi_2(y)}f(x,y)dx$ 存在，则

\[\iint\limits_Df(x,y)dxdy=\int_c^d\left[\int_{\psi_1(y)}^{\psi_2(y)}f(x,y)dx\right]dy \]
\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.8] \draw[->] (-0.10,0) -- (3,0) node[anchor=west]{$x$}; \draw[->] (0,-0.10) -- (0,2.1) node[anchor=south]{$y$}; \node at (0,0) [anchor=north east] {$O$}; \coordinate (a1) at (2.8, 0.35); \coordinate (a2) at (1.15, 0.35); \coordinate (a3) at (0.0, 0.35); \coordinate (b1) at (2.65, 1.93); \coordinate (b2) at (1.145, 1.93); \coordinate (b3) at (0.0, 1.93); \draw[dashed] (a3) -- (a1); \draw[dashed] (b3) -- (b1); \draw (a1) .. controls (2.0, 1.4) .. (b1); \draw (a2) .. controls (1.5, 1.5) .. (b2); \node at (a3) [anchor=east] {$c$}; \node at (b3) [anchor=east] {$d$}; \node at (1.8, 1.0) {$D$}; \node at (1.25, 1.35) [anchor=east] {$x=\psi_1(y)$}; \node at (2.3, 1.3) [anchor=west] {$x=\psi_2(y)$}; \end{tikzpicture}

例 3. (例7.1.3) 计算累次积分

\[\int_0^1dx\int_x^{\sqrt x}\frac{\sin y}y dy \]
\begin{tikzpicture}[x=4cm, y=4cm, global scale=0.7] \draw[->] (-0.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.1) node[above] {$y$}; \draw[blue] plot[domain=0:1] (\x, {sqrt(\x)}); \draw[blue] plot[domain=0:1] (\x, \x); \draw[dashed, red] (0.5,1) -- (0.5,0) node[below] {$x_0$}; \draw[dashed, blue] (0.5,0.5) -- (0, 0.5) node[left] {$x_0$}; \draw[dashed, blue] (0.5,{sqrt(0.5)}) -- (0, {sqrt(0.5)}) node[left] {$\sqrt{x_0}$}; \end{tikzpicture}

例 4. 画出计算积分区域，改写计算顺序

\[\int_0^2dy\int_{\frac{y}2}^yf(x,y)dx+\int_2^4dy\int_{\frac{y}2}^2f(x,y)dx \]
\begin{tikzpicture}[global scale=0.8] \draw[->] (-0.1,0) -- (2.4,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,4.4) node[above] {$y$}; \draw[red] plot[domain=0:2] ({0.5*\x}, \x); \draw[red] plot[domain=0:2] (\x, \x); \draw[blue] plot[domain=2:4] ({0.5*\x}, \x); \draw[blue] plot[domain=2:4] (2, \x); \draw[dashed] (2,2) -- (2,0) node[below] {$2$}; \draw[dashed] (2,4) -- (0,4) node[left] {$4$}; \draw[dashed, red] (2,1) -- (0,1) node[left] {$y_0$}; \draw[dashed, red] (1,1) -- (1,0) node[below] {$y_0$}; \draw[dashed, red] (0.5,1) -- (0.5,0) node[below] {$\frac{y_0}2$}; \end{tikzpicture}

例 5. 交换积分顺序

\[\int_0^{\frac{\pi}2}d\phi \int_0^{a\sqrt{\sin(2\phi)}} f(\phi,\rho)d\rho,\, a>0 \]
\begin{tikzpicture}[scale=1.8] \draw[->] (-0.1,0) -- (1.8,0) node[right] {$\phi$}; \draw[->] (0,-0.1) -- (0,1.1) node[above] {$\rho$}; \draw[blue] plot[domain=0:0.5*pi] (\x, {sqrt(sin(2*\x r))}); \draw[dashed, red] (1.6, 1) -- (0,1) ; % node[left] {$\rho_0$}; \draw[dashed, red] (1.6, 0.65) -- (0,0.65) node[left] {$\rho$}; %\draw[dashed, red] (1,1) -- (1,0) node[below] {$y_0$}; %\draw[dashed, red] (0.5,1) -- (0.5,0) node[below] {$\frac{y_0}2$}; \end{tikzpicture}
\[\int_0^a d\rho \int_{\frac12\arcsin\frac{\rho^2}{a^2}}^{\frac{\pi}2-\frac12\arcsin\frac{\rho^2}{a^2}}f(\phi, \rho)d\phi \]

例 6. 画出计算积分区域，改写计算顺序

\[\int_{-1}^0dy\int_{-2\sqrt{y+1}}^{2\sqrt{y+1}}f(x,y)dx+\int_0^8dy\int_{-2\sqrt{1+y}}^{2-y}f(x,y)dx \]

 

例 7. (例7.1.5) 计算由两个圆柱面$x^2+y^2=a^2$与$x^2+z^2=a^2$所围成的立体的体积

\begin{tikzpicture} [scale=0.9] \begin{axis}[axis lines=middle,xlabel=$x$, axis equal=true, %unit vector ratio=1 1, ylabel=$y$, samples=40, xmax=2.1, ymax=8.2] \addplot[domain=-2:2, red!90!black] {x*x/4-1}; %\addplot[domain=0:2, red!90!black] {x}; \addplot[domain=-6:-2, blue!90!black] {x*x/4-1}; \addplot[domain=-6:2, blue!90!black] {2-x}; \end{axis} \end{tikzpicture}
% 两个柱面相交 \begin{tikzpicture}[x=(215:2em/sqrt 2), y=(0:2em), z=(90:2em), thick,font=\small] %coordinates \draw [-stealth, black!75, thin] (0,0,0) -- (2.6,0,0) node [above] {$x$}; \draw [-stealth, black!75, thin] (0,0,0) -- (0,3.5,0) node [right] {$y$}; \draw [-stealth, black!75, thin] (0,0,0) -- (0,0,3.5) node [left] {$z$}; \def\R{1.5} %cyclinders \draw[densely dashed] (\R,0,0)--(\R,0,2*\R)--(0,0,2*\R)--(0,\R,2*\R)--(0,\R,\R)--(0,2*\R,\R)--(0,2*\R,0)--(\R,2*\R,0)--cycle; \draw[densely dashed] plot[variable=\t,domain=0:pi*0.5,smooth]({\R*cos(\t r)},{\R*sin(\t r)},2*\R); \draw[densely dashed] plot[variable=\t,domain=0:pi*0.5,smooth]({\R*cos(\t r)},2*\R,{\R*sin(\t r)}); \draw[densely dashed] plot[variable=\t,domain=0:pi*0.5,smooth]({\R*cos(\t r)},{\R*sin(\t r)},0)-- (0,\R,\R)-- plot[variable=\t,domain=0:pi*0.5,smooth]({\R*sin(\t r)},0,{\R*cos(\t r)}); \node[above right] at(0,{\R/2},{2*\R}) {$x^2+y^2=a^2$}; \node[below] at ({\R},{2*\R},0) {$x^2+z^2=a^2$}; %domain1 \draw[thick,red](\R,0,0)--(0,0,0)--(0,\R,0); %intersection \draw[thick] plot[variable=\t,domain=0:pi*0.5,smooth]({\R*cos(\t r)},{\R*sin(\t r)},{\R*sin(\t r)})-- (0,0,\R)-- plot[variable=\t,domain=0:pi*0.5,smooth]({\R*sin(\t r)},0,{\R*cos(\t r)})-- plot[variable=\t,domain=0:pi*0.5,smooth]({\R*cos(\t r)},{\R*sin(\t r)},0)--(0,\R,\R); % \draw[<-] ({\R/2},{\R/2},{\R}) to[out=60,in=185] (0,{\R*1.5},{\R*1.5}) node[right] {$z=\sqrt{a^2-x^2}$}; %domain2 \draw[thick,red] plot[variable=\t,domain=0:pi*0.5,smooth] ({\R*cos(\t r)},{\R*sin(\t r)},0); \node[red,font=\footnotesize] at({0.5*\R*cos(40)},{0.5*\R*sin(40)},0) {$D$}; \end{tikzpicture}
\[\begin{aligned} V=&8\iint\limits_{\substack{x^2+y^2\leq a^2, \\ x\geq0, y\geq0}}\sqrt{a^2-x^2}dxdy \\ =&8\int_0^a\int_0^{\sqrt{a^2-x^2}}\sqrt{a^2-x^2}dydx \\ =&8\int_0^a(a^2-x^2)dx \end{aligned} \]

int2d-ex-7-1-5 int2d-ex-7-1-5-2

极坐标代换

例 8. 计算积分

\[\iint\limits_D\sqrt{x^2+y^2}dxdy \]

其中$D$是以原点为圆心的单位圆盘$x^2+y^2\leq 1$
\begin{tikzpicture}[thick,font=\small,scale=1.3,inner sep=2pt] \def\a{1.0} \def\b{0.3} \pgfmathsetmacro{\c}{(\a)^2} %Dxy %\uncover<2->{ \draw [blue,thick,fill=black!35, opacity=0.7] (0,0) ellipse ({\a} and {\b}); \draw [blue,thin,densely dashed] (-\a,0)--(-\a,\c); \draw [blue,thin,densely dashed] (\a,0)--(\a,\c); \node[below right,blue,font={\footnotesize}] at(-30:0.6) {$D$}; %} %coordinates \draw [-stealth,black!75,thin] (-1.1,0) -- (1.1,0) node [right] {$y$}; \draw [-stealth,black!75,thin] (0,0) -- (0,{\c+0.6}) node [left] {$z$}; \draw [-stealth,black!75,thin] (35:0.1) -- (215:1) node [left] {$x$}; %suface \draw [right color=black!80, left color=white, opacity=0.7] plot[variable=\t,domain=-\a:\a,smooth] (\t,{abs(\t)})-- (\a,\c) arc (0:-180:{\a} and {\b}); %\draw [top color=white, bottom color=black!80, opacity=0.7] (0,\c) ellipse ({\a} and {\b}); \draw[opacity=0.7] (0,\c) ellipse ({\a} and {\b}); %\node[right] at(0,\c) {$z\!=\!1$}; \node[right] at(0.5,{0.5*\c}) {$z\!=\!\sqrt{x^2+y^2}$}; \end{tikzpicture}

解. 用累次积分来解

\[\begin{aligned} \iint\limits_D\sqrt{x^2+y^2}dxdy=4\int_0^1dx\int_0^{\sqrt{1-x^2}}\sqrt{x^2+y^2}dy \\ %=2\int_0^1[y\sqrt{x^2+y^2}+x^2\ln(y+\sqrt{x^2+y^2})]|_0^{\sqrt{1-x^2}}dx \\ %=2\int_0^1(\sqrt{1-x^2}+x^2\ln\frac{1+\sqrt{1-x^2}}x)dx \end{aligned} \]
\[\begin{aligned} =&4\int_0^1dx\int_0^{\sqrt{1-x^2}}\sqrt{x^2+y^2}dy \\ =&2\int_0^1\left(\left.\left[y\sqrt{x^2+y^2}+x^2\ln(y+\sqrt{x^2+y^2})\right]\right|_{y=0}^{\sqrt{1-x^2}}\right)dx \\ =&2\int_0^1\left(\sqrt{1-x^2}+x^2\ln\frac{1+\sqrt{1-x^2}}x\right)dx \end{aligned} \]

不容易求解

\[\begin{aligned} &\int_0^1 x^2 \ln\frac{1+\sqrt{1-x^2}}x dx \\ =&\int_0^1 \ln\frac{1+\sqrt{1-x^2}}x d\frac{x^3}3 \\ =&\frac{x^3}3\ln\frac{1+\sqrt{1-x^2}}x\bigg|_0^1 +\int_0^1 \frac{x^3}3\frac{-1}{x\sqrt{1-x^2}}dx \\ =&\frac13\int_0^{\frac{\pi}2}\frac{-\sin^2t}{\cos t}\cos t dt =-\frac13\cdot\frac{\pi}4 \end{aligned} \]

换用极坐标： 对区域圆$D$的弧度和径向做分割

\[\begin{aligned} T_r: &0=r_0<r_1<\cdots<r_n=1 \\ T_\theta: &0=\theta_0<\theta_1<\cdots<\theta_m=2\pi \end{aligned} \]
\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.7] \draw[->] (-1.5,0) -- (1.5,0) node[anchor=west]{$x$}; \draw[->] (0,-1.5) -- (0,1.5) node[anchor=south]{$y$}; %\node at (0,0) [anchor=north east] {$O$}; \foreach \r in {1,...,5}{ \draw[green, dashed] (0,0) circle [radius=\r*0.25]; } \foreach \r in {1,...,19}{ \draw[dashed, blue, rotate=\r*20] (0,0)--(1.4,0); } %\draw[green, thick] (60:0.75) arc [start angle=60, end angle=30, radius=0.75]; \filldraw[black!30, draw=green] (40:1) arc [start angle=40, end angle=60, radius=1] -- (60:0.75) arc [start angle=60, end angle=40, radius=0.75]; \draw[-latex] (50:1.5) node [above]{$rd\theta$} --(50:1.02); \draw[-latex] (73:1.4) node [above]{$dr$} --(62:0.82); \node[red] at (40:0.75) [below] {$M_{ij}$}; \fill (40:0.75) circle (1pt); \end{tikzpicture}

则每一小块的面积

\[\sigma(D_{ij})\approx r_i\Delta r_i\Delta \theta_i \]

Riemann和为

\[\begin{aligned} S(T)=&\sum_{i,j=1}^{m,n}f(M_{ij})\sigma(D_{ij}) \\ %S(T)=&\sum_{i,j=1}^{m,n}(\sqrt{x^2+y^2}\big|_{M_{ij}})\sigma(D_{ij}) \\ \approx & \sum_{i,j=0}^{m,n}f(M_{ij})r_i\Delta r_i\Delta \theta_j %\approx & \sum_{i,j=0}^{m,n}r^2_i\Delta r_i\Delta \theta_j \end{aligned} \]

可以取$M_{ij}=(r_i,\theta_j)\in D_{ij}$

Riemann和为

\[\begin{aligned} S(T) =& \sum_{i,j=1}^{m,n}r_i\cdot r_i\Delta r_i\Delta \theta_j =& \sum_{i,j=1}^{m,n}r^2_i\Delta r_i\Delta \theta_j %=&2\pi \sum_{i=0}^n r^2_i\Delta r_i \end{aligned} \]

取极限后，有

\[\begin{aligned} \iint_D \sqrt{x^2+y^2}dxdy =&\int_0^{2\pi}\int_0^1 r^2drd\theta \\ =&2\pi\int_0^1 r^2dr=\frac{2\pi}3 \end{aligned} \]
% 极坐标下的二重积分 \usetikzlibrary{snakes} \begin{tikzpicture} [global scale=0.7] %coordinates \draw[-stealth,thin,gray] (0,0)--(5.8,0) node[right,font={\small}] {$x$}; \draw[-stealth,thin,gray] (0,0)--(0,5.2) node[below right,font={\small}] {$y$}; %domain \draw[fill=gray!30] (1,1) to[bend left=105] (4,3.2) to[bend right=-75] node[pos=0.1,below right,font={\small}] {$D$} (1,1); %partition %\visible<2->{ \foreach \b in {10,20,...,70}{ \draw[densely dashed,gray,thin] (0,0)--(\b:5.5); } %} %\visible<3->{ \foreach \t in {0.5,1,...,5.5}{ \draw[densely dashed,gray,thin] (10:\t) arc (10:70:\t); } %} %one piece %\visible<4->{ \draw[red,thick,fill=blue!30,opacity=0.6] (40:3)--(40:3.5)-- (40:3.5) arc (40:50:3.5)-- (50:3.5)--(50:3)-- (50:3) arc (50:40:3)--cycle; \node[font={\footnotesize}] at (45:3.25) {${\Delta\sigma}$}; %} %\visible<7->{ \draw[blue,thick,snake=brace,mirror snake,raise snake=2pt](10:0.05)--(10:2.95) node[pos=0.5,below=0.1,font={\small}] {$r$}; \draw[blue,thick,snake=brace,mirror snake,raise snake=2pt](10:3)--(10:3.5) node[pos=0.5,below=0.1,font={\small}] {$\Delta r$}; \draw[red,thick,snake=brace,raise snake=3pt](50:3)--(50:3.5) node[pos=0.5,above left,font={\footnotesize}] {$\Delta r$}; %} %\visible<8->{ \draw[blue,thick,<->](0:5.5) arc (0:39.5:5.5); \node[blue,right,font={\small}] at (20:5.5) {$\theta$}; \draw[blue,thick,<->](40:5.5) arc (40:50:5.5); \node[blue,right,font={\small}] at (47:5.5) {$\Delta\theta$}; % some bug in old tikz %} %\visible<9->{ \draw[red,thick,snake=brace,raise snake=2pt](40:3)--(50:3) node[pos=0.5,below left,font={\footnotesize}] {$r\Delta\theta$}; %} \end{tikzpicture}

对于函数

\[f(x,y), (x,y)\in D \]

取极坐标变换

\[\begin{cases} x=r\cos\theta \\ y=r\sin\theta \end{cases}, (r,\theta)\in D' \]

则有

\[\iint_D f(x,y)dxdy = \iint_{D'} f(r\cos\theta,r\sin\theta) r dr d\theta \]
% 极坐标下的二重积分 %\usetikzlibrary{snakes} \begin{tikzpicture} %coordinates \draw[-stealth,thin,gray] (0,0)--(5.8,0) node[right,font={\small}] {$x$}; \draw[-stealth,thin,gray] (0,0)--(0,5.2) node[below right,font={\small}] {$y$}; %domain \draw[fill=gray!30] (1,1) to[bend left=105] (4,3.2) to[bend right=-75] node[pos=0.1,below right,font={\small}] {$D$} (1,1); %partition %\visible<2->{ \foreach \b in {10,20,...,70}{ \draw[densely dashed,gray,thin] (0,0)--(\b:5.5); %node[above right,font={\tiny}] {$\theta=\b ^\circ$}; } %} %\visible<3->{ \foreach \t in {0.5,1,...,5.5}{ \draw[densely dashed,gray,thin] (10:\t) arc (10:70:\t); %node[left,font={\tiny}] {$\rho=\t$}; } %} %one piece %\visible<4->{ \draw[red,thick,fill=blue!30,opacity=0.6] (40:3)--(40:3.5)-- (40:3.5) arc (40:50:3.5)-- (50:3.5)--(50:3)-- (50:3) arc (50:40:3)--cycle; \node[font={\footnotesize}] at (45:3.25) {${\Delta\sigma}$}; %} %\visible<7->{ \draw[blue,thick,snake=brace,mirror snake,raise snake=2pt](10:0.05)--(10:2.95) node[pos=0.5,below,font={\small}] {$\rho$}; \draw[blue,thick,snake=brace,mirror snake,raise snake=2pt](10:3)--(10:3.5) node[pos=0.5,below,font={\small}] {$\Delta\rho$}; \draw[red,thick,snake=brace,raise snake=3pt](50:3)--(50:3.5) node[pos=0.5,above left,font={\footnotesize}] {$\Delta\rho$}; %} %\visible<8->{ \draw[blue,thick,<->](0:5.5) arc (0:39.5:5.5); \node[blue,right,font={\small}] at (20:5.5) {$\theta$}; % some bug in old tikz \draw[blue,thick,<->](40:5.5) arc (40:50:5.5);% node[right,font={\footnotesize}] {$\Delta\theta$}; \node[blue,right,font={\small}] at (47:5.5) {$\Delta\theta$}; % some bug in old tikz %} %\visible<9->{ \draw[red,thick,snake=brace,raise snake=2pt](40:3)--(50:3) node[pos=0.5,below left,font={\footnotesize}] {$\rho\Delta\theta$}; %} \end{tikzpicture}
\begin{tikzpicture}[thick,font={\small}, global scale=0.6] %coordinates \draw[-stealth,thin,gray] (0,0)--(5.5,0);% node[right] {$x$}; \draw[-stealth,thin,gray] (0,0)--(0,4.8);% node[right] {$y$}; %domain \draw[fill=gray!30] (1,1) to[bend left=105] (4,3.2) to[bend right=-75] (1,1); \node[right] at (2.5,2){$D$}; %compute the domain \path[name path=curve 1,draw=none] (1,1) to[bend left=105](4,3.2); \path[name path=curve 2,draw=none] (4,3.2) to[bend right=-75](1,1); \def\a{64.7} \def\b{19.7} %\visible<2->{ \draw (0,0) -- (\a:5) node[below left=0pt and 4pt]{$\theta=\beta$}; \draw[fill] (\a:2.05) circle(0.8pt); \draw (0,0) -- (\b:5) node[below=4pt]{$\theta=\alpha$}; \draw[fill] (\b:2.56) circle(0.8pt); %} %\visible<3->{ \draw[name path=line](0,0)--(43:6); \draw[->,color=blue](0:0.8) arc (0:43:0.8); \node[color=blue] at (22:0.5) {$\theta$}; % some bug in old tikz \fill [red,name intersections={of=line and curve 1}] (intersection-1) circle (1.5pt) node[right]{${r=\phi_2(\theta)}$}; \fill [red,name intersections={of=line and curve 2}] (intersection-1) circle (1.5pt) node[right]{${r=\phi_1(\theta)}$}; %} \end{tikzpicture}

用极坐标表示积分区域，可以

  • 先求$\theta$的范围
  • 再求$r$的函数

$\begin{aligned}D=\Big\{(r,\theta)\mid & {\alpha\leq\theta\leq\beta},\, \\& {{\phi_{1}(\theta)\leq r \leq\phi_{2}(\theta)}}\Big\}\end{aligned}$

\[\iint_D f(x,y)dxdy =\int_\alpha^\beta\left[ \int_{\phi_1(\theta)}^{\phi_2(\theta)} f(r\cos\theta, r\sin\theta) r dr\right] d\theta \]

例 9. 求

\[\iint\limits_{x^2+y^2\leq x}f(\frac{y}x)dxdy \]
\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.5] \draw[->] (0,-1.1)--(0,1.1); \draw[->] (-0.1,0)--(2.3,0); \draw (1,0) circle (1); \draw[->, red] (0,0) -- node[sloped, below, black] {$\cos\theta$} (-30:{2*cos(30)}); \draw[dashed,blue] (-30:{2*cos(30)})--(2,0); \node[below] at (-30:{2*cos(30)}) {$(r,\theta)$}; \fill (1,0) circle (1pt) node[font=\small, below] {$0.5$}; \draw[red] (-30:0.3) arc [start angle=-30, end angle=0, radius=0.3] node [right, pos=0.5] {$\theta$}; \end{tikzpicture}

 

解. 可以看到，$\theta\in[-\frac{\pi}2, \frac{\pi}2]$, $r\in[0, \cos\theta]$

例 10. 求 $\displaystyle\iint\limits_{\pi^2\leq x^2+y^2\leq 4 \pi^2}\sin(\sqrt{x^2+y^2}) dxdy$

例 11. 球体$x^{2}+y^{2}+z^{2}\leq 4a^2$被圆柱面$x^2+y^2=2ax$ 所截得部分的体积

\begin{tikzpicture}[x={(-.2cm,-.5cm)},y={(1cm,0cm)},z={(0cm,1cm)},samples=40,scale=1.45] \draw[densely dashed] (0,0,0)--(1,0,0); %,(2,0,0)}; \draw[->] (1,0,0) -- (2,0,0) node[left,below] {$x$}; \draw[densely dashed] (0,0,0)--(0,1,0); %,(2,0,0)}; \draw[->] (0,1,0) -- (0,1.5,0) node[right] {$y$}; \draw[densely dashed] (0,0,0)--(0,0,1); %,(2,0,0)}; \draw[->] (0,0,1) -- (0,0,1.5) node[above] {$z$}; % 柱面 \draw[densely dashed,color=cyan] plot[domain=0:2*pi] ({(1/2)*cos(\x r)+(1/2)},{(1/2)*sin(\x r)},0); \draw[densely dashed,color=cyan] plot[domain=0:2*pi] ({(1/2)*cos(\x r)+(1/2)},{(1/2)*sin(\x r)},.7); \draw[densely dashed,color=cyan] (.5,.5,0) -- (.5,.5,.7) (.5,-.5,0) -- (.5,-.5,.7); % 球面 \draw[thick] plot[domain=0:pi] ({sin(\x r)},{cos(\x r)},0); \draw[thick, densely dashed] plot[domain=pi:2*pi] ({sin(\x r)},{cos(\x r)},0); \draw[thick] plot[domain=0:pi] (0,{cos(\x r)},{sin(\x r)}); % 两个曲面的交线 \draw[thick,color=red] plot[domain=0:2*pi] ({(1/2)*cos(\x r)+(1/2)},{(1/2)*sin(\x r)},{sqrt((1-cos(\x r))/2)}); \end{tikzpicture}

例 12. (例7.1.10) 求

\[\iint\limits_{ x^2+y^2\leq R^2}e^{-(x^2+y^2)} dxdy , \quad \int_{0}^{+\infty}e^{-x^2}dx \]

例 13. 求

\[\iint\limits_D(x+y)dxdy \]

其中$D$由 $x^2+y^2=x+y$围成

二重积分的换元公式

  • $D$为$Oxy$平面上的有界区域，$f(x,y)$为$D$上的可积函数。
  • $D'$为$Ouv$平面上的区域，$\Phi: D'\to D$为一一的$C^1$映射，
    \[\Phi(u,v)=\begin{cases} x=x(u,v) \\ y=y(u,v) \end{cases}, \quad (u,v)\in D' \]
  • $\mbox{det} J\Phi = \frac{\partial(x,y)}{\partial(u,v)}\neq0$， 则$f(x(u,v),y(u,v))$为$D'$上的可积函数

对$O'uv$平面上有界区域$D'$进行矩形分割

\[T': u_0<u_1<\cdots<u_n, \quad v_0<v_1<\cdots<v_m \]

它们对应的u曲线和v曲线把$Oxy$平面上的区域$D$分割成小区域$D_{ij}$，

\begin{tikzpicture}[scale=1.4] \draw[->] (-0.15,0) -- (1.5,0) node[anchor=west]{$u$}; \draw[->] (0,-0.15) -- (0,1.5) node[anchor=south]{$v$}; \node at (0,0) [anchor=north east] {$O'$}; \draw[->] (2.85,0) -- (4.5,0) node[anchor=west]{$x$}; \draw[->] (3,-0.15) -- (3,1.5) node[anchor=south]{$y$}; \node at (3,0) [anchor=north east] {$O$}; % 区域边界 \draw[thick] plot[smooth cycle] coordinates {(0.2,0.351) (0.71, 0.342) (1.05, 0.24) (1.02,1.0) (0.6,1.1)}; \draw[thick] plot[smooth cycle] coordinates {(2.8,0.351) (3.51, 0.242) (3.9, 0.24) (4.42,1.0) (3.81,1.1)}; \draw[->] (1.6,0.6)..controls (1.9,0.8) and (2.2,0.8)..(2.5,0.7); % u-曲线， v-曲线 \foreach \y in {0.35,0.6,...,1.3}{ \draw[red, dashed] (-0.1,\y)--(1.4, \y); } \foreach \x in {0.3,0.55,...,1.2}{ \draw[blue, dashed] (\x, -0.1)--(\x, 1.4); } \fill[black!40] (0.55, 0.35)--+(0.25,0)--+(0.25,0.25)--+(0,0.25)--cycle; % uuline \path[shift = {(2.7,0)}, yellow] [name path=uuline1] (0.1, 0.4)..controls (0.6, 0.4) and (1.1, 0.3)..(1.4, 0.1); \path[shift = {(2.7+0.125,0.25)}] [name path=uuline2] (0.1, 0.4)..controls (0.6, 0.4) and (1.1, 0.3)..(1.4, 0.1); % vvline \path[shift = {(2.8+0.25,-0.125)}] [name path=vvline1] (0.1,0.4) .. controls (0.3, 0.8) and (0.7, 1.3) .. (1.1, 1.4); \path[shift = {(2.8+0.5,-0.25)}] [name path=vvline2] (0.1,0.4) .. controls (0.3, 0.8) and (0.7, 1.3) .. (1.1, 1.4); \path [name intersections={of=uuline1 and vvline1, by=c1}]; %\draw (c1) circle [radius=0.2em]; \path [name intersections={of=uuline1 and vvline2, by=c2}]; %\draw (c2) circle [radius=0.2em]; \path [name intersections={of=uuline2 and vvline2, by=c3}]; %\draw (c3) circle [radius=0.2em]; \path [name intersections={of=uuline2 and vvline1, by=c4}]; %\draw (c4) circle [radius=0.2em]; %\draw (c1)--(c2)--(c3); \fill[black!40] (c1)--(c2)--(c3)--(c4)--cycle; % uuline \foreach \y in {0, 0.25,0.5,0.75}{ \draw[red,dashed, shift={(2.7+0.5*\y,\y)}] (0.1, 0.4)..controls (0.6, 0.4) and (1.1, 0.3)..(1.4, 0.1); } % vvline \foreach \x in {0, 0.25, 0.5, 0.75}{ \draw[blue,dashed, shift={(2.8+\x, -0.5*\x)}] (0.1,0.4) .. controls (0.3, 0.8) and (0.7, 1.3) .. (1.1, 1.4); } \end{tikzpicture}
  • 由于映射是一一对应，所以同族的曲线彼此不相交，而一条$u$曲线和一条$v$曲线也只有一个交点。
\begin{tikzpicture}[x=2.5cm, y=2.5cm, global scale=0.7] \draw[->] (2.85,0) -- (4.5,0) node[anchor=west]{$x$}; \draw[->] (3,-0.15) -- (3,1.5) node[anchor=south]{$y$}; \node at (3,0) [anchor=north east] {$O$}; % 区域边界 \draw[thick] plot[smooth cycle] coordinates {(3.2,0.351) (3.71, 0.242) (4.05, 0.24) (4.02,1.0) (3.16,1.1)}; % % uuline \path[shift = {(2.7,0)}, yellow] [name path=uuline1] (0.1, 0.4)..controls (0.6, 0.4) and (1.1, 0.3)..(1.4, 0.1); \path[shift = {(2.7+0.125,0.25)}] [name path=uuline2] (0.1, 0.4)..controls (0.6, 0.4) and (1.1, 0.3)..(1.4, 0.1); % vvline \path[shift = {(2.8+0.25,-0.125)}] [name path=vvline1] (0.1,0.4) .. controls (0.3, 0.8) and (0.7, 1.3) .. (1.1, 1.4); \path[shift = {(2.8+0.5,-0.25)}] [name path=vvline2] (0.1,0.4) .. controls (0.3, 0.8) and (0.7, 1.3) .. (1.1, 1.4); \path [name intersections={of=uuline1 and vvline1, by=c1}]; \node at (c1) [anchor=north east, font=\tiny] {$M_1$}; \fill (c1) circle [radius=0.1em]; \path [name intersections={of=uuline1 and vvline2, by=c2}]; \node at (c2) [anchor=north west, font=\tiny] {$M_2$}; \fill (c2) circle [radius=0.1em]; \path [name intersections={of=uuline2 and vvline2, by=c3}]; \node at (c3) [anchor=south west, font=\tiny] {$M_3$}; \fill (c3) circle [radius=0.1em]; \path [name intersections={of=uuline2 and vvline1, by=c4}]; \node at (c4) [anchor=south east, font=\tiny] {$M_4$}; \fill (c4) circle [radius=0.1em]; %\draw (c1)--(c2)--(c3); \fill[black!40] (c1)--(c2)--(c3)--(c4)--cycle; % uuline \foreach \y in {0, 0.25,0.5,0.75}{ \draw[red,dashed, shift={(2.7+0.5*\y,\y)}] (0.1, 0.4)..controls (0.6, 0.4) and (1.1, 0.3)..(1.4, 0.1); } % vvline \foreach \x in {0, 0.25, 0.5, 0.75}{ \draw[blue,dashed, shift={(2.8+\x, -0.5*\x)}] (0.1,0.4) .. controls (0.3, 0.8) and (0.7, 1.3) .. (1.1, 1.4); } \end{tikzpicture}

$D'$中的一个方块

\[[u,u+\Delta u]\times[v,v+\Delta v] \]

对应到$D$中，就是一个曲边四边形$M_1M_2M_3M_4$，它的面积近似一个平行四边形的面积，

把平面向量看作空间向量，则

\[\begin{aligned} \Delta \sigma &\approx \left|\overrightarrow{M_1M_2}\times\overrightarrow{M_1M_4}\right| \approx \left| \begin{vmatrix} \vec i & \vec j & \vec k \\ x'_u \Delta u & y'_u \Delta u & 0 \\ x'_v \Delta v & y'_v \Delta v & 0 \\ \end{vmatrix} \right|\\ &= \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\Delta u\Delta v \end{aligned} \]

曲边四边形的四个顶点坐标为

\[\begin{aligned} M_1 = (x_1,y_1)= & (x(u,v), y(u,v)) \\ M_2 = (x_2,y_2)= & (x(u+\Delta u,v), y(u+\Delta u,v)) \\ M_3 = (x_3,y_3)= & (x(u+\Delta u,v+\Delta v), y(u+\Delta u,v+\Delta v)) \\ M_4 = (x_4,y_4)= & (x(u,v+\Delta v), y(u,v+\Delta v)) \\ \end{aligned} \]

从而

\[\begin{aligned} \overrightarrow{M_1M_2} =& (x(u+\Delta u,v)-x(u,v), y(u+\Delta u,v)-y(u,v))\\ =& (\frac{\partial x}{\partial u}\Delta u+o(\Delta u), \frac{\partial y}{\partial u}\Delta u+o(\Delta u)) \end{aligned} \]
\[\begin{aligned} \overrightarrow{M_1M_4} =& (x(u, v++\Delta v)-x(u,v), y(u, v+\Delta v)-y(u,v))\\ =& (\frac{\partial x}{\partial v}\Delta v+o(\Delta v), \frac{\partial y}{\partial v}\Delta v+o(\Delta v)) \end{aligned} \]

略去高阶无穷小后，有

\[\begin{aligned} &\left|\overrightarrow{M_1M_2}\times\overrightarrow{M_1M_4}\right| \approx \left| \begin{vmatrix} \vec i & \vec j & \vec k \\ x'_u \Delta u & y'_u \Delta u & 0 \\ x'_v \Delta v & y'_v \Delta v & 0 \\ \end{vmatrix} \right|\\ =&\left|\left|\begin{matrix} \frac{\partial x}{\partial u} &\frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} &\frac{\partial y}{\partial v} \end{matrix}\right|\Delta u\Delta v\right| =\left|\left|\begin{matrix} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} &\frac{\partial y}{\partial v} \end{matrix}\right|\Delta u\Delta v\right| \\ =&\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\Delta u\Delta v \\ =&\left|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right| \Delta u\Delta v \end{aligned} \]

取$M_{ij}=\Phi(u_i,v_j)=(x(u_i,v_j),y(u_i,v_j))$，则Riemann和

\[\begin{aligned} &\sum_{i,j}f(x_{ij},y_{ij})\Delta D_{ij}\approx \\ &\quad \sum_{ij}f(x(u_i,v_j),y(u_i,v_j)) \left|\frac{\partial (x,y)}{\partial (u,v)}\bigg|_{(u_i,v_j)}\right|\Delta u_i\Delta v_j \end{aligned} \]

取极限后，上式的两边分别为积分

\[\iint\limits_Df(x,y)dxdy , \iint\limits_{D'}f(x(u,v),y(u,v))\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dudv \]

定理 10.
设$D$, $D'$为由分段光滑曲线围成的区域，

  • $\Phi: D'\to D$， $\Phi(u,v)=(x(u,v),y(u,v))$为$C^1$的一一映射，
  • 且$\frac{\partial (x,y)}{\partial (u,v)}\neq0$。
  • 若$f(x,y)$为$D$上的可积函数，

则

\[\iint\limits_Df(x,y)dxdy= \iint\limits_{D'}f(x(u,v),y(u,v))\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dudv \]
  • 公式说明了区域$D$的面积元素$dxdy$与$D'$的面积元素$dudv$之间有如下关系：
    \[dxdy=\left|{\frac{\partial (x,y)}{\partial (u,v)}}\right|dudv \]
  • 如果变换为极坐标变量，$x=r\cos\theta$, $y=r\sin\theta$，则
    \[\frac{\partial (x,y)}{\partial (r,\theta)}=\left|{\begin{aligned} & \cos\theta & -r\sin\theta \\ & \sin\theta & r\cos\theta \end{aligned}}\right|=r \]
    因此有
    \[\iint\limits_Df(x,y)dxdy= \iint\limits_{D'}f(r\cos\theta,r\sin\theta) r drd\theta \]

换元的关键在于，$D'$上的积分易于求解。

  • 计算区域变得简单，或者积分易于得到。

例 14. 求椭球$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+z^2\leq 1$的体积

\[V=2\iint\limits_D\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} dxdy \]

其中$D$由 $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$围成

解. 例 14

采用参数方程

\[\begin{cases} x= a r \cos\theta \\ y= b r \sin\theta \end{cases}, r\in[0,1], \theta\in[0,2\pi] \]

例 15. 求积分$\displaystyle\iint\limits_D\sqrt{\frac{x^2}{a^2}+\frac{y^2}{b^2}} dxdy$ 其中$D$在第一象限中，由$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $y=0$, $y=x$围成

例 16. 求 $\displaystyle\iint\limits_{\Omega} f(x,y)dxdy$

（1） $\Omega$由$\sqrt x+\sqrt y=\sqrt a$， $x=0$, $y=0$ 围成

\begin{tikzpicture} [scale=0.6] \begin{axis}[axis lines=middle,width=5cm, height={}, scale only axis=true,axis equal=true, xtick=\empty, trig format plots=rad,variable=t] \addplot[red,thick,domain=0:2,samples=200] {(sqrt(2)-sqrt(t))^2}; \end{axis} \end{tikzpicture}

（2） $\Omega$由$y=\frac1x$， $y=\frac2x$, $y=x+1$, $y=x-1$ 围成

\begin{tikzpicture} [scale=1.3] \draw[->] (0,0)--(2,0) ; \draw[->] (0,0)--(0,2) ; \draw[domain=0.5:2,variable=\t, smooth, blue, name path=l1] plot (\t, {1/\t}) ;%node[right] {$y=\frac1x$}; \draw[domain=0.3:2,variable=\t, smooth, blue, name path=l1] plot (\t, {0.5/\t}); \draw[domain=0:1,variable=\t, smooth, red, name path=l1] plot (\t, {\t+1}); \draw[domain=1:2,variable=\t, smooth, red, name path=l1] plot (\t, {\t-1}); \end{tikzpicture}

例 17. 求$\displaystyle\iint\limits_{\Omega} f(x,y)dxdy$， $\displaystyle \iint\limits_{\Omega} \sin\frac{y}{x+y}dxdy$

（3） $\Omega$由$x+y=1$， $x=0$, $y=0$ 围成

\begin{tikzpicture} [scale=1.8] \draw[->] (0,0) node[below left] {$O$}--(1.1,0) ; \draw[->] (0,0)--(0,1.1) ; \draw[blue] (1,0) node[below] {$1$} -- (0,1) node[left] {$1$}; \end{tikzpicture}
  1. 令$\begin{cases} x+y=\xi \\ y-x=\eta \end{cases}$
  2. 令$\begin{cases} x+y=\xi \\ y=\xi\eta \end{cases}$

例 18. 求积分 $\displaystyle \iint_D x^2y^2 dxdy$，其中$D$由$y^2=px$, $y^2=qx$, $x^2=ay$, $x^2=by$所围成（$0<p<q$, $0<a<b$）

\begin{tikzpicture} [scale=1.4] \draw[->] (0,0)--(2,0) node[right] {$x$}; \draw[->] (0,0)--(0,2) node[right] {$y$}; \draw[domain=0.5:1.1,variable=\t,smooth, red] plot (\t, {\t*\t}) node[above] {$y=q x^2$}; \draw[domain=0.5:1.8,variable=\t,smooth, red] plot[dashed] (\t, {0.5*\t*\t}) node[above] {$y=p x^2$}; \draw[domain=0.5:1.8,variable=\t,smooth, blue] plot (\t, {sqrt(\t)}) node[right] {$x=b y^2$} plot (\t, {sqrt(0.5*\t)}) node[right] {$x=a y^2$}; \end{tikzpicture}

积分$\displaystyle \iint_D dxdy$即为这4条抛物线围成的面积。

例 19. 求

\[\int_a^b dx\int_{\alpha x}^{\beta x} f(x,y)dy , 0<a<b , 0<\alpha<\beta \]
\begin{tikzpicture} [scale=0.8] \begin{axis}[axis lines=middle,width=5cm,% height={}, scale only axis=true, axis equal=true, %ymax = 3, xtick=\empty, ytick=\empty, %trig format plots=rad, variable=t] \addplot[red,thick,domain=-0.1:3,samples=50] (0.5,{t}); \addplot[red,thick,domain=-0.1:3,samples=50] (2.0,{t}); \addplot[blue,thick,domain=-0.1:2.5,samples=50] {0.4*t}; %\draw (2.5, 1.0) node[ right] {$\alpha x$}; \addplot[blue,thick,domain=-0.1:2.2,samples=50] {1.4*t}; %\draw (2.2, 3.0) node[below right] {$\beta x$}; \end{axis} \end{tikzpicture}

例 20. 求

\[\iint\limits_{|x|+|y|\leq1}f(x+y)dxdy \]
\begin{tikzpicture} [scale=0.8] \begin{axis}[axis lines=middle,width=5cm,% height={}, scale only axis=true, axis equal=true, %ymax = 3, xtick=\empty, ytick=\empty, %trig format plots=rad, variable=t] \addplot[blue, thick, domain=0:1,samples=2] (t, {1-t}); \addplot[blue, thick, domain=0:1,samples=2] (t,{t-1}) node[below] {$1$}; \addplot[blue, thick, domain=-1:0,samples=2] (t,{t+1}); \addplot[blue, thick, domain=-1:0,samples=2] {-1-t} node[right] {$-1$}; \end{axis} \end{tikzpicture}

20.

令

\[\begin{cases} x+y=u \\ x-y=v \end{cases} , u\in[-1,1], v\in[-1,1] \]

则

\[\begin{cases} x=\frac{u+v}2 \\ y=\frac{u-v}2 \end{cases} \Rightarrow \frac{\partial(x,y)}{\partial(u,v)} =\begin{vmatrix} \frac12 & \frac12 \\ \frac12 & \frac{-1}2 \end{vmatrix} =-\frac12 \]

例 21. 计算曲线围成的平面区域的面积

（1） $x^2+2y^2=3$, $xy=1$

\begin{tikzpicture} [scale=1.4] \draw[->] (0,0)--(2,0) node[right] {$x$}; \draw[->] (0,0)--(0,2) node[right] {$y$}; \coordinate (orig) at (0,0); \draw[domain=0:0.5*pi,variable=\t,smooth, red, name path=l1] plot ({sqrt(3)*cos(\t r)}, {sqrt(1.5)*sin(\t r)}); \draw[domain=0.5:1.8,variable=\t,smooth, blue, name path=l2] plot[dashed] (\t, {1/\t}); \path [name intersections={of=l1 and l2}]; \draw[dashed] (intersection-1) -- (intersection-1 |- orig) node[below] {$x_2$}; \draw[dashed] (intersection-2) -- (intersection-2 |- orig) node[below] {$x_1$}; \end{tikzpicture}

(2) $(x-y)^2+y^2=a^2$

\begin{tikzpicture} [scale=1.4] \draw[->] (0,0)--(1.2,0) node[right] {$x$}; \draw[->] (0,0)--(0,1.2) node[right] {$y$}; \coordinate (orig) at (0,0); \draw[domain=0:2*pi,variable=\t,smooth, red] plot ({cos(\t r)+sin(\t r)}, {sin(\t r)}); \draw[domain=0:2*pi,variable=\t,smooth, blue] plot ({cos(\t r)}, {sin(\t r)}); \end{tikzpicture}

广义二重积分

定义 4.
设$f(x,y)$是定义在有界区域$D$及$\partial D$上的非负函数。

  • 在$\partial D$上的某些点的邻域中，$f(x,y)$无界（这种点叫作函数的瑕点）。
  • 假定$f(x,y)$在$D$内的任何闭区域上可积。

做$D$中任一有界闭域列$\{D_n\}$， 使得

\[D_n\subset D_{n+1}, \quad \bigcup\limits_n^{\infty}D_n=D \]

如果

\[\lim_{n\to\infty}\iint\limits_{D_n}f(x,y)dxdy \]

存在有限，且与闭域列$\{D_n\}$的取法无关，那么称瑕积分是收敛的， 并规定瑕积分的值为

\[\iint\limits_Df(x,y)dxdy=\lim_{n\to\infty}\iint\limits_{D_n}f(x,y)dxdy \]

否则称$f(x,y)$在$D$上的瑕积分发散。

例 22. (例7.1.15) $D=[0,1]\times[0,1]$，计算 $\displaystyle\iint\limits_D\frac{y}{\sqrt{x}}dxdy$

解. 令$D_n=[\frac1n,1]\times[0,1]$，则

\[\begin{aligned} \iint_{D_n}\frac{y}{\sqrt{x}}dxdy =&\int_{\frac1n}^1 \int_0^1 \frac{y}{\sqrt x} dy dx \\ =&\int_{\frac1n}^1 \frac1{\sqrt x}dx \int_0^1 {y} dy =1-\frac1{\sqrt n} \end{aligned} \]

定义 5.
设$f(x,y)$是定义在无界闭区域$D$上的非负函数， 且在$D$内的任意有界闭区域上可积。做$D$中任一有界闭域列$\{D_n\}$，使得$D_n\subset D_{n+1}$及$\bigcup\limits_n^{\infty}D_n=D$，如果

\[\lim_{n\to\infty}\iint\limits_{D_n}f(x,y)dxdy \]

存在有限，且与闭域列$\{D_n\}$的取法无关，那么称无穷积分是收敛的， 并规定无穷积分的值为

\[\iint\limits_Df(x,y)dxdy=\lim_{n\to\infty}\iint\limits_{D_n}f(x,y)dxdy \]

否则称$f(x,y)$在$D$上的无穷积分发散。

例 23. (例7.1.16) $D$为第一象限，求

\[\iint\limits_D\frac1{(1+x+y)^3}dxdy \]

解. 取$D_n=[0,n]\times[0,n]$，则

\[\begin{aligned} \iint_{D_n} \frac1{(1+x+y)^3}dxdy =&\int_0^n dx\int_0^1 \frac1{(1+x+y)^3}dy \\ =&\int_0^n \frac12\frac{-1}{(1+x+y)^2}\bigg|_{y=0}^n dx \end{aligned} \]
\[\begin{aligned} =&\int_0^n \frac12\frac{-1}{(1+x+y)^2}\bigg|_{y=0}^n dx \\ =&\int_0^n \frac12\left[\frac1{(1+x)^2}-\frac1{(1+x+n)^2}\right] dx \\ =&\frac12\left(1-\frac1{n+1}+\frac1{2n+1}-\frac1{n+1}\right) \end{aligned} \]

定义 6.
在$f(x,y)$可正可负的情形下，令

\[\begin{aligned} p(x,y)=\frac{|f(x,y)|+f(x,y)}2 \\ q(x,y)=\frac{|f(x,y)|-f(x,y)}2 \end{aligned} \]

若$p(x,y)$和$q(x,y)$在$D$上的瑕积分(或无穷积分)都收敛，则称$f(x,y)$在$D$上的瑕积分（或无穷积分）绝对收敛，并规定瑕积分（无穷积分）的值为

\[\iint\limits_Df(x,y)dx=\iint\limits_Dp(x,y)dx-\iint\limits_Dq(x,y)dx \]

目录

谢谢

例 24. 本节读完

24.