2. 二重积分的换元

多变量函数的重积分

张瑞
中国科学技术大学数学科学学院

二重积分的换元

例 1. 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$x^2+y^2\leq 1$

解. 用累次积分来解

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

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}[scale=1.4] \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[red] (0,0) circle [radius=\r*0.25]; } \foreach \r in {1,...,19}{ \draw[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[->] (50:1.5) node [anchor=west]{$rd\theta$} --(50:1.02); \draw[->] (33:1.4) node [anchor=west]{$dr$} --(38:0.82); \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

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

Riemann和为

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

取极限后，有

\[\begin{aligned} \iint_D \sqrt{x^2+y^2}dxdy =&\int_0^{2\pi}\int_0^1 r^2drd\theta \\ =&2\pi\int_0^1 r^2dr=\frac{2\pi}3 \end{aligned} \]

坐标曲线和面积元素

前述的例子中，做了坐标变换

\[x=r\cos \theta, \quad y=r\sin\theta \]

它给出了$O'r\theta$平面上区域$D'=[0,1]\times[0,2\pi)$到$Oxy$平面上单位圆$D$的映射。 在这个变换下，积分的面积元素变换关系为

\[d\sigma = dxdy = rdrd\theta \]

一般地，设

\[x=x(u,v), \quad y=y(u,v), \quad (u,v)\in D' \]

是从$O'uv$平面上的有界区域$D'$到$Oxy$平面上的区域$D$的坐标变换。

它的面积元素的变换关系是什么？

确定$u=u_0$($D'$中平行与$v$轴的直线)，则

\[x=x(u_0,v), y=y(u_0,v) \]

是$D$中一条曲线，称为v曲线。类似可以得到u曲线。

\begin{tikzpicture}[scale=1.6] \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 {(3.2,0.351) (3.71, 0.242) (4.05, 0.24) (4.02,1.0) (3.16,1.1)}; \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。
  • 在$O'uv$平面上对$D'$的矩形分割，就给出了$Oxy$平面上对$D$的一个由$u$曲线和$v$曲线网构成的分割。
\begin{tikzpicture}[scale=1.9] \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，

\[\begin{aligned} \Delta \sigma &\approx \left|\overrightarrow{M_1M_2}\times\overrightarrow{M_1M_4}\right| \\ &\approx \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\Delta u\Delta v \end{aligned} \]

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

\[\begin{aligned} \left|\overrightarrow{M_1M_2}\times\overrightarrow{M_1M_4}\right| =&\left|\left|\begin{matrix} \frac{\partial x}{\partial u} &\frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} &\frac{\partial y}{\partial v} \end{matrix}\right|\Delta u\Delta v\right| \\ =&\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\Delta u\Delta v \\ =&\left|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right| \Delta u\Delta v \end{aligned} \]

当$\frac{\partial(x,y)}{\partial(u,v)}=0$时，$D'$上的方块映射到$D$中时可能退化，因此，设

\[x=x(u,v), \quad y=y(u,v), \quad (u,v)\in D' \]

是$O'uv$平面上$D'$到$Oxy$平面中$D$的一一可微映射，且

\[\frac{\partial(x,y)}{\partial(u,v)}\neq 0, \quad (u,v)\in D' \]

则有$D$的面积元素等式

\[d\sigma = dxdy = \left|\frac{\partial(x,y)}{\partial(u,v)}\right| dudv \]

二重积分的换元

设$f(x,y)$是$D$上连续的二元函数。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$D_{ij}$，有

\[\sigma(D_{ij})\approx \left|\frac{\partial(x,y)}{\partial(u,v)}\right|_{u_i,v_j} \Delta u_i\Delta v_j \]

因此， $f(x,y)$在$Oxy$平面区域$D$上的Riemann和可以表示为

\[\sum_{i,j}f(\xi_{ij},\eta_{ij})\sigma({D_{i,j}}) \approx \sum_{i,j} f(\xi_{ij},\eta_{ij}) \left|\frac{\partial(x,y)}{\partial(u,v)}\right|_{(u_i,v_j)}\Delta u_i\Delta v_j \]

式子的右端是函数

\[f(x(u,v), y(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right| \]

在区域$D'$上的Riemann和。当$|T'|\to0$时，得到

定理 1.
设$D$是$Oxy$平面中可测的有界闭区域，$f(x,y)$在$D$上连续。变换

\[\Phi: D'\to D, \quad \Phi(u,v)=(x(u,v),y(u,v)) \]

为$C^1$(有连续偏导数)的一一映射，且

\[\frac{\partial (x,y)}{\partial (u,v)}\neq0 \]

则

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

如果变换为极坐标变量，$x=r\cos\theta$, $y=r\sin\theta$，则

\[\frac{\partial (x,y)}{\partial (r,\theta)} =\left|{\begin{aligned} & \cos\theta & \sin\theta \\ & -r\sin\theta & r\cos\theta \end{aligned}}\right|=r \]

因此有

\[\iint\limits_Df(x,y)dxdy= \iint\limits_{D'}f(r\cos\theta,r\sin\theta) r drd\theta \]

关键在于，$D'$上的积分易于求解：

积分区域变得简单（规整）或者被积函数变得简单，或者两者都是。

例 2. 求

\[\iint\limits_{x^2+y^2\leq x}f(\frac{y}x)dxdy \]

解. 计算区域如图

\begin{tikzpicture}[scale=1.2] \draw[->] (0,-1)--(0,1.4); \draw[->] (-0.1,0)--(2.3,0); \draw (1,0) circle (1); \draw[->, red] (0,0) -- node[sloped, below, black] {$r\cos\theta$} (-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 [below right] {$\theta$}; \end{tikzpicture}

例 3. 求

\[\iint\limits_{\pi^2\leq x^2+y^2\leq 4 \pi^2}\sin(\sqrt{x^2+y^2}) dxdy \]

例 4. (例10.2.7) 求

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

例 5. 求

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

其中$D$由 $x^2+y^2=x+y$围成

例 6. 求

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

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

例 7. 求

\[\iint\limits_{\Omega} f(x,y)dxdy \]

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

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

（3） $\Omega$由$x+y=1$， $x=0$, $y=0$ 围成

例 8. 求

\[\int_a^b dx\int_{\alpha x}^{\beta x} f(x,y)dy , 0<a<b , 0<\alpha<\beta \]

例 9. 求

\[\iint\limits_{|x|+|y|\leq1}f(x+y)dxdy \]

谢谢

在曲线上取两点$M$和$M'$，其横坐标分别为$x$与$x+dx$， 则两点的距离为

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