二重积分的概念与性质

定义：

$$

\iint\limits_{D}f(x,y)d \sigma=\lim\limits_{\lambda \to 0}\sum\limits_{i=1}^{n}f(x_{i},y_{i})\Delta \sigma_{i}

$$

 

性质

性质1（不等式）：

1. 在$D$上若$f(x,y)\leq g(x,y)$，则$$\iint\limits_{D}f(x,y)d \sigma \leq \iint\limits_{D}g(x,y)d \sigma$$

2. 若在$D$上有$m \leq f(x,y)\leq M$，则

         $$mS \leq \iint\limits_{D}f(x,y)d \sigma \leq MS$$

3. $$\left|\iint\limits_{D}f(x,y)d \sigma\right|\leq \iint\limits_{D}\left|f(x,y)\right|d \sigma$$

 

性质2（中值定理）：设函数$f(x,y)$在闭区域$D$上连续，$S$为区域$D$的面积，则在$D$上至少存在一点$(\xi ,\eta )$，使得

$$

\iint\limits_{D}f(x,y)d \sigma=f(\xi ,\eta )\cdot S

$$

 

二重积分计算

利用直角坐标计算

先$y$后$x$

$$

\iint\limits_{D}f(x,y)d \sigma=\int_{a}^{b}dx \int_{\phi_{2}(x)}^{\phi_{1}(x)}f(x,y)dy

$$

 

先$x$后$y$

$$

\iint\limits_{D}f(x,y)d \sigma=\int_{c}^{d}dy \int_{\psi {2}(y)}^{\psi{1}(y)}f(x,y)dx

$$

 

利用极坐标计算

先$\rho$后$\theta$

$$

\iint\limits_{D}f(x,y)d \sigma=\int_{\alpha}^{\beta}d \theta \int_{\phi_{1}(\theta )}^{\phi_{2}(\theta )}f(\rho \cos \theta ,\rho \sin \theta )\rho d \rho

$$

常用于相同$\theta$对应的不同$\rho$

 

适合用极坐标计算的二重积分的特征

1. 适合用极坐标计算的被积函数

    $$f(\sqrt{x^{2}+y^{2}}),f(\frac{y}{x}),f(\frac{x}{y})$$

2. 适合用极坐标的积分域

    $$\begin{aligned} x^{2}&+y^{2}\leq R^{2}\r^{2}\leq x^{2}&+y^{2}\leq R^{2}\x^{2}&+y^{2}\leq 2ax\x^{2}&+y^{2}\leq 2by\end{aligned}$$

    如果圆心既不在坐标原点也不在坐标轴，考虑平移+极坐标，即

    $$令x-x_{0}=\rho \sin \theta ,y-y_{0}=\rho \sin \theta $$

    则有

    $$\int_{0}^{2\pi}d \theta \int_{0}^{R}(\quad )\rho d \rho$$

 

利用对称性和奇偶性计算

若积分域$D$关于$y$轴对称，则

$$

\iint\limits_{D}f(x,y)d \sigma=\left{\begin{aligned}&2\iint\limits_{D_{x \geq 0}}f(x,y)d \sigma& f(-x,y)=f(x,y)\&0&f(-x,y)=-f(x,y)\end{aligned}\right.

$$

 

若积分域$D$关于$x$轴对称，则

$$

\iint\limits_{D}f(x,y)d \sigma=\left{\begin{aligned}&2\iint\limits_{D_{y \geq 0}}f(x,y)d \sigma&f(x,-y)=f(x,y)\&0&f(x,-y)=-f(x,y)\end{aligned}\right.

$$

 

利用变量对称性计算

若$D$关于$y=x$对称，则

$$

\iint\limits_{D}f(x,y)d \sigma=\iint\limits_{D}f(y,x)d \sigma

$$

 

常考题型方法与技巧

累次积分交换次序及计算

例1：交换累次积分$\begin{aligned} \int_{0}^{1}dx \int_{x^{2}}^{2-x}f(x,y)dy\end{aligned}$的次序

 

 

 

$$

原式=\int_{0}^{1}dy \int_{0}^{\sqrt{y}}f(x,y)dx+\int_{1}^{2}dy \int_{0}^{2-y}f(x,y)dx

$$

 

例2：累次积分$\begin{aligned} \int_{0}^{\frac{\pi}{2}}d \theta \int_{0}^{\cos \theta }f(\rho \cos \theta ,\rho \sin \theta )\rho d \rho\end{aligned}$化为直角坐标

 

$$

\begin{aligned}

\rho&=\cos \theta \Rightarrow \rho^{2}=\rho \cos \theta \Rightarrow x^{2}+y^{2}=x\

\rho&=0\

\theta &\in (0, \frac{\pi}{2})

\end{aligned}

$$

 

因此有

$$

\begin{aligned}

原式&=\int_{0}^{\frac{1}{2}}dy \int_{\frac{1}{2}- \sqrt{\frac{1}{4}-y^{2}}}^{\frac{1}{2}+\sqrt{\frac{1}{4}-y^{2}}}f(x,y)dy\

原式&=\int_{0}^{1}dx \int_{0}^{\sqrt{x-x^{2}}}f(x,y)dx

\end{aligned}

$$

 

不同坐标系累次积分相互转化，先画区域，然后重新定上下限即可

 

例3：积分$\begin{aligned} \int_{0}^{2}dx \int_{0}^{\sqrt{2x-x^{2}}}\sqrt{x^{2}+y^{2}}dy\end{aligned}$的值等于（）

 

累次积分不好算，考虑交换积分次序，换坐标系

 

$$

\begin{aligned}

原式&=\int_{0}^{\frac{\pi}{2}}d \theta \int_{0}^{2 \cos \theta }\rho \cdot \rho d \rho\

&=\frac{8}{3}\int_{0}^{\frac{\pi}{2}}\cos ^{3}\theta d \theta \

&=\frac{8}{3} \frac{2}{3}=\frac{16}{9}

\end{aligned}

$$

 

二重积分计算

例4：设$D:\left{(x,y)|x^{2}+y^{2} \leq 1\right}$，则$\begin{aligned} \iint\limits_{D}(x^{2}-y)dxdy=()\end{aligned}$

 

$$

\begin{aligned}

原式&=\iint\limits_{D}x^{2}dxdy\

&=\iint\limits_{D}y^{2}dxdy\

&=\frac{1}{2}\iint\limits_{D}(x^{2}+y^{2})dxdy\

&=\frac{1}{2}\int_{0}^{2\pi}d \theta \int_{0}^{1}\rho^{2}\rho d \rho\

&=\frac{1}{2}\cdot 2\pi \cdot \frac{1}{4}= \frac{\pi}{4}

\end{aligned}

$$

 

对称性常用于平方项化极坐标，本题就是一个例子，还有类似

设$D:\left{(x,y)|x^{2}+y^{2} \leq 1\right}$，则

$$\begin{aligned} \iint\limits_{D}(2x-3y)^{2}dxdy&=\iint\limits_{D}(4x^{2}-12xy+9y^{2})dxdy\&=\iint\limits_{D}(4x^{2}+9y^{2})dxdy\&=\iint\limits_{D}(4y^{2}+9x^{2})dxdy\&=\frac{13}{2}\iint\limits_{D}(x^{2}+y^{2})dxdy\&=\frac{13}{4}\pi\end{aligned}$$

 

例5：设$D$是$xOy$平面上以$(1,1),(-1,1),(-1,-1)$为顶点的三角形区域，$D_{1}$是$D$在第一象限的部分，说明$\begin{aligned} \iint\limits_{D}(xy+\cos x \sin y )dxdy=2 \iint\limits_{D_{1}}\cos x \sin y dxdy\end{aligned}$

 

将该直角三角形沿$y=-x$划分成两部分，显然$y=-x$以上关于$y$轴对称，$y=-x$以下关于$x$轴对称，因此有

$$

\begin{aligned}

原式&=\iint\limits_{D}\cos x \sin ydxdy\

&=2 \iint\limits_{D_{1}}\cos x \sin y dxdy

\end{aligned}

$$

 

例6：设平面区域$D$由曲线$y=\sqrt{3(1-x^{2})}$与直线$y=\sqrt{3}x$及$y$轴围成，计算二重积分$\begin{aligned} \iint\limits_{D}x^{2}dxdy=()\end{aligned}$

 

$$

\begin{aligned}

原式&=\int_{0}^{\frac{1}{\sqrt{2}}}dx \int_{\sqrt{3}x}^{\sqrt{3(1-x^{2})}}x^{2}dy\

&=\int_{0}^{\frac{1}{\sqrt{2}}}x^{2}[\sqrt{3(1-x^{2})}-\sqrt{3}x] dx\

&=\sqrt{3}\int_{0}^{\frac{1}{\sqrt{2}}}x^{2}\sqrt{1-x^{2}}dx-\sqrt{3}\int_{0}^{\frac{1}{\sqrt{2}}}x^{3}dx\

&=\sqrt{3}\int_{0}^{\frac{1}{\sqrt{2}}}x^{2}\sqrt{1-x^{2}}dx - \frac{\sqrt{3}}{16}\

&\overset{x=\sin t}{=}\sqrt{3}\int_{0}^{\frac{\pi}{4}}\sin ^{2}t \cos ^{2}tdt- \frac{\sqrt{3}}{16}\

&=\frac{\sqrt{3}}{4}\int_{0}^{\frac{\pi}{4}}(\sin 2t)^{2}dt - \frac{\sqrt{3}}{16}\

&\overset{2t =u}{=}\frac{\sqrt{3}}{8}\int_{0}^{\frac{\pi}{2}}\sin ^{2}udu-\frac{\sqrt{3}}{16}\

&=\frac{\sqrt{3}}{32}\pi-\frac{\sqrt{3}}{16}

\end{aligned}

$$

 

例7：已知平面域$D=\left{(x,y)|x^{2}+y^{2}\leq 2y\right}$计算二重积分$\begin{aligned} I=\iint\limits_{D}(x+1)^{2}dxdy\end{aligned}$

 

$$

\begin{aligned}

I&=\iint\limits_{D}(x^{2}+2x+1)dxdy\

&=\iint\limits_{D}(x^{2}+1)dxdy\

&=2\int_{0}^{\frac{\pi}{2}}d \theta \int_{0}^{2\sin

 \theta }\rho^{2}\cos ^{2}\theta \rho d \rho+\pi\

&此处d \theta 的上限不用\pi而用 \frac{\pi}{2}用的是对称性\

&而且 \frac{\pi}{2}更方便用点火公式\

&x^{2}+y^{2}-2y=0\Rightarrow \rho^{2}=2 \rho \sin \theta \Rightarrow \rho=2\sin \theta \

&=8 \int_{0}^{\frac{\pi}{2}}\sin ^{4}\theta \cos ^{2}\theta d \theta +\pi\

&=8\int_{0}^{\frac{\pi}{2}}\sin ^{4}\theta (1-\sin ^{2}\theta )d \theta +\pi\

&=8\left(\frac{3}{16}\pi- \frac{15}{96}\pi\right)+\pi\

&=\frac{5}{4}\pi

\end{aligned}

$$

 

例8：计算二重积分$\begin{aligned} \iint\limits_{D}\left|x^{2}+y^{2}-1\right|d \sigma\end{aligned}$，其中$D=\left{(x,y)|0\leq x \leq 1,0\leq y \leq 1\right}$

 

一元带括号的积分是根据正负分区间，多元也类似。多元分区域，去绝对值，求积分的时候常用减法，因为往往分出来的两块，一块好算，一块不好算，不好算的可以用整体-好算的，即原式变为2×好算的+整体

 

 

如图，将$D$分成$D_{1}$与$D_{2}$两部分

$$

\begin{aligned}

原式&=\iint\limits_{D_{1}}(1-x^{2}-y^{2})d \sigma+\iint\limits_{D_{2}}(x^{2}+y^{2}-1)d \sigma\

&=\iint\limits_{D_{1}}(1-x^{2}-y^{2})d \sigma+[\iint\limits_{D}(x^{2}+y^{2}-1)d \sigma-\iint\limits_{D_{1}}(x^{2}+y^{2}-1)d \sigma ]\

&=2\iint\limits_{D_{1}}(1-x^{2}-y^{2})d \sigma+\iint\limits_{D}(x^{2}+y^{2}-1)d \sigma\

&=2\int_{0}^{\frac{\pi}{2}}d \theta \int_{0}^{1}(1-\rho^{2})\rho d \rho+\int_{0}^{1}dx \int_{0}^{1}(x^{2}+y^{2}-1)dy\

&= \frac{\pi}{4}- \frac{1}{3}

\end{aligned}

$$

 

例8：设平面域$D=\left{(x,y)|1\leq x^{2}+y^{2}\leq 4,x \geq 0,y \geq 0\right}$，计算$\begin{aligned} \iint\limits_{D}\frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y}dxdy\end{aligned}$

 

观察被积函数显然用直角坐标和极坐标都不好做，因此想其他方法

 

$$

\begin{aligned}

\iint\limits_{D}\frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y }dxdy&=\iint\limits_{D}\frac{y \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y }dxdy\

&=\frac{1}{2}\left[\iint\limits_{D}\frac{y \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y }dxdy+\iint\limits_{D}\frac{y \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y }dxdy\right]\

&=\frac{1}{2}\iint\limits_{D}\sin (\pi \sqrt{x^{2}+y^{2}})dxdy\

&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}d \theta \int_{1}^{2}\sin (\pi \rho)\rho d \rho\

&=-\frac{3}{4}

\end{aligned}

$$

 

也可以用

$$

\begin{aligned}

\iint\limits_{D}\frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y}dxdy&=\int_{0}^{\frac{\pi}{2}}\frac{\cos \theta }{\cos \theta +\sin \theta }d \theta \cdot \int_{1}^{2}\rho \sin (\pi \rho)d \rho\

由于\int_{0}^{\frac{\pi}{2}}\frac{\cos \theta }{\cos \theta +\sin \theta }d \theta &=\int_{0}^{\frac{\pi}{2}}\frac{\sin \theta }{\cos \theta +\sin \theta }d \theta \

&=\frac{1}{2}\int_{0}^{\frac{\pi}{2} }\frac{\cos \theta +\sin \theta }{\cos \theta +\sin \theta }d \theta \

&=\frac{\pi}{4}\

\int_{1}^{2}\rho \sin (\pi \rho)d \rho&=\frac{1}{\pi}(-\rho \cos \pi \rho+ \frac{1}{\pi} \sin \pi \rho)\Big|_{1}^{2}\

&=- \frac{3}{\pi}\

故\iint\limits_{D}\frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y}dxdy&=- \frac{3}{4}

\end{aligned}

$$

 

这里用到了区间再现公式，即

$$\int_{a}^{b}f(x)dx \overset{x=a+b-t}{=}\int_{a}^{b}f(a+b-t)dt$$

观察左右两式，显然积分区域不变

 

也可以不用公式，用正常思路

$$

\iint\limits_{D}\frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y}dxdy=\int_{0}^{\frac{\pi}{2}}\frac{\cos \theta }{\cos \theta +\sin \theta }d \theta \cdot \int_{1}^{2}\rho \sin (\pi \rho)d \rho

$$

显然满足$R(-\sin \theta ,-\cos \theta )=-R(\sin \theta ,\cos \theta )$，因此，令$u=\tan \theta$

$$

\begin{aligned}

\int_{0}^{\frac{\pi}{2}}\frac{\cos \theta }{\cos \theta +\sin \theta }d \theta&=\int_{0}^{\frac{\pi}{2}}\frac{\cos ^{2}\theta }{1+\tan \theta }d \tan \theta \

&=\int_{0}^{\frac{\pi}{2}}\frac{1}{(\tan ^{2}\theta +1)(1+ \tan \theta )}d \tan \theta \

&\overset{u=\tan \theta }{=}\int_{0}^{+\infty}\frac{1}{(1+u)(1+u^{2})}du\

&=\int_{0}^{+\infty} \left[\frac{A}{1+u}+ \frac{Bu+C}{1+u^{2}}\right]du\

&\Rightarrow \left{\begin{aligned}&A+B=0\&B+C=0\&A+C=1\end{aligned}\right.解得\left{\begin{aligned}&A = \frac{1}{2}\&B =- \frac{1}{2}\&C = \frac{1}{2}\end{aligned}\right.\

&=\frac{1}{2}\int_{0}^{+\infty} \frac{1}{1+u}du+ \frac{1}{2}\int_{0}^{+\infty}\frac{-u+1}{u^{2}+1}du\

&=\frac{1}{2}\ln (1+u)\Big|{u=+\infty}^{}- \frac{1}{2}\int{0 }^{+\infty}\frac{\frac{1}{2}d(u^{2}+1)-du}{u^{2}+1}\

&=\frac{1}{2}\ln (1+u)\Big|{u=+\infty}^{}- \frac{1}{4}\ln (u^{2}+1)\Big|{u=+\infty}^{}+ \frac{1}{2}\arctan u \Big|_{u=+\infty}^{}\

&=\frac{1}{4}\ln \left(\frac{u^{2}+2u+1}{u^{2}+1}\right)\Big|_{u=+\infty}^{}+ \frac{\pi}{4}\

&=\frac{1}{4}\ln \left(1+ \frac{2u}{u^{2}+1}\right)\Big|_{u=+\infty}^{}+\frac{\pi}{4}\

&=0+\frac{\pi}{4}=\frac{\pi}{4}

\end{aligned}

$$

后面计算同上

 

例9：设$D_{k}$是圆域$D=\left{(x,y)|x^{2}+y^{2}\leq 1\right}$在第$k$象限的部分，记 $\begin{aligned} I_{k}=\iint\limits_{D_{k}}(y-x)dxdy(k=1,2,3,4)\end{aligned}$，说明$I_{k}$的正负

 

由于在第二象限$y-x>0$，因此$I_{2}>0$，同理$I_{4}<0$

对于$I_{1}$，有

$$

\begin{aligned}

I_{1}&=\iint\limits_{D_{1}}(y-x)d \sigma\

&=\iint\limits_{D_{1}}(x-y) d \sigma\

\iint\limits_{D_{1}}(y-x)d \sigma&=-\iint\limits_{D_{1}}(y-x)d \sigma=0

\end{aligned}

$$

同理$I_{3}=0$

 

例10：已知平面域$D=\left{(x,y)||x|+|y|\leq \frac{\pi}{2}\right}$，记$\begin{aligned} I_{1}=\iint\limits_{D}\sqrt{x^{2}+y^{2}}d \sigma,I_{2}=\iint\limits_{D}\sin \sqrt{x^{2}+y^{2}}d \sigma,I_{3}=\iint\limits_{D}(1+\cos \sqrt{x^{2}+y^{2}})d \sigma\end{aligned}$，说明$I_{3}<I_{2}<I_{1}$

 

由于被积区域相同，因此函数值大则$I$大，令$\sqrt{x^{2}+y^{2}}=r$显然有

$$

r> \sin r \geq \sin ^{2}r=1-\cos ^{2}r \geq 1-\cos r

$$

 

画个图也行，不难