【高等数学基础进阶】二重积分

二重积分的概念与性质

定义：
$$\underset{D}{\iint }f(x,y)d\sigma =\underset{\lambda \to 0}{\lim }\sum _{i=1}^{n}f(x_i,y_i)\Delta {\sigma }_i$$

几何意义

二重积分$\begin{array}{r}\underset{D}{\iint }f(x,y)d\sigma \end{array}$是一个数，当$f(x,y)\ge 0$时，其值等于以区域$D$为底，以曲面$z=f(x,y)$为曲顶的曲顶柱体的体积；当$f(x,y)\le 0$时，二重积分的值为负值，其绝对值等于上述圆顶柱体的体积

性质

性质1（不等式）：

1. 在$D$上若$f(x,y)\le g(x,y)$，则$$\underset{D}{\iint }f(x,y)d\sigma \le \underset{D}{\iint }g(x,y)d\sigma$$
2. 若在$D$上有$m\le f(x,y)\le M$，则
   $$mS\le \underset{D}{\iint }f(x,y)d\sigma \le MS$$
3. $$\left|\underset{D}{\iint }f(x,y)d\sigma \right|\le \underset{D}{\iint }\left|f(x,y)\right|d\sigma$$

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

二重积分计算

利用直角坐标计算

先$y$后$x$
$$\underset{D}{\iint }f(x,y)d\sigma ={\int }_a^{b}dx{\int }_{{\varphi }_2(x)}^{{\varphi }_1(x)}f(x,y)dy$$

先$x$后$y$
$$\underset{D}{\iint }f(x,y)d\sigma ={\int }_c^{d}dy{\int }_{{\psi }_2(y)}^{{\psi }_1(y)}f(x,y)dx$$

利用极坐标计算

先$\rho$后$\theta$
$$\underset{D}{\iint }f(x,y)d\sigma ={\int }_{\alpha }^{\beta }d\theta {\int }_{{\varphi }_1(\theta )}^{{\varphi }_2(\theta )}f(\rho \cos \theta ,\rho \sin \theta )\rho d\rho$$
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
   x 2 + y 2 ≤ R 2 r 2 ≤ x 2 + y 2 ≤ R 2 x 2 + y 2 ≤ 2 a x x 2 + y 2 ≤ 2 b y
   x2+y2≤R2r2≤x2+y2≤R2x2+y2≤2axx2+y2≤2by" role="presentation" style="position: relative;">x2r2≤x2x2x2+y2≤R2+y2≤R2+y2≤2ax+y2≤2by$$\begin{array}{rl}{x}^2& +y^2\le R^2\\ r^2\le x^2& +y^2\le R^2\\ x^2& +y^2\le 2ax\\ x^2& +y^2\le 2by\end{array}$$
   x2r2≤x2x2x2​+y2≤R2+y2≤R2+y2≤2ax+y2≤2by​
   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
   $${\int }_0^{2\pi }d\theta {\int }_0^R()\rho d\rho$$

利用对称性和奇偶性计算

若积分域$D$关于$y$轴对称，则
$$\underset{D}{\iint }f(x,y)d\sigma =\{\begin{array}{rlr}& 2\underset{D_{x\ge 0}}{\iint }f(x,y)d\sigma & f(-x,y)=f(x,y)\\ & 0& f(-x,y)=-f(x,y)\end{array}$$

若积分域$D$关于$x$轴对称，则
$$\underset{D}{\iint }f(x,y)d\sigma =\{\begin{array}{rlr}& 2\underset{D_{y\ge 0}}{\iint }f(x,y)d\sigma & f(x,-y)=f(x,y)\\ & 0& f(x,-y)=-f(x,y)\end{array}$$

利用变量对称性计算

若$D$关于$y=x$对称，则
$$\underset{D}{\iint }f(x,y)d\sigma =\underset{D}{\iint }f(y,x)d\sigma$$

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$\begin{array}{r}{\int }_0^{1}dx{\int }_{x^2}^{2-x}f(x,y)dy\end{array}$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$\begin{array}{r}{\int }_0^{\frac{\pi }{2}}d\theta {\int }_0^{\cos \theta }f(\rho \cos \theta ,\rho \sin \theta )\rho d\rho \end{array}$化为直角坐标

$$\begin{array}{rl}\rho & =\cos \theta \Rightarrow {\rho }^2=\rho \cos \theta \Rightarrow x^2+y^2=x\\ \rho & =0\\ \theta & \in (0,\frac{\pi }{2})\end{array}$$

因此有
$$\begin{array}{rl}原式& ={\int }_0^{\frac{1}{2}}dy{\int }_{\frac{1}{2}-\sqrt{\frac{1}{4}-y^2}}^{\frac{1}{2}+\sqrt{\frac{1}{4}-y^2}}f(x,y)dy\\ 原式& ={\int }_0^{1}dx{\int }_0^{\sqrt{x-x^2}}f(x,y)dx\end{array}$$

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$\begin{array}{r}{\int }_0^{2}dx{\int }_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2}dy\end{array}$的值等于（）

累次积分不好算，考虑交换积分次序，换坐标系

$$\begin{array}{rl}原式& ={\int }_0^{\frac{\pi }{2}}d\theta {\int }_0^{2\cos \theta }\rho \cdot \rho d\rho \\ & =\frac{8}{3}{\int }_0^{\frac{\pi }{2}}{\cos }^3\theta d\theta \\ & =\frac{8}{3}\frac{2}{3}=\frac{16}{9}\end{array}$$

二重积分计算

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

$$\begin{array}{rl}原式& =\underset{D}{\iint }{x}^{2}dxdy\\ & =\underset{D}{\iint }{y}^{2}dxdy\\ & =\frac{1}{2}\underset{D}{\iint }(x^2+y^2)dxdy\\ & =\frac{1}{2}{\int }_0^{2\pi }d\theta {\int }_0^1{\rho }^2\rho d\rho \\ & =\frac{1}{2}\cdot 2\pi \cdot \frac{1}{4}=\frac{\pi }{4}\end{array}$$

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$D:\left\{(x,y)|x^2+y^2\le 1\right\}$，则
$$\begin{array}{rl}\underset{D}{\iint }(2x-3y{)}^{2}dxdy& =\underset{D}{\iint }(4x^2-12xy+9y^2)dxdy\\ & =\underset{D}{\iint }(4x^2+9y^2)dxdy\\ & =\underset{D}{\iint }(4y^2+9x^2)dxdy\\ & =\frac{13}{2}\underset{D}{\iint }(x^2+y^2)dxdy\\ & =\frac{13}{4}\pi \end{array}$$

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$\begin{array}{r}\underset{D}{\iint }(xy+\cos x\sin y)dxdy=2\underset{D_1}{\iint }\cos x\sin ydxdy\end{array}$

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
$$\begin{array}{rl}原式& =\underset{D}{\iint }\cos x\sin ydxdy\\ & =2\underset{D_1}{\iint }\cos x\sin ydxdy\end{array}$$

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$\begin{array}{r}\underset{D}{\iint }{x}^{2}dxdy=()\end{array}$

$$\begin{array}{rl}原式& ={\int }_0^{\frac{1}{\sqrt{2}}}dx{\int }_{\sqrt{3}x}^{\sqrt{3(1-x^2)}}{x}^{2}dy\\ & ={\int }_0^{\frac{1}{\sqrt{2}}}{x}^2[\sqrt{3(1-x^2)}-\sqrt{3}x]dx\\ & =\sqrt{3}{\int }_0^{\frac{1}{\sqrt{2}}}{x}^2\sqrt{1-x^2}dx-\sqrt{3}{\int }_0^{\frac{1}{\sqrt{2}}}{x}^{3}dx\\ & =\sqrt{3}{\int }_0^{\frac{1}{\sqrt{2}}}{x}^2\sqrt{1-x^2}dx-\frac{\sqrt{3}}{16}\\ & \stackrel{x=\sin t}{=}\sqrt{3}{\int }_0^{\frac{\pi }{4}}{\sin }^{2}t{\cos }^{2}tdt-\frac{\sqrt{3}}{16}\\ & =\frac{\sqrt{3}}{4}{\int }_0^{\frac{\pi }{4}}(\sin 2t{)}^{2}dt-\frac{\sqrt{3}}{16}\\ & \stackrel{2t=u}{=}\frac{\sqrt{3}}{8}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}udu-\frac{\sqrt{3}}{16}\\ & =\frac{\sqrt{3}}{32}\pi -\frac{\sqrt{3}}{16}\end{array}$$

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

$$\begin{array}{rl}I& =\underset{D}{\iint }(x^2+2x+1)dxdy\\ & =\underset{D}{\iint }(x^2+1)dxdy\\ & =2{\int }_0^{\frac{\pi }{2}}d\theta {\int }_0^{2\sin \theta }{\rho }^2{\cos }^2\theta \rho d\rho +\pi \\ & 此处d\theta 的上限不用\pi 而用\frac{\pi }{2}用的是对称性\\ & 而且\frac{\pi }{2}更方便用点火公式\\ & x^2+y^2-2y=0\Rightarrow {\rho }^2=2\rho \sin \theta \Rightarrow \rho =2\sin \theta \\ & =8{\int }_0^{\frac{\pi }{2}}{\sin }^4\theta {\cos }^2\theta d\theta +\pi \\ & =8{\int }_0^{\frac{\pi }{2}}{\sin }^4\theta (1-{\sin }^2\theta )d\theta +\pi \\ & =8\left(\frac{3}{16}\pi -\frac{15}{96}\pi \right)+\pi \\ & =\frac{5}{4}\pi \end{array}$$

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

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
$$\begin{array}{rl}原式& =\underset{D_1}{\iint }(1-x^2-y^2)d\sigma +\underset{D_2}{\iint }(x^2+y^2-1)d\sigma \\ & =\underset{D_1}{\iint }(1-x^2-y^2)d\sigma +[\underset{D}{\iint }(x^2+y^2-1)d\sigma -\underset{D_1}{\iint }(x^2+y^2-1)d\sigma ]\\ & =2\underset{D_1}{\iint }(1-x^2-y^2)d\sigma +\underset{D}{\iint }(x^2+y^2-1)d\sigma \\ & =2{\int }_0^{\frac{\pi }{2}}d\theta {\int }_0^1(1-{\rho }^2)\rho d\rho +{\int }_0^{1}dx{\int }_0^1(x^2+y^2-1)dy\\ & =\frac{\pi }{4}-\frac{1}{3}\end{array}$$

例8：设平面域$D=\left\{(x,y)|1\le x^2+y^2\le 4,x\ge 0,y\ge 0\right\}$，计算$\begin{array}{r}\underset{D}{\iint }\frac{x\sin (\pi \sqrt{x^2+y^2})}{x+y}dxdy\end{array}$

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

$$\begin{array}{rl}\underset{D}{\iint }\frac{x\sin (\pi \sqrt{x^2+y^2})}{x+y}dxdy& =\underset{D}{\iint }\frac{y\sin (\pi \sqrt{x^2+y^2})}{x+y}dxdy\\ & =\frac{1}{2}\left[\underset{D}{\iint }\frac{y\sin (\pi \sqrt{x^2+y^2})}{x+y}dxdy+\underset{D}{\iint }\frac{y\sin (\pi \sqrt{x^2+y^2})}{x+y}dxdy\right]\\ & =\frac{1}{2}\underset{D}{\iint }\sin (\pi \sqrt{x^2+y^2})dxdy\\ & =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}d\theta {\int }_1^2\sin (\pi \rho )\rho d\rho \\ & =-\frac{3}{4}\end{array}$$

也可以用
$$\begin{array}{rl}\underset{D}{\iint }\frac{x\sin (\pi \sqrt{x^2+y^2})}{x+y}dxdy& ={\int }_0^{\frac{\pi }{2}}\frac{\cos \theta }{\cos \theta +\sin \theta }d\theta \cdot {\int }_1^2\rho \sin (\pi \rho )d\rho \\ 由于{\int }_0^{\frac{\pi }{2}}\frac{\cos \theta }{\cos \theta +\sin \theta }d\theta & ={\int }_0^{\frac{\pi }{2}}\frac{\sin \theta }{\cos \theta +\sin \theta }d\theta \\ & =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{\cos \theta +\sin \theta }{\cos \theta +\sin \theta }d\theta \\ & =\frac{\pi }{4}\\ {\int }_1^2\rho \sin (\pi \rho )d\rho & =\frac{1}{\pi }(-\rho \cos \pi \rho +\frac{1}{\pi }\sin \pi \rho ){|}_1^2\\ & =-\frac{3}{\pi }\\ 故\underset{D}{\iint }\frac{x\sin (\pi \sqrt{x^2+y^2})}{x+y}dxdy& =-\frac{3}{4}\end{array}$$

这里用到了区间再现公式，即
$${\int }_a^{b}f(x)dx\stackrel{x=a+b-t}{=}{\int }_a^{b}f(a+b-t)dt$$
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
$$\underset{D}{\iint }\frac{x\sin (\pi \sqrt{x^2+y^2})}{x+y}dxdy={\int }_0^{\frac{\pi }{2}}\frac{\cos \theta }{\cos \theta +\sin \theta }d\theta \cdot {\int }_1^2\rho \sin (\pi \rho )d\rho$$
显然满足$R(-\sin \theta ,-\cos \theta )=-R(\sin \theta ,\cos \theta )$，因此，令$u=\tan \theta$
$$\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}\frac{\cos \theta }{\cos \theta +\sin \theta }d\theta & ={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^2\theta }{1+\tan \theta }d\tan \theta \\ & ={\int }_0^{\frac{\pi }{2}}\frac{1}{({\tan }^2\theta +1)(1+\tan \theta )}d\tan \theta \\ & \stackrel{u=\tan \theta }{=}{\int }_0^{+\infty }\frac{1}{(1+u)(1+u^2)}du\\ & ={\int }_0^{+\infty }\left[\frac{A}{1+u}+\frac{Bu+C}{1+u^2}\right]du\\ & \Rightarrow \{\begin{array}{rl}& A+B=0\\ & B+C=0\\ & A+C=1\end{array}解得\{\begin{array}{rl}& A=\frac{1}{2}\\ & B=-\frac{1}{2}\\ & C=\frac{1}{2}\end{array}\\ & =\frac{1}{2}{\int }_0^{+\infty }\frac{1}{1+u}du+\frac{1}{2}{\int }_0^{+\infty }\frac{-u+1}{u^2+1}du\\ & =\frac{1}{2}\ln (1+u){|}_{u=+\infty }^{}-\frac{1}{2}{\int }_0^{+\infty }\frac{\frac{1}{2}d(u^2+1)-du}{u^2+1}\\ & =\frac{1}{2}\ln (1+u){|}_{u=+\infty }^{}-\frac{1}{4}\ln (u^2+1){|}_{u=+\infty }^{}+\frac{1}{2}\arctan u{|}_{u=+\infty }^{}\\ & =\frac{1}{4}\ln \left(\frac{u^2+2u+1}{u^2+1}\right){|}_{u=+\infty }^{}+\frac{\pi }{4}\\ & =\frac{1}{4}\ln \left(1+\frac{2u}{u^2+1}\right){|}_{u=+\infty }^{}+\frac{\pi }{4}\\ & =0+\frac{\pi }{4}=\frac{\pi }{4}\end{array}$$
后面计算同上

例9：设$D_k$是圆域$D=\left\{(x,y)|x^2+y^2\le 1\right\}$在第$k$象限的部分，记 $\begin{array}{r}{I}_k=\underset{D_k}{\iint }(y-x)dxdy(k=1,2,3,4)\end{array}$，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
$$\begin{array}{rl}{I}_1& =\underset{D_1}{\iint }(y-x)d\sigma \\ & =\underset{D_1}{\iint }(x-y)d\sigma \\ \underset{D_1}{\iint }(y-x)d\sigma & =-\underset{D_1}{\iint }(y-x)d\sigma =0\end{array}$$
同理$I_3=0$

例10：已知平面域$D=\left\{(x,y)||x|+|y|\le \frac{\pi }{2}\right\}$，记$\begin{array}{r}{I}_1=\underset{D}{\iint }\sqrt{x^2+y^2}d\sigma ,I_2=\underset{D}{\iint }\sin \sqrt{x^2+y^2}d\sigma ,I_3=\underset{D}{\iint }(1+\cos \sqrt{x^2+y^2})d\sigma \end{array}$，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
$$r>\sin r\ge {\sin }^{2}r=1-{\cos }^{2}r\ge 1-\cos r$$

画个图也行，不难