7.3 习题课讲解

7.3.1 确定积分区域的不等式表达、累次积分

解¹ 画出积分域的图像，如7.3.1所示，由图可知该区域可表示为$$\begin{array}{}\text{(2)}& 0\le y\le 1,\sqrt{y}\le x\le 3-2y\end{array}$$ 故原累次积分等于$$\begin{array}{}\text{(3)}& {\int }_0^1\mathrm{d}y{\int }_{\sqrt{y}}^{3-2y}f(x,y)\mathrm{d}x\end{array}$$ $◻$

解² 利用示性函数计算可得$$\begin{array}{}\text{(4)}& \begin{array}{rl}& {\int }_0^1\mathrm{d}x{\int }_0^{x^2}f(x,y)\mathrm{d}y+{\int }_1^3\mathrm{d}x{\int }_0^{\frac{3-x}{2}}f(x,y)\mathrm{d}y\\ & ={\int }_{{\mathbb{R}}^2}f(x,y)(1_{0\le x\le 1,0\le y\le x^2}+1_{1\le x\le 3,0\le y\le \frac{3-x}{2}})\mathrm{d}x\mathrm{d}y\\ & ={\int }_{{\mathbb{R}}^2}f(x,y)(1_{\sqrt{y}\le x\le 1}+1_{0\le y,1\le x\le min\{3,3-2y\}=3-2y})\mathrm{d}x\mathrm{d}y\\ & ={\int }_{{\mathbb{R}}^2}f(x,y)(1_{\sqrt{y}\le x\le 1}+1_{0\le y\le 1,1\le x\le 3-2y})\mathrm{d}x\mathrm{d}y\\ & ={\int }_{{\mathbb{R}}^2}f(x,y)\left(1_{0\le y\le 1,\sqrt{y}\le x\le 3-2y}\right)\mathrm{d}x\mathrm{d}y={\int }_0^1\mathrm{d}y{\int }_{\sqrt{y}}^{3-2y}f(x,y)\mathrm{d}x\end{array}\end{array}$$ 其中容易被忽视的是：由$\sqrt{y}\le 3-2y$解得$(2\sqrt{y}+3)(\sqrt{y}-1)\le 0$，即$0\le y\le 1$。 $◻$

注 解²使用了示性函数(Characteristic function)，其定义为$$\begin{array}{}\text{(5)}& 1_{(x,y)\in D}:{\mathbb{R}}^2\to \mathbb{R},1_{(x,y)\in D}=\{\begin{array}{ll}1,& (x,y)\in D\\ 0,& (x,y)\in {\mathbb{R}}^2\setminus D.\end{array}\end{array}$$ $1_{(x,y)\in D}$也可以简写为$1_D$。示性函数有以下性质：$$\begin{array}{}\text{(6)}& 1_{\varnothing }=0,1_{A^c}=1-1_A,1_{A\cap B}=1_A\cdot 1_B,1_{A\cup B}=1_A+1_B-1_A\cdot 1_B\end{array}$$ 即它可以把集合运算变成函数的代数运算。利用示性函数，我们可以把有界区域上的积分改写为全空间上的积分，从而使得重积分（累次积分）更容易转化为（另一次序的）累次积分，如$$\begin{array}{}\text{(7)}& {\int }_{D}f(x,y)\mathrm{d}x\mathrm{d}y={\int }_{{\mathbb{R}}^2}f(x,y)1_D\mathrm{d}x\mathrm{d}y={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}x{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f(x,y)1_{(x,y)\in D}\mathrm{d}y\end{array}$$ 对给定的$x$，记$D_x=\{y\mid (x,y)\in D\}$；记$A=\{x\mid D_x\ne \varnothing \}$。于是上述累次积分最终写成$$\begin{array}{}\text{(8)}& {\int }_A\mathrm{d}x{\int }_{D_x}f(x,y)\mathrm{d}y\end{array}$$ 通常$D$由不等式表达。集合$D_x$就是视$x$为给定的参数，求解关于变量$y$的不等式得到的解集。

当空间维数较高，或者区域边界很难通过画图分辨的时候，可以使用示性函数确定积分变量的范围。

解 由$x=0$、$y=0$、$x+y=1$知下界为$x\ge 0,y\ge 0$、上界为$x+y\le 1$，由此得到$$\begin{array}{}\text{(9)}& 0\le x\le 1-y,0\le y\le 1\end{array}$$ 再由$z=0$和$z=1+x+y$得到$$\begin{array}{}\text{(10)}& 0\le z\le 1+x+y\end{array}$$ 因此区域可表示为$$\begin{array}{}\text{(11)}& D=\{(x,y,z)\mid x\ge 0,y\ge 0,x+y\le 1,0\le z\le 1+x+y\}\end{array}$$ 按累次积分顺序，依次写出$x,y,z$满足的不等式：$$\begin{array}{}\text{(12)}& max\{0,z-1-y\}\le x\le 1-y\end{array}$$ 由此得到$$\begin{array}{}\text{(13)}& max\{0,z-1-y\}\le 1-y\end{array}$$ 即 $y\le 1,z\le 2$，再结合$D$的其他不等式得到$$\begin{array}{}\text{(14)}& 0\le y\le 1,0\le z\le 2\end{array}$$ 写成给定顺序的累次积分为$$\begin{array}{}\text{(15)}& \begin{array}{rl}& {\int }_0^2\mathrm{d}z{\int }_0^1\mathrm{d}y{\int }_{max\{0,z-1-y\}}^{1-y}\cdots \mathrm{d}x\\ & ={\int }_0^2\mathrm{d}z{\int }_{max\{0,z-1\}}^1\mathrm{d}y{\int }_0^{1-y}\cdots \mathrm{d}x+{\int }_0^2\mathrm{d}z{\int }_0^{min\{1,z-1\}}\mathrm{d}y{\int }_{z-1-y}^{1-y}\cdots \mathrm{d}x\\ & ={\int }_0^1\mathrm{d}z{\int }_0^1\mathrm{d}y{\int }_0^{1-y}\cdots \mathrm{d}x+{\int }_1^2\mathrm{d}z{\int }_{z-1}^1\mathrm{d}y{\int }_0^{1-y}\cdots \mathrm{d}x+{\int }_1^2\mathrm{d}z{\int }_0^{z-1}\mathrm{d}y{\int }_{z-1-y}^{1-y}\cdots \mathrm{d}x\end{array}\end{array}$$ 利用示性函数法（通过示性函数求解积分变量的范围）：$$\begin{array}{}\text{(16)}& \begin{array}{rl}& {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}z{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}y{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\cdots I_{x\ge 0,y\ge 0,x+y\le 1,0\le z\le 1+x+y}\mathrm{d}x\\ & ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}z{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}y{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\cdots I_{max\{0,z-1-y\}\le x\le 1-y,y\ge 0,0\le z}\mathrm{d}x\\ & ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}z{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}y{\int }_{max\{0,z-1-y\}}^{1-y}\cdots I_{max\{0,z-1-y\}\le 1-y,y\ge 0,0\le z}\mathrm{d}x\\ & ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}z{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{I}_{0\le 1-y,z-1-y\le 1-y,y\ge 0,0\le z}\mathrm{d}y{\int }_{max\{0,z-1-y\}}^{1-y}\cdots \mathrm{d}x\\ & ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{d}z{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{I}_{y\le 1,z\le 2,y\ge 0,0\le z}\mathrm{d}y{\int }_{max\{0,z-1-y\}}^{1-y}\cdots \mathrm{d}x\\ & ={\int }_0^2\mathrm{d}z{\int }_0^1\mathrm{d}y{\int }_{max\{0,z-1-y\}}^{1-y}\cdots \mathrm{d}x\end{array}\end{array}$$ 也可以利用图形和割补法（图7.3.2）：$$\begin{array}{}\text{(17)}& {\int }_0^2\mathrm{d}z{\int }_0^1\mathrm{d}y{\int }_0^{1-y}\cdots \mathrm{d}x-{\int }_1^2\mathrm{d}z{\int }_0^{z-1}\mathrm{d}y{\int }_0^{z-1-y}\cdots \mathrm{d}x\end{array}$$ 更好的累次积分方式为$$\begin{array}{}\text{(18)}& {\int }_0^1\mathrm{d}x{\int }_0^{1-x}\mathrm{d}y{\int }_0^{1+x+y}\cdots \mathrm{d}z\end{array}$$ $◻$

解 本题结论不唯一。如图7.3.3所示，区域1：由于$$\begin{array}{}\text{(19)}& 0\le z\le a-x-y\end{array}$$ 从而$x+y\le a$。由于$$\begin{array}{}\text{(20)}& 0\le x\le min\{a-y,\sqrt{R^2-y^2}\},0\le y\le R\end{array}$$ 结合$(x+y{)}^2\le 2(x^2+y^2)\le 2R^2<a^2$可得$$\begin{array}{}\text{(21)}& 0\le x\le \sqrt{R^2-y^2},0\le y\le R,0\le z\le a-x-y\end{array}$$ 故相应的累次积分为$$\begin{array}{}\text{(22)}& {\int }_{x^2+y^2\le R^2,x\ge 0,y\ge 0}\mathrm{d}x\mathrm{d}y{\int }_0^{a-x-y}\cdots \mathrm{d}z\end{array}$$ 采用以$z$轴为对称轴的圆柱坐标系，则可表示为$$\begin{array}{}\text{(23)}& {\int }_0^{R}r\mathrm{d}r{\int }_0^{\pi /2}\mathrm{d}\theta {\int }_0^{a-r\cos \theta -r\sin \theta }f(r^2,z)\mathrm{d}z\end{array}$$

区域2：由于$$\begin{array}{}\text{(24)}& 0\le z\le a-x-y,x^2+y^2\ge R^2,x+y\le a\end{array}$$ 相应的累次积分为$$\begin{array}{}\text{(25)}& {\int }_{x^2+y^2\ge R^2,x+y\le a}\mathrm{d}x\mathrm{d}y{\int }_0^{a-x-y}\cdots \mathrm{d}z\end{array}$$ 采用以$z$轴为对称轴的圆柱坐标系，则可表示为：$$\begin{array}{}\text{(26)}& {\int }_0^{\frac{\pi }{2}}\mathrm{d}\theta {\int }_R^{\frac{a}{\cos \theta +\sin \theta }}r\mathrm{d}r{\int }_0^{a-r\cos \theta -r\sin \theta }f(r^2,z)\mathrm{d}z\end{array}$$ $◻$

解 如图7.3.4所示，利用示性函数计算可得$$\begin{array}{}\text{(28)}& \begin{array}{rl}& {\int }_{0\le x\le 1,0\le y\le 1-x,0\le z\le x+y}f(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z\\ & ={\int }_{0\le x\le 1,x\le 1-y,z-y\le x,y\ge 0,z\ge 0}f(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z\\ & ={\int }_{y\ge 0,z\ge 0,max\{0,z-y\}\le 1-y}\mathrm{d}y\mathrm{d}z{\int }_{max\{0,z-y\}}^{1-y}f(x,y,z)\mathrm{d}x\\ & ={\int }_{0\le y\le 1,0\le z\le 1}\mathrm{d}y\mathrm{d}z{\int }_{max\{0,z-y\}}^{1-y}f(x,y,z)\mathrm{d}x\\ & ={\int }_0^1\mathrm{d}y{\int }_0^1\mathrm{d}z{\int }_{max\{0,z-y\}}^{1-y}f(x,y,z)\mathrm{d}x\\ & ={\int }_0^1\mathrm{d}y{\int }_y^1\mathrm{d}z{\int }_{z-y}^{1-y}f(x,y,z)\mathrm{d}x+{\int }_0^1\mathrm{d}y{\int }_0^y\mathrm{d}z{\int }_0^{1-y}f(x,y,z)\mathrm{d}x\end{array}\end{array}$$ 另一种写法是$$\begin{array}{}\text{(29)}& {\int }_0^1\mathrm{d}z{\int }_z^1\mathrm{d}y{\int }_0^{1-y}f(x,y,z)\mathrm{d}x+{\int }_0^1\mathrm{d}z{\int }_0^z\mathrm{d}y{\int }_{z-y}^{1-y}f(x,y,z)\mathrm{d}x\end{array}$$ $◻$

7.3.2 重积分换元

解 (1) 令$u=x+y$、$v=x-y$，则$$\begin{array}{}\text{(30)}& det\frac{\partial (u,v)}{\partial (x,y)}=\left|\begin{array}{cc}1& 1\\ 1& -1\end{array}\right|=-2\end{array}$$ 计算可得$$\begin{array}{}\text{(31)}& {\int }_D(x+y)\sin (x-y)\mathrm{d}x\mathrm{d}y={\int }_{[0,\pi {]}^2}u\sin v|det\frac{\partial (x,y)}{\partial (u,v)}|\mathrm{d}u\mathrm{d}v=\frac{1}{2}{\int }_0^{\pi }u\mathrm{d}u{\int }_0^{\pi }\sin v\mathrm{d}v=\frac{{\pi }^2}{2}\end{array}$$

(2) 令$u=\frac{y}{x+y}$、$v=x+y$，则$y=uv$、$x=v-uv$。由于$$\begin{array}{}\text{(32)}& v\le 1,v-uv\ge 0,uv\ge 0\end{array}$$ 故有$0\le v\le 1,0\le u\le 1$，因此$$\begin{array}{}\text{(33)}& \mathrm{d}x\mathrm{d}y=|det\frac{\partial (x,y)}{\partial (u,v)}|\mathrm{d}u\mathrm{d}v=|det\left(\begin{array}{cc}-v& 1-u\\ v& u\end{array}\right)|\mathrm{d}u\mathrm{d}v=v\mathrm{d}u\mathrm{d}v\end{array}$$ 计算可得$$\begin{array}{}\text{(34)}& {\int }_D{\mathrm{e}}^{\frac{y}{x+y}}\mathrm{d}x\mathrm{d}y={\int }_{[0,1{]}^2}{\mathrm{e}}^{u}v\mathrm{d}u\mathrm{d}v=\frac{1}{2}(\mathrm{e}-1)\end{array}$$ $◻$

解¹ 积分区域可表示为$$\begin{array}{}\text{(36)}& x^2+y^2\le 1,x\ge 0,y\ge 0,\sqrt{x^2+y^2}\le z\le \sqrt{2-x^2-y^2}\end{array}$$ 利用柱坐标系可表示为$$\begin{array}{}\text{(37)}& 0\le r\le 1,0\le \theta \le \frac{\pi }{2},r\le z\le \sqrt{2-r^2}\end{array}$$ 计算可得$$\begin{array}{}\text{(38)}& \begin{array}{rl}& {\int }_0^1\mathrm{d}x{\int }_0^{\sqrt{1-x^2}}\mathrm{d}y{\int }_{\sqrt{x^2+y^2}}^{\sqrt{2-x^2-y^2}}{z}^2\mathrm{d}z={\int }_0^{\frac{\pi }{2}}\mathrm{d}\theta {\int }_0^{1}r\mathrm{d}r{\int }_r^{\sqrt{2-r^2}}{z}^2\mathrm{d}z\\ & =\frac{\pi }{2}{\int }_0^1\frac{r{(2-r^2)}^{\frac{3}{2}}}{3}-\frac{r^4}{3}\mathrm{d}r=\frac{\pi (2\sqrt{2}-1)}{15}\end{array}\end{array}$$ $◻$

解² 利用球坐标系可表示为$$\begin{array}{}\text{(39)}& R\sin \theta \le 1,0\le \phi \le \frac{\pi }{2},\sin \theta \le \cos \theta ,0\le R\le 2\end{array}$$ 亦即$$\begin{array}{}\text{(40)}& R\sin \theta \le 1,0\le \phi \le \frac{\pi }{2},0\le \theta \le \frac{\pi }{4},0\le R\le \sqrt{2}\end{array}$$ 计算可得$$\begin{array}{}\text{(41)}& \begin{array}{rl}& {\int }_0^{\frac{\pi }{2}}\mathrm{d}\phi {\int }_0^{\frac{\pi }{4}}\mathrm{d}\theta {\int }_0^{min\{\sqrt{2},\frac{1}{\sin \theta }\}}(R\cos \theta {)}^2{R}^2\sin \theta \mathrm{d}R\\ & =\frac{\pi }{2}\cdot \frac{4\sqrt{2}}{5}\cdot \frac{1-\frac{\sqrt{2}}{4}}{3}=\frac{\pi (2\sqrt{2}-1)}{15}\end{array}\end{array}$$ $◻$

解 用柱坐标系可得$$\begin{array}{}\text{(43)}& F(t)={\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^t\mathrm{d}r{\int }_0^h[z^2+f\left(r^2\right)]r\mathrm{d}z=\frac{\pi t^2{h}^3}{3}+\pi hG\left(t^2\right)\end{array}$$ 其中$G$是$f$的原函数，满足$G(0)=0$。用L’Hôpital法则可得$$\begin{array}{}\text{(44)}& \underset{t\to 0^{+}}{lim}\frac{F(t)}{t^2}=\frac{\pi h^3}{3}+\pi h\underset{t\to 0^{+}}{lim}\frac{G\left(t^2\right)}{t^2}=\frac{\pi h^3}{3}+\pi h{G}^{\prime }(0)=\frac{\pi h^3}{3}+\pi hf(0)\end{array}$$ $◻$

解 如图7.3.6所示，区域关于$(x,y)\to (-x,-y)$对称，被积函数中$x+y$关于这个对称是奇函数，所以$$\begin{array}{}\text{(46)}& I={\int }_{\mathrm{\Omega }}z\mathrm{d}x\mathrm{d}y\mathrm{d}z\end{array}$$ 球坐标系：$0\le r\le 1$、$0\le \theta \le \frac{\pi }{4}$且$\cos \theta \ge \sin \theta$，计算可得$$\begin{array}{}\text{(47)}& I={\int }_0^{\pi /4}\mathrm{d}\theta {\int }_0^{2\pi }\mathrm{d}\phi {\int }_0^{1}r\cos \theta r^2\sin \theta \mathrm{d}r=\frac{\pi }{8}\end{array}$$ 柱坐标系（绕$z$轴）：$r\le z\le \sqrt{1-r^2}$，计算可得$$\begin{array}{}\text{(48)}& I={\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^{\frac{\sqrt{2}}{2}}r\mathrm{d}r{\int }_r^{\sqrt{1-r^2}}z\mathrm{d}z=\frac{\pi }{8}\end{array}$$ 你会选择直角坐标系吗？为什么？ $◻$

解 如图7.3.7所示，这是绕$z$轴旋转的曲面，取柱坐标$$\begin{array}{}\text{(49)}& x=r\cos \theta ,y=r\sin \theta ,z=z\end{array}$$ 代入曲面方程得到$r^4+z^4=z^2$。为让$(x,y)$有界，故$\mathrm{\Omega }$的表示为$$\begin{array}{}\text{(50)}& r^4\le z^2-z^4,z^2\le 1\end{array}$$ 从而$$\begin{array}{}\text{(51)}& -1\le z\le 1,0\le r\le \sqrt[4]{z^2(1-z^2)},0\le \theta \le 2\pi \end{array}$$ 计算可得$$\begin{array}{}\text{(52)}& |\mathrm{\Omega }|={\int }_0^{2\pi }\mathrm{d}\theta {\int }_{-1}^1\mathrm{d}z{\int }_0^{{\left(z^2(1-z^2)\right)}^{1/4}}r\mathrm{d}r=2\pi {\int }_0^{1}z\sqrt{1-z^2}\mathrm{d}z=\frac{2\pi }{3}\end{array}$$ 如果使用直角坐标系，则$$\begin{array}{}\text{(53)}& \begin{array}{rl}|\mathrm{\Omega }|& ={\int }_{-1}^1\mathrm{d}z{\int }_{x^2+y^2\le \sqrt{z^2-z^4}}\mathrm{d}x\mathrm{d}y={\int }_{-1}^1\pi \sqrt{z^2(1-z^2)}\mathrm{d}z\\ & =2\pi {\int }_0^{1}z\sqrt{1-z^2}\mathrm{d}z=\pi {\int }_1^{0}u\mathrm{d}(1-u^2)=\frac{2\pi }{3},u=\sqrt{1-z^2}\end{array}\end{array}$$ $◻$

解 由于$A$对称正定，因此存在可逆矩阵$P$，使得$A=P^{\mathrm{T}}P$。在线性变换$\mathit{y}=P\mathit{x}$下，有$$\begin{array}{}\text{(54)}& \mathrm{d}{y}_1\mathrm{d}{y}_2\mathrm{d}{y}_3=|detP|\mathrm{d}{x}_1\mathrm{d}{x}_2\mathrm{d}{x}_3=\sqrt{detA}\mathrm{d}{x}_1\mathrm{d}{x}_2\mathrm{d}{x}_3\end{array}$$ 注意到$$\begin{array}{}\text{(55)}& {\mathit{y}}^{\mathrm{T}}\mathit{y}={\mathit{x}}^{\mathrm{T}}{P}^{\mathrm{T}}P\mathit{x}={\mathit{x}}^{\mathrm{T}}A\mathit{x}=1\end{array}$$ 于是$$\begin{array}{}\text{(56)}& |V|={\int }_V\mathrm{d}{x}_1\mathrm{d}{x}_2\mathrm{d}{x}_3={\int }_{{\mathit{y}}^{\mathrm{T}}\mathit{y}\le 1}\frac{1}{\sqrt{detA}}\mathrm{d}{y}_1\mathrm{d}{y}_2\mathrm{d}{y}_3=\frac{4\pi }{3\sqrt{detA}}\end{array}$$ $◻$

解 令$$\begin{array}{}\text{(57)}& u=3x-y-z,v=-x+3y-z,w=-x-y+3z⟹det\frac{\partial (u,v,w)}{\partial (x,y,z)}=16\end{array}$$ 于是所求体积为$$\begin{array}{}\text{(58)}& |V|={\int }_V\mathrm{d}x\mathrm{d}y\mathrm{d}z={\int }_{|u|,|v|,|w|\le 1}|det\frac{\partial (x,y,z)}{\partial (u,v,w)}|\mathrm{d}u\mathrm{d}v\mathrm{d}w=\frac{1}{16}{2}^3=\frac{1}{2}\end{array}$$ $◻$

解 作正交变换$$\begin{array}{}\text{(60)}& \left(\begin{array}{c}u\\ v\\ w\end{array}\right)=P\left(\begin{array}{c}x\\ y\\ z\end{array}\right),u=\frac{ax+by+cz}{h}\end{array}$$ 则$$\begin{array}{}\text{(61)}& u^2+v^2+w^2=x^2+y^2+z^2\le 1,|det\frac{\partial (u,v,w)}{\partial (x,y,z)}|=|detP|=1\end{array}$$ 计算可得$$\begin{array}{}\text{(62)}& \begin{array}{rl}& {\int }_{x^2+y^2+z^2\le 1}f(ax+by+cz)\mathrm{d}x\mathrm{d}y\mathrm{d}z={\int }_{u^2+v^2+w^2\le 1}f(hu)\mathrm{d}u\mathrm{d}v\mathrm{d}w\\ & ={\int }_{-1}^{1}f(hu)\mathrm{d}u{\int }_{v^2+w^2\le 1-u^2}\mathrm{d}v\mathrm{d}w=\pi {\int }_{-1}^{1}f(hu)(1-u^2)\mathrm{d}u.\end{array}\end{array}$$ $◻$

解¹ 在极坐标下，利用方向导数、微分（及微分的形式不变性）和偏导数的关系可得$$\begin{array}{}\text{(64)}& \begin{array}{rl}& \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}-\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}=\mathrm{d}f(x,y)(\genfrac{}{}{0ex}{}{\frac{y}{\sqrt{x^2+y^2}}}{-\frac{x}{\sqrt{x^2+y^2}}})\\ & =\mathrm{d}f(r\cos \theta ,r\sin \theta )(\genfrac{}{}{0ex}{}{\sin \theta }{-\cos \theta })=\frac{1}{r}\frac{\partial f(r\cos \theta ,r\sin \theta )}{\partial \theta }\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(65)}& I=-{\int }_0^R\mathrm{d}r{\int }_0^{2\pi }\frac{\partial }{\partial \theta }f(r\cos \theta ,r\sin \theta )\mathrm{d}\theta =-{{\int }_0^{R}f(r\cos \theta ,r\sin \theta )|}_{\theta =0}^{2\pi }\mathrm{d}r=0\end{array}$$ $◻$

解² 由Green公式可得$$\begin{array}{}\text{(66)}& \begin{array}{rl}I& =-{\int }_D[\frac{\partial }{\partial x}\left(\sqrt{x^2+y^2}\frac{\partial f}{\partial y}\right)-\frac{\partial }{\partial y}\left(\sqrt{x^2+y^2}\frac{\partial f}{\partial x}\right)]\mathrm{d}x\mathrm{d}y\\ & =-{\oint }_{\partial D}\sqrt{x^2+y^2}(\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y)=-R{\oint }_{\partial D}\mathrm{d}f=0\end{array}\end{array}$$ $◻$

7.3.3 *重积分在概率中的应用

解 设$Z:=X+Y$，依“概率不变原则”可得$$\begin{array}{}\text{(72)}& \mathrm{d}P=f_{X,Z}(x,z)\mathrm{d}\mu (x,z)=f_{X,Y}(x,y)\mathrm{d}\mu (x,y)\end{array}$$ 因此$$\begin{array}{}\text{(73)}& f_{X,Z}(x,z)=f_{X,Y}(x,y)|det\frac{\partial (x,y)}{\partial (x,z)}|=f_X(x)f_Y(z-x)\end{array}$$ 故有$$\begin{array}{}\text{(74)}& f_Z(z)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{f}_X(x)f_Y(z-x)\mathrm{d}x=(f_X\ast f_Y)(z)\end{array}$$ 其中$f_X\ast f_Y$表示$f_X$和$f_Y$的卷积。同理可得$$\begin{array}{}\text{(75)}& f_{X-Y}(z)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{f}_X(x)f_Y(x-z)\mathrm{d}x\end{array}$$ $◻$