Eufisky - The lost book

含奇点的第二型曲面积分计算

谢惠民下册上的一道题:求

$$I=\iint_{\Sigma}{\left( x^3+\frac{1}{x} \right) dydz+\left( y^2-xz \right) dzdx+\left( z^3+\frac{z}{x^2} \right) dxdy},$$其中$\Sigma$是球面$x^2+y^2+z^2=2z$,取外侧.

解.注意到球面上的圆$x=0,y^2+z^2=2z$是上述积分的奇点,我们考察两半球$\Sigma_1:(x-\varepsilon)^2+y^2+(z-1)^2=1,x\geq\varepsilon$和$\Sigma_2:(x+\varepsilon)^2+y^2+(z-1)^2=1,x\leq -\varepsilon$, 其中$\varepsilon$为足够小的正数.并记$\Gamma_1$为圆盘$x=\varepsilon,y^2+(z-1)^2=1$,而$\Gamma_2$为圆盘$x=-\varepsilon,y^2+(z-1)^2=1$.

利用球的极坐标方程

\[x=\varepsilon+r\sin \varphi \cos \theta ,y=r\sin \varphi \sin \theta ,z=r\cos \varphi +1,\quad 0\leq \varphi\leq \pi,-\pi/2\leq\theta\leq \pi/2,0\leq r\leq 1\]

以及

\[x=-\varepsilon+r\sin \varphi \cos \theta ,y=r\sin \varphi \sin \theta ,z=r\cos \varphi +1,\quad 0\leq \varphi\leq \pi,\pi/2\leq\theta\leq 3\pi/2,0\leq r\leq 1\]

由Gauss公式可知

\begin{align*}I_{11}&=\iiint_{D_1}{\left( 3x^2+2y+3z^2 \right) dxdydz}\\&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{d\theta \int_0^{\pi}{d\varphi \int_0^1{\left[ 3\left( \varepsilon +r\sin \varphi \cos \theta \right) ^2+2r\sin \varphi \sin \theta +3\left( r\cos \varphi +1 \right) ^2 \right] r^2\sin \varphi dr}}}\\&=2\left( \varepsilon ^2+1 \right) \pi +\frac{3\pi}{2}\varepsilon +\frac{4\pi}{5}.\end{align*}
\begin{align*}I_{12}&=\iint_{\Gamma _1}{\left( x^3+\frac{1}{x} \right) dydz+\left( y^2-xz \right) dzdx+\left( z^3+\frac{z}{x^2} \right) dxdy}\\&=\iint_{\Gamma _1}{\left( \varepsilon ^3+\frac{1}{\varepsilon} \right) dydz}=\left( \varepsilon ^3+\frac{1}{\varepsilon} \right) \pi .\end{align*}
\begin{align*}I_{21}&=\iiint_{D_2}{\left( 3x^2+2y+3z^2 \right) dxdydz}\\&=\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{d\theta \int_0^{\pi}{d\varphi \int_0^1{\left[ 3\left( -\varepsilon +r\sin \varphi \cos \theta \right) ^2+2r\sin \varphi \sin \theta +3\left( r\cos \varphi +1 \right) ^2 \right] r^2\sin \varphi dr}}}\\&=2\left( \varepsilon ^2+1 \right) \pi +\frac{3\pi}{2}\varepsilon +\frac{4\pi}{5}\end{align*}
\begin{align*}I_{22}&=\iint_{\Gamma _2}{\left( x^3+\frac{1}{x} \right) dydz+\left( y^2-xz \right) dzdx+\left( z^3+\frac{z}{x^2} \right) dxdy}\\&=\iint_{\Gamma _2}{\left( -\varepsilon ^3-\frac{1}{\varepsilon} \right) dydz}=-\left( \varepsilon ^3+\frac{1}{\varepsilon} \right) \pi.\end{align*}
因此$$I‘=I_{11}+I_{21}-I_{12}-I_{22}=4\left( \varepsilon ^2+1 \right) \pi +3\pi \varepsilon +\frac{8\pi}{5}\rightarrow \frac{28}{5}\pi,$$即$I=\frac{28}{5}\pi$.
设$J$为关于$x\left( t \right) $和$t$的连续函数,满足
$$\frac{\partial J}{\partial t}=\frac{1}{4}\left( \frac{\partial J}{\partial x} \right) ^2-x^2-\frac{1}{2}x^4,\qquad \text{其中}J\left[ x\left( 1 \right) ,1 \right] =0$$
求$J\left[ x\left( t \right) ,t \right]$.
 
关于 I will not change, no matter how U change … 
笙歌姐,这句话何解?
 
文科生:“不论你怎么移情别恋,我是不会变心的”理科生:“电流不随电压的变化而变化。”

I am here,because U are here.

$$IR\cdot \frac{\varepsilon S}{4\pi kd}\cdot \lim_{n\rightarrow \infty}\frac{\prod_{k=1}^n{k^k}}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}\cdot k\ln W$$
Glaisher-Kinkelin constant

二重积分计算

求$$\iint_D \sin x\sin y(\sin x+\sin y)e^{\sin x\sin y}d\sigma,$$其中$D:0<x<\pi/2,0<y<\pi/2$.

解.首先把待求积分写成

\begin{align*}&\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin x\sin y\left( \sin x+\sin y \right) \text{e}^{\sin x\sin y}\text{d}x\text{d}y}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\sin y\text{e}^{\sin x\sin y}\text{d}x\text{d}y}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\cos y\text{e}^{\sin x\cos y}\text{d}x\text{d}y}}\\=&2\iint_S{x\text{e}^x\text{d}x\text{d}y},\end{align*}
其曲面$S$是单位球面在第一象限的部分.
考虑到平面$x=0$处距离为$x$的宽度为$\mathrm{d}x$的球面窄条的面积,相当于是底边长为$y=\sqrt{1-x^2}$,宽为$\mathrm{d}s=\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\mathrm{d}x=\frac{\mathrm{d}x}{\sqrt{1-x^2}}$,
因此
$$I=2\int_0^1{x\text{e}^x\frac{\pi}{2}\sqrt{1-x^2}\frac{\text{d}x}{\sqrt{1-x^2}}}=\pi \int_0^1{x\text{e}^x\text{d}x}=\pi.$$
解法二.利用分部积分及对称性可知
\begin{align*}&\iint_D{\sin x\sin y\left( \sin x+\sin y \right) e^{\sin x\sin y}d\sigma}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\sin ye^{\sin x\sin y}dxdy}}=2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\left( 1-\cos ^2x \right) \sin ye^{\sin x\sin y}dxdy}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\cos x\frac{\partial e^{\sin x\sin y}}{\partial x}dxdy}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\left[ \cos x\left. e^{\sin x\sin y} \right|_{0}^{\pi /2}-\left( \int_0^{\frac{\pi}{2}}{\left( -\sin x \right) e^{\sin x\sin y}dx} \right) \right] dy}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\left( -1 \right) dy}-2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin xe^{\sin x\sin y}dxdy}}\\=&\pi .\end{align*}

 

两道积分习题

1.设$f:[0,1]\to\mathbb{R}$连续,求极限$$\lim\limits_{n\rightarrow \infty}\int_0^1\int_0^1\cdots\int_0^1 f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)dx_1dx_2\cdots dx_n.$$

解法一.设$|f|$最大值为$M$.对任何$\varepsilon>0$,存在$\delta>0$,使得当$|x-1/2|<\delta$时,有$$\left|f(x)-f(\frac{1}{2})\right|<\varepsilon.$$

\begin{align*}&\int_{[0,1]^n}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\\leq &\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\+&\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|<\delta}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\\leq&2M\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}dx_1dx_2\cdots dx_n+\varepsilon\\\leq&\frac{2M}{\delta^2}\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|^2dx_1dx_2\cdots dx_n+\varepsilon\\\leq&\frac{2M}{\delta^2}\int_{[0,1]^n}\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|^2dx_1dx_2\cdots dx_n+\varepsilon\\=&\frac{M}{6n\delta^2}+\varepsilon.\end{align*}

因此$$\limsup_{n\rightarrow\infty}\int_{[0,1]^n}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\leq \varepsilon.$$

令$\varepsilon\rightarrow0$即可.

 

解法二.由科尔莫格罗夫强大数定律得$$\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}\mathop \to \limits^{a.s.} E\left( {{X_i}} \right) = \frac{1}{2}\left( {n \to + \infty } \right).$$

又因为$f(x)$连续有界,由控制收敛定理可知

$$\mathop {\lim }\limits_{n \to \infty } E\left( {f\left( {\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = E\left( {\mathop {\lim }\limits_{n \to \infty } f\left( {\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = E\left( {f\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = f\left( {\frac{1}{2}} \right).$$


2.求证$$\int_0^1\prod_{n=1}^\infty(1-x^n)dx=\frac{4\pi\sqrt3}{\sqrt{23}}\frac{\sinh\frac{\pi\sqrt{23}}3}{\cosh\frac{\pi\sqrt{23}}2}.$$

解.注意到Pentagonal number theorem,我们知$$\int_{0}^{1}\prod_{n\geq1}\left(1-x^{n}\right)dx=\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}\int_{0}^{1}x^{k\left(3k-1\right)/2}dx=\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}\frac{2}{3k^{2}-k+2}.$$再利用求和公式可知$$\sum_{n\in\mathbb{Z}}\left(-1\right)^{n}f\left(n\right)=-\sum\left\{ \pi\csc\left(\pi z\right)f(z) \textrm{ 在 } f\left(z\right)\textrm{ 的极点上的留数}\right\}.$$而极点为$z=\frac{1}{6}\left(1\pm i\sqrt{23}\right)$,由此求得.

2013武大数分压轴题

(13年武大数分)求$\displaystyle I = \iint\limits_\Sigma  {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{3}{2}}}{{\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} + \frac{{{z^2}}}{{{c^4}}}} \right)}^{ - \frac{1}{2}}}dS} $,其中$\sum$为椭球面: $\displaystyle \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1(a,b,c>0)$.


下面是自己的解答:

令$x = a\sin \varphi \cos \theta ,y = b\sin \varphi \sin \theta ,z = c\cos \varphi $,其中$0\leq \theta\leq 2\pi,0\leq\varphi \leq\pi$,经计算得到
\[\frac{{\partial \left( {y,z} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = bc{\sin ^2}\varphi \cos \theta ,\frac{{\partial \left( {z,x} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = ac{\sin ^2}\varphi \sin \theta ,\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = ab\sin \varphi \cos \varphi ,\]
所以
\begin{align*}EG - {F^2} &= {\left( {\frac{{\partial \left( {y,z} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2} + {\left( {\frac{{\partial \left( {z,x} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2} + {\left( {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2}\\& = {\left( {abc} \right)^2}{\sin ^2}\varphi \left( {\frac{{{{\sin }^2}\varphi {{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\varphi {{\sin }^2}\theta }}{{{b^2}}} + \frac{{{{\cos }^2}\varphi }}{{{c^2}}}} \right).\end{align*}
而这时被积函数化为
\begin{align*}&{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}{\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} + \frac{{{z^2}}}{{{c^4}}}} \right)^{ - \frac{1}{2}}}\\= &{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)^{ - \frac{3}{2}}}\\&{\left( {\frac{{{{\sin }^2}\varphi {{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\varphi {{\sin }^2}\theta }}{{{b^2}}} + \frac{{{{\cos }^2}\varphi }}{{{c^2}}}} \right)^{ - \frac{1}{2}}}.\end{align*}
因此
\[I = abc\iint\limits_{\left[ {0,\pi } \right] \times \left[ {0,2\pi } \right]} {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}\sin \varphi d\varphi d\theta } \]
 
注意到这么一个事实,当$M+Nx^2$不取$0$且$M\neq 0$时,我们有
\[\int {{{\left( {M + N{x^2}} \right)}^{ - 3/2}}dx}  = \frac{1}{M} \cdot \frac{x}{{\sqrt {M + N{x^2}} }} + C.\]
 
\begin{align*}I &= abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}\sin \varphi d\varphi } \\&=  - abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}d\left( {\cos \varphi } \right)} \\&=  - abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left[ {\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right) + \left( {{c^2} - {a^2}{{\cos }^2}\theta  - {b^2}{{\sin }^2}\theta } \right){{\cos }^2}\varphi } \right]}^{ - \frac{3}{2}}}d\left( {\cos \varphi } \right)} \\&= abc\int_0^{2\pi } {d\theta } \int_{ - 1}^1 {{{\left[ {\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right) + \left( {{c^2} - {a^2}{{\cos }^2}\theta  - {b^2}{{\sin }^2}\theta } \right){x^2}} \right]}^{ - \frac{3}{2}}}dx} \\&= abc\int_0^{2\pi } {\frac{2}{{\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right)c}}d\theta }  = 4ab\int_0^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta } .\end{align*}
\begin{align*}&\int_0^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  = \int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  + \int_{\frac{\pi }{2}}^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta } \\= &\int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  + \int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta }}d\theta } \\= &\int_0^{ + \infty } {\frac{1}{{{a^2} + {b^2}{x^2}}}d}  + \int_0^{ + \infty } {\frac{1}{{{a^2}{x^2} + {b^2}}}dx}  = \frac{1}{{ab}}\left. {\arctan \left( {\frac{b}{a}x} \right)} \right|_0^{ + \infty } + \frac{1}{{ab}}\left. {\arctan \left( {\frac{a}{b}x} \right)} \right|_0^{ + \infty }\\= &\frac{\pi }{{ab}}.\end{align*}
进而得到
\[I = 4ab\int_0^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  = 4\pi .\]

另外有更好的方法:(Hansschwarzkopf)

注意到$\Sigma$ 在点$(x,y,z)$处的单位外法向量是
$$n=\frac{\left(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}\right)}{\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}},$$
且$1=x\cdot \frac{x}{a^2}+y\cdot\frac{y}{b^2}+z\cdot\frac{z}{c^2}$.
从而原积分可写成第二型曲面积分
$$\iint\limits_\Sigma \frac{xdydz+yd zd x+zdxdy}{\sqrt{(x^2+y^2+z^2)^3}}.$$
作小球面$S_\varepsilon: x^2+y^2+z^2=\varepsilon^2$. 运用Gauss公式可知
$$\iint\limits_\Sigma \frac{xd yd z+yd zd x+zd xd y}{\sqrt{(x^2+y^2+z^2)^3}} =\iint\limits_{S_\varepsilon} \frac{xdyd z+ydzd x+zd xd y}{\sqrt{(x^2+y^2+z^2)^3}}=4\pi.$$ 即
$$\iint\limits_\Sigma\frac{d S}{\sqrt{(x^2+y^2+z^2)^3}\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}}=4\pi.$$

 

谢惠民一道全微分题

前几天徐半仙问了我谢惠民下册P325页上一道难度稍大的全微分题目,利用今晚美好的独处时间(笑哭),做了下,解答如下:

对于以下一阶微分形式$\omega $,求函数$M(x,y)\neq0$, 使得在适当的区域内$M\omega $为全微分,并求其原函数:

(1) $\displaystyle \omega = \left[ { - y\sqrt {{x^2} + {y^2} + 1} - x\left( {{x^2} + {y^2}} \right)} \right]dx + \left[ {x\sqrt {{x^2} + {y^2} + 1} - y\left( {{x^2} + {y^2}} \right)} \right]dy$;

 

(2) $\displaystyle \omega = x\left[ {{{\left( {ay + bx} \right)}^3} + a{y^3}} \right]dx + y\left[ {{{\left( {ay + bx} \right)}^3} + b{x^3}} \right]dy$.

解:(1)取\[M = \frac{1}{{\left( {{x^2} + {y^2}} \right)\sqrt {{x^2} + {y^2} + 1} }},\]我们有

\[M\omega = \left( { - \frac{y}{{{x^2} + {y^2}}} - \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dx + \left( {\frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dy.\]

则有$P = - \frac{y}{{{x^2} + {y^2}}} - \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }},Q = \frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}$,且\[\frac{{\partial P}}{{\partial y}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} + \frac{{xy}}{{{{\left( {{x^2} + {y^2} + 1} \right)}^{3/2}}}} = \frac{{\partial Q}}{{\partial x}}.\]

此时原函数为

\begin{align*}\varphi \left( {x,y} \right) = &\int_{{x_0}}^x {P\left( {x,{y_0}} \right)dx} + \int_{{y_0}}^y {Q\left( {x,y} \right)dy} + C'\\= &\int_{{x_0}}^x {\left( { - \frac{{{y_0}}}{{{x^2} + y_0^2}} - \frac{x}{{\sqrt {{x^2} + y_0^2 + 1} }}} \right)dx} + \int_{{y_0}}^y {\left( {\frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dy} + C'\\= &\left( { - \arctan \frac{x}{{{y_0}}} - \sqrt {{x^2} + y_0^2 + 1} + \arctan \frac{{{x_0}}}{{{y_0}}} + \sqrt {x_0^2 + y_0^2 + 1} } \right)\\&+ \left( {\arctan \frac{y}{x} - \sqrt {{x^2} + {y^2} + 1} - \arctan \frac{{{y_0}}}{x} + \sqrt {{x^2} + y_0^2 + 1} } \right) + C'\\=& \arctan \frac{y}{x} - \sqrt {{x^2} + {y^2} + 1} + C.\end{align*}

 

值得一提的是:本题的积分因子是通过Wolfram Alpha求解出ODE,然后分别对$x,y$求偏导得来的.

 

(2)丁同仁书上一定理:


齐次方程$P(x,y)dx+Q(x,y)dy=0$有积分因子$M=\frac{1}{xP+yQ}$.


 

定理的证明:作变换$y=ux$,则由$P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0$是齐次方程,我们有$$P\left( {x,ux} \right)dx + Q\left( {x,ux} \right)\left( {udx + xdu} \right) = \left[ {{x^m}P\left( {1,u} \right) + u{x^m}Q\left( {1,u} \right)} \right]dx + {x^{m + 1}}Q\left( {1,u} \right)du = 0.$$

 

方程两边同乘\[\frac{1}{{xP + yQ}} = \frac{1}{{{x^{m + 1}}\left[ {P\left( {1,u} \right) + uQ\left( {1,u} \right)} \right]}},\]则有

\[\frac{1}{x}dx + \frac{{Q\left( {1,u} \right)}}{{P\left( {1,u} \right) + uQ\left( {1,u} \right)}}du = 0.\]显然此方程为全微分方程.证毕.

 

取\[M = \frac{1}{{xP + yQ}} = \frac{1}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}}.\]

则有

\[P' = \frac{{x{{\left( {ay + bx} \right)}^3} + ax{y^3}}}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}},Q' = \frac{{y{{\left( {ay + bx} \right)}^3} + b{x^3}y}}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}}.\]

 

我猜此时一定成立\[\frac{{\partial P'}}{{\partial y}} = \frac{{\partial Q'}}{{\partial x}}.\]

 

事实上

\begin{align*}\frac{{\partial P'}}{{\partial y}} = & - 2xy{\left( {ay + bx} \right)^6} - 2{x^3}y{\left( {ay + bx} \right)^4} + \left( {5a{x^3}{y^2} + ax{y^4}} \right){\left( {ay + bx} \right)^3}\\& - 3{a^2}x{y^3}\left( {{x^2} + {y^2}} \right){\left( {ay + bx} \right)^2} + a{x^3}{y^4}\left( {ay + bx} \right) - {a^2}{x^3}{y^5}\\\frac{{\partial Q'}}{{\partial x}} = &- 2xy{\left( {ay + bx} \right)^6} - 2x{y^3}{\left( {ay + bx} \right)^4} + \left( {5b{x^2}{y^3} + b{x^4}y} \right){\left( {ay + bx} \right)^3}\\& - 3{b^2}{x^3}y\left( {{x^2} + {y^2}} \right){\left( {ay + bx} \right)^2} + b{x^4}{y^3}\left( {ay + bx} \right) - {b^2}{x^5}{y^3}.\end{align*}

 

于是

\begin{align*}\varphi \left( {x,y} \right) &= \int_{{x_0}}^x {P'\left( {x,{y_0}} \right)dx} + \int_{{y_0}}^y {Q'\left( {x,y} \right)dy} + C'\\&= \frac{1}{2}\ln \left[ {\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)} \right] - \frac{3}{2}\ln \left( {ay + bx} \right) + C.\end{align*}

 

事实上,我们还可取\[M = \frac{1}{{{{\left( {ay + bx} \right)}^3}}},\]由此得到

\[\varphi \left( {x,y} \right) = \frac{{{x^2} + {y^2}}}{2} + \frac{{{x^2}{y^2}}}{{2{{\left( {ay + bx} \right)}^2}}} + C.\]


多元里的两道问题

“数学是你们的选择,你们随时都可以放弃。但当数学仍是你们的选择时,就必须为此负责。”
——S.Lang对他学生上课前说的话

多重积分计算的一些题

(1)设$f$在$D:x^2+y^2\leq1$上二阶连续可微,且\[\Delta f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=x^2+y^2,\]求\[\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.\]


cool解:(Hansschwarzkopf)根据Gauss公式

\begin{align*}&\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\&=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}s\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\frac{\partial f}{\partial n}\mathrm{d}s=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}\Delta f\mathrm{d}x\mathrm{d}y\\&=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}(x^2+y^2)\mathrm{d}x\mathrm{d}y=\int_0^1\frac{\pi r^4}{2}\mathrm{d}r =\frac{\pi}{10} .\end{align*}
 
(2)设$f$在$D:x^2+y^2\leq1$上二阶连续可微,且\[\Delta f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\exp{(-x^2-y^2)},\]求

\[\iint\limits_D\left(x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.\]


cool解:(Hansschwarzkopf)根据Gauss公式

\begin{align*}&\iint\limits_D\left(x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 = r^2}\left(x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}\right)\mathrm{d}s\\ &=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 =r^2}r\frac{\partial f}{\partial n}\mathrm{d}s=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}r\Delta f\mathrm{d}x\mathrm{d}y\\ &=\int_0^1\mathrm{d}r\iint\limits_{x^2+y^2 \leqslant r^2}r\exp{(-x^2-y^2)}\mathrm{d}x\mathrm{d}y=\int_0^1\pi r(1-e^{-r^2})\mathrm{d}r =\frac{\pi }{2e} .\end{align*}