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*}
而
\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*}
又
\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*}
和
\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
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