Eufisky - The lost book

## 含奇点的第二型曲面积分计算

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

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

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

\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*}

$$\frac{\partial J}{\partial t}=\frac{1}{4}\left( \frac{\partial J}{\partial x} \right) ^2-x^2-\frac{1}{2}x^4,\qquad \text{其中}J\left[ x\left( 1 \right) ,1 \right] =0$$

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

\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*}

$$I=2\int_0^1{x\text{e}^x\frac{\pi}{2}\sqrt{1-x^2}\frac{\text{d}x}{\sqrt{1-x^2}}}=\pi \int_0^1{x\text{e}^x\text{d}x}=\pi.$$

\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$$\lim\limits_{n\rightarrow \infty}\int_0^1\int_0^1\cdots\int_0^1 f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)dx_1dx_2\cdots dx_n.$$

\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*}

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

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

## 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: $\displaystyle \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1(a,b,c>0)$.

$\frac{{\partial \left( {y,z} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = bc{\sin ^2}\varphi \cos \theta ,\frac{{\partial \left( {z,x} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = ac{\sin ^2}\varphi \sin \theta ,\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = ab\sin \varphi \cos \varphi ,$

\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*}

\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*}

$I = abc\iint\limits_{\left[ {0,\pi } \right] \times \left[ {0,2\pi } \right]} {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}\sin \varphi d\varphi d\theta }$

$\int {{{\left( {M + N{x^2}} \right)}^{ - 3/2}}dx} = \frac{1}{M} \cdot \frac{x}{{\sqrt {M + N{x^2}} }} + C.$

\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*}

$I = 4ab\int_0^\pi {\frac{1}{{{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta }}d\theta } = 4\pi .$

$$n=\frac{\left(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}\right)}{\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}},$$

$$\iint\limits_\Sigma \frac{xdydz+yd zd x+zdxdy}{\sqrt{(x^2+y^2+z^2)^3}}.$$

$$\iint\limits_\Sigma \frac{xd yd z+yd zd x+zd xd y}{\sqrt{(x^2+y^2+z^2)^3}} =\iint\limits_{S_\varepsilon} \frac{xdyd z+ydzd x+zd xd y}{\sqrt{(x^2+y^2+z^2)^3}}=4\pi.$$ 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

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

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

\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*}

（2）丁同仁书上一定理：

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

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

\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*}

\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*}

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

## 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$\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.$

\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

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

\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*}