武汉⼤学2011年数学分析试题解答
武汉⼤学2011年数学分析试题解答
1：计算题
（1）解：原极限\text{=}\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt[n]{n}}{n}\cdot {{n}^{1-\alpha }}={{e}^{-1}}\underset{n\to \infty }
{\mathop{\lim }}\,{{n}^{1-\alpha }}=\left\{\begin{array} +\infty, & \hbox{$0<\alpha<1$;} \\ e^{-1}, & \hbox{$\alpha=1$;} \\ 0, & \hbox{$\alpha>1$.} \end{array} \right.
(解释⼀下：\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt[n]{n}}{n}={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}
{n}\sum\limits_{i=1}^{n}{In\frac{i}{n}}}}={{e}^{\int\limits_{0}^{1}{Inxdx}}}={{e}^{-1}}（来源于数学分析上册第⼆章课后习题）
（2）解：考虑等价⽆穷⼩：\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{\frac{1}{2}{{x}^{2}}}=1，则
1-\cos \sqrt{\tan x-\sin x}=2{{\sin }^{2}}\frac{\sqrt{\tan x-\sin x}}{2}\sim \frac{1}{2}(\tan x-\sin x)=\frac{\sin x}{2\cos x}(1-\cos x)\sim \frac{1}{4} {{x}^{3}}
另⼀⽅⾯：\sqrt[3]{1+{{x}^{3}}}-\sqrt[3]{1-{{x}^{3}}}=\frac{2{{x}^{3}}}{{{(\sqrt[3]{1+{{x}^{3}}})}^{2}}+\sqrt[3]{(1+{{x}^{3}})(1-{{x}^{3}})}+{{(\sqrt[3]{1-{{x}^{3}}})}^{2}}}\sim \frac{2{{x}^{3}}}{3}
从⽽原式 =\frac{3}{8}
（3）法⼀：
解：原式=\int{\frac{1+\cos x}{\sqrt{1+\cos x}}dx=}\int{\frac{2{{\cos }^{2}}\frac{x}{2}}{\sqrt{1+\cos x}}dx=}\int{\frac{2{{\cos }^{2}}\frac{x}{2}(1-{{\sin }^{2}}\frac{x}{2})}{{{\cos }^{2}}\frac{x}{2}\sqrt{1+\cos x}}dx}
=\int{\frac{1+\cos x-\sin x\sin \frac{x}{2}\cos \frac{x}{2}}{{{\cos }^{2}}\frac{x}{2}\sqrt{1+\cos x}}dx=}\int{\sqrt{1+\cos x}{{\sec }^{2}}\frac{x} {2}+\frac{-\sin x}{\sqrt{1+\cos x}}\tan \frac{x}{2}dx}
=2\int{d(\sqrt{1+\cos x}\tan \frac{x}{2})=2}\sqrt{1+\cos x}\tan \frac{x}{2}+C
法⼆：由于\int{\sqrt{\text{1+}\cos x}dx}=\sqrt{2}\int{\left| \cos \frac{x}{2} \right|}dx
考虑到\int{\left| x \right|}dx=\frac{{{x}^{2}}}{2}sgn x+C
于是\int{\sqrt{\text{1+}\cos x}dx}=2\sqrt{2}\sin \frac{x}{2}sgn (\cos \frac{x}{2})+C（C为常数）
法三：由于\int{\sqrt{\text{1+}\cos x}dx}=\sqrt{2}\int{\left| \cos \frac{x}{2} \right|}dx
\overset{t=\frac{x}{2}}{\mathop{=}}\,2\sqrt{2}\int{\left| \cos t \right|}dt
⽽
\int {\left| {\cos t} \right|} dt=\left\{\begin{array}{ll} \sin t + {c_k}, & \hbox{$- \frac{\pi }{2} + 2k\pi \le t \le \frac{\pi }{2} + 2k\pi$;} \\ - \sin t + {d_k}, & \hbox{$\frac{\pi }{2} + 2k\pi \le t \le \frac{{3\pi }}{2} + 2k\pi$.} \end{array} \right.+C为连续函数，其中C为常数
于是
\left\{\begin{array}{ll} {c_k} + 1 = - 1 + {d_k} \\ {c_{k + 1}} - 1 = 1 + {d_k} \end{array} \right.
，令{{c}_{0}}=0
则{{c}_{k}}=4k,{{d}_{k}}=4k+2
于是
\int {\left| {\cos t} \right|} dt=\left\{\begin{array}{ll} \sin t + 4k, & \hbox{$ - \frac{\pi }{2} + 2k\pi \le t \le \frac{\pi }{2} + 2k\pi$;} \\ - \sin t + 4k + 2, & \hbox{$\frac{\pi }{2} + 2k\pi \le t \le \frac{{3\pi }}{2} + 2k\pi$.} \end{array} \right.+C
即
\int {\sqrt {{\rm{1 + }}\cos x} dx}=\left\{\begin{array}{ll} 2\sqrt 2 (\sin \frac{x}{2} + 4k), & \hbox{$ - \pi + 4k\pi \le x \le \pi + 4k\pi$;} \\ 2\sqrt 2 ( - \sin \frac{x}{2} + 4k + 2), & \hbox{$\pi + 4k\pi \le x \le 3\pi + 4k\pi$.} \end{array} \right.+C，其中C为常数
（4）解：F(x,y)=x\int_{\frac{y}{x}}^{xy}{zf(z)dz-y\int_{\frac{y}{x}}^{xy}{f(z)dz}}
则F_{x}^{'}=\int_{\frac{y}{x}}^{xy}{zf(z)dz+x[xyf(xy)\cdot y-\frac{y}{x}f(\frac{y}{x})\cdot \frac{-y}{{{x}^{2}}}]-y[f(xy)\cdot y-f(\frac{y}{x})\cdot \frac{-y} {{{x}^{2}}}]}
=\int_{\frac{y}{x}}^{xy}{zf(z)dz+({{x}^{2}}-1){{y}^{2}}f(xy)}
F_{xx}^{''}=xyf(xy)\cdot y-\frac{y}{x}f(\frac{y}{x})\cdot \frac{-y}{{{x}^{2}}}+2x{{y}^{2}}f(xy)+({{x}^{2}}-1){{y}^{2}}f'(xy)\cdot y
=3x{{y}^{2}}f(xy)+\frac{{{y}^{2}}}{{{x}^{3}}}f(\frac{y}{x})+({{x}^{2}}-1){{y}^{3}}f'(xy)
（5）解：原式=\int\limits_{0}^{1}{dy\int\limits_{-1}^{{{y}^{2}}}{({{y}^{2}}-x)dx+}}\int\limits_{0}^{1}{dy\int\limits_{{{y}^{2}}}^{1}{(-
{{y}^{2}}+x)dx}}=\frac{6}{5}
已知f(x),g(x)在[a,b]上连续，在(a,b)上可微，且g'(x)在(a,b)上⽆零点，证明：\exists \xi \in (a,b),st\frac{f'(\xi )}{g'(\xi )}=\frac{f(b)-g(\xi )}{g(\xi )-g(a)}
如果有思路的话，欢迎补充！
证明：作辅助函数F\left( x \right)=f\left( x \right)g\left( x \right)-g\left( b \right)f\left( x \right)-f\left( a \right)g\left( x \right)
虽然F\left( x \right) 在[a,b]上连续，在(a,b)上可微, F\left( a \right)=F\left( b \right)=-f\left( a \right)g\left( b \right)
由罗尔中值定理，存在\xi \in \left( a,b \right)使得{F}'\left( \xi \right)=0
即{f}'\left( \xi \right)g\left( \xi \right)+f\left( \xi \right){g}'\left( \xi \right)-g\left( b \right){f}'\left( \xi \right)-f\left( a \right){g}'\left( \xi \right)=0
整理\left[ f\left( a \right)-f\left( \xi \right) \right]{g}'\left( \xi \right)-{f}'\left( \xi \right)\left[ g\left( \xi \right)-g\left( b \right) \right]=0
即\frac{f\left( a \right)-f\left( \xi \right)}{g\left( \xi \right)-g\left( b \right)}=\frac{{f}'\left( \xi \right)}{{g}'\left( \xi \right)} ，证得
3：（⽅法⼀）
证明：\left| \frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}-b \right|=\left|
\frac{{{a}_{1}}({{b}_{n}}-b)+{{a}_{2}}({{b}_{n-1}}-b)+\cdots +{{a}_{n}}({{b}_{1}}-b)}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}} \right| \le \frac{{{a}_{1}}\left| {{b}_{n}}-b \right|+\cdots +{{a}_{n-N}}\left| {{b}_{N+1}}-b \right|}{{{a}_{1}}+\cdots +{{a}_{n}}}+\frac{{{a}_{n-N+1}}\left| {{b}_{N}}-b \right|+\cdots +{{a}_{n}}\left| {{b}_{1}}-b \right|}{{{a}_{1}}+\cdots +{{a}_{n}}}
\le \underset{N+1\le k\le n}{\mathop \max }\,\left| {{b}_{k}}-b \right|+\underset{N+1\le k\le n}{\mathop \m
ax }\,\left| {{b}_{k}}-b \right|\cdot \frac{N}{n-N+1}=I_{1}^{n}+I_{2}^{n}
从⽽对\forall \varepsilon >0，先取定N使得I_{1}^{n}<\frac{\varepsilon }{2}，后让n充分⼤即有I_{2}^{n}<\frac{\varepsilon }{2}，于是有结论成⽴。
（⽅法⼆）
证明：设{{t}_{nk}}=\frac{{{a}_{n-k+1}}}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}},k=1,2,\cdots ,n,n=1,2,\cdots
则{{t}_{nk}}>0且\sum\limits_{k=1}^{n}{{{t}_{nk}}}=1
再由{{a}_{n}}\ge 0,{{a}_{n}}\le {{a}_{n-1}}\Rightarrow {{a}_{1}}+{{a}_{2}}+\cdots \text{+}{{a}_{n}}\ge {{a}_{1}}+{{a}_{2}}+\cdots \text{+}{{a}_{n-k+1}}\ge (n-k+1){{a}_{n-k+1}}
于是0\le {{t}_{nk}}\le \frac{{{a}_{n-k+1}}}{(n-k+1){{a}_{n-k+1}}}=\frac{1}{(n-k+1)}\to 0(n\to +\infty )
由迫敛性知：\underset{n\to +\infty }{\mathop{\lim }}\,{{t}_{nk}}=0
再由\underset{n\to +\infty }{\mathop{\lim }}\,{{b}_{n}}=b，可知
\exists M>0,s.t对任意的n\in {{N}^{*}},\left| {{a}_{n}}-a \right|<M
同时，对任意的\varepsilon >0,\exists {{N}_{1}}\in {{N}^{*}}，当n>{{N}_{1}}时，有\left| {{a}_{n}}-a \right|<\frac{\varepsilon }{2}
固定{{N}_{1}}，由\underset{n\to +\infty }{\mathop{\lim }}\,{{t}_{nk}}=0可知：
\exists {{N}_{2}}\in {{N}^{*}}，当n>{{N}_{2}}时，有\left| {{a}_{n}}-a \right|<\frac{\varepsilon }{2{{N}_{1}}M},k=1,2,\cdots ,{{N}_{1}}
令N=\max \{{{N}_{1}},{{N}_{2}}\}，当n>N时，有
\left| \frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}}}-b \right|=\left|
\sum\limits_{k=1}^{n}{{{t}_{nk}}{{b}_{k}}-b} \right|=\left| \sum\limits_{k=1}^{n}{{{t}_{nk}}{{b}_{k}}-\sum\limits_{k=1}^{n}{{{t}_{nk}}}b} \right|
\le \sum\limits_{k=1}^{n}{{{t}_{nk}}}\left| {{b}_{k}}-b \right|<M({{t}_{n1}}+{{t}_{n2}}+\cdots +{{t}_{n{{N}_{1}}}})+\frac{\varepsilon }{2}({{t}_{n, {{N}_{1}}+1}}+\cdots +{{t}_{nn}})<\varepsilon
即\underset{n\to +\infty }{\mathop{\lim }}\,\frac{{{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\cdots +{{a}_{n}}{{b}_{1}}}{{{a}_{1}}+{{a}_{2}}+\cdots + {{a}_{n}}}=b
（⽅法三，书上并⽆介绍此公式，不建议使⽤）：
直接利⽤Stolz进⾏计算即可
4：证明：{{a}_{k}}=\frac{1}{2\pi }\int\limits_{-\pi }^{\pi }{f(x)\cos kxdx=}\frac{1}{2k\pi }\int\limits_{-\pi }^{\pi }{f(x)d(sinkx)=}\frac{-1}{2k\pi
}\int\limits_{-\pi }^{\pi }{f'(x)sinkxdx}
=\frac{-1}{2{{k}^{2}}\pi }\int\limits_{-k\pi }^{k\pi }{f'(\frac{t}{k})sintdt}
则
{{a}_{2n}}=\frac{-1}{8{{n}^{2}}\pi }\int\limits_{-2n\pi }^{2n\pi }{f'(\frac{t}{2n})\operatorname{sint}dt=}\frac{-1}{8{{n}^{2}}\pi }\sum\limits_{m=-n}^{n-1}{[\int\limits_{-2m\pi }^{(2m+1)\pi }{f'(\frac{t}{2n})\operatorname{sint}dt+\int\limits_{(2m+1)\pi }^{2(m+1)\pi }{f'(\frac{t}
{2n})\operatorname{sint}dt]=}}}
\ge \frac{-1}{8{{n}^{2}}\pi }\sum\limits_{m=-n}^{n-1}{[\int\limits_{2m\pi }^{(2m+1)\pi }{f'(\frac{t}{2n})\sin tdt+}}\int\limits_{(2m+1)\pi }^{2(m+1)\pi } {\operatorname{sintdt}]}
⽽
\sin t\left\{\begin{array}{ll} > 0, & \hbox{$ 2m\pi < x < (2m + 1)\pi$;} \\ < 0, & \hbox{$(2m + 1)\pi < x < 2(m + 1)\pi$.} \end{array} \right.
且f为凸函数
从⽽原式\ge \frac{-1}{8{{n}^{2}}\pi }\sum\limits_{m=-n}^{n-1}{f'(\frac{2m+1}{2n}\pi )\int\limits_{2m\pi }^{2(m+1)\pi }{\sin tdt=0}}
另⼀⽅⾯：
{{a}_{2n+1}}=\frac{-1}{2{{(2n+1)}^{2}}\pi }\int\limits_{-(2n+1)\pi }^{(2n+1)\pi }{f'(\frac{t}{2n+1})\sin tdt}
=\frac{-1}{2{{(2n+1)}^{2}}\pi }\sum\limits_{m=-n}^{n}{[\int\limits_{(2m-1)\pi }^{2m\pi }{f'(\frac{t}{2n+1})\sin tdt+\int\limits_{2m\pi }^{(2m+1)\pi } {f'(\frac{t}{2n+1})\sin tdt}}}
\le \frac{-1}{2{{(2n+1)}^{2}}\pi }\sum\limits_{m=-n}^{n}{[\int\limits_{(2m+1)\pi }^{2m\pi }{f'(\frac{2m}{2n+1})\sin tdt}}+\int\limits_{2m\pi
}^{(2m+1)\pi }{f'(\frac{2m}{2n+1})\sin tdt}]
==\frac{-1}{2{{(2n+1)}^{2}}\pi }\sum\limits_{m=-n}^{n}{f'(\frac{2m}{2n+1}\pi )}\int\limits_{(2m-1)\pi }^{(2m+1)\pi }{\sin tdt=0}
5：证明：
（1）设{{f}_{k}}(x)=\sum\limits_{n=1}^{k}{{{u}_{n}}(x),}
因为
\left| f(x)-f(y) \right|\le \left| f(x)-{{f}_{K}}(x) \right|+\left| {{f}_{K}}(x)-{{f}_{K}}(y) \right|+\left| {{f}_{K}}(y)-f(y) \right|
\le 2\cdot \sup \left| \left| {{f}_{K}}(z)-f(z) \right| \right|+\left| {{f}_{K}}(x)-{{f}_{K}}(y) \right|\equiv I_{1}^{K}+I_{2}^{K}
从⽽对\forall \varepsilon >0，可先选定K使I_{1}^{K}<\frac{\varepsilon }{2}，后取\delta >0使\left| x-y \right|<\delta \Rightarrow I_{2}^{K}
<\frac{\varepsilon }{2}
（2）若条件改为逐点连续，则结论不成⽴。为此，仅需构造出{{f}_{k}}(x)\to f(x)
其中每个{{f}_{k}}(x)⼀致收敛，但f(x)不⼀致收敛，甚⾄不连续的例⼦很多，例如：
{f_k}(x)\left\{\begin{array}{ll} - 1, & \hbox{$x \le \frac{{ - 1}}{k}$;} \\ kx, & \hbox{$\frac{{ - 1}}{k} < x < \frac{1}{k}$;} \\ 1, & \hbox{$x \ge \frac{1}{k}$.} \end{array} \right.
，恒有{{f}_{k}}(x)\to sgn (x)
6：证明：
（1）\int\limits_{c}^{d}{f(x,y)dy}⼀致收敛
\Leftrightarrow \exists F(x),\forall \varepsilon >0,\exists \delta >
\left\{\begin{array}{ll} x \in [a,b] \\ \delta ' \in (0,\delta ) \end{array} \right.
其有Cauchy准则：
\forall \varepsilon >0,\exists \delta >
\left\{\begin{array}{ll} x \in [a,b] \\ \delta ',\delta '' \in (0,\delta ) \end{array} \right.
\Rightarrow \left| \int\limits_{c+\delta '}^{c+\delta ''}{f(x,y)dy} \right|<\varepsilon
（3）仅需注意到\left| \int\limits_{c+\delta '}^{c+\delta }{f(x,y)g(x,y)dy} \right|\le \underset{[a,b]\times [c,d]}{\mathop{\max }}\,\left| g(x,y)
\right|\cdot \int\limits_{c+\delta '}^{c+\delta ''}{\left| f(x,y) \right|}dy
后利⽤Cauchy即可得证
7：（1）解：g=f(\frac{x}{z})\mp zf(\frac{z}{x})+f(\frac{y}{z})\mp zf(\frac{z}{y})
{{g}_{x}}=\frac{1}{z}f''(\frac{x}{z})\mp \frac{{{z}^{2}}}{{{x}^{2}}}f''(\frac{z}{x})
{{g}_{xx}}=\frac{1}{{{z}^{2}}}f''(\frac{x}{z})\mp \frac{2{{z}^{2}}}{{{x}^{3}}}f'(\frac{z}{x})\mp \frac{{{z}^{3}}}{{{x}^{4}}}f''(\frac{z}{x})
{{g}_{yy}}=\frac{1}{{{z}^{2}}}f''(\frac{y}{z})\mp \frac{{{z}^{2}}}{{{y}^{3}}}f'(\frac{z}{y})\mp \frac{{{z}^{3}}}{{{y}^{4}}}f''(\frac{z}{y})
{{g}_{z}}=\frac{-1}{{{z}^{2}}}f'(\frac{x}{z})\mp f(\frac{z}{x})\mp \frac{z}{x}f'(\frac{z}{x})
=\frac{-y}{{{z}^{2}}}f'(\frac{y}{z})\mp f(\frac{z}{y})\mp \frac{z}{y}f'(\frac{z}{y})
{{g}_{zz}}=\frac{2x}{{{z}^{3}}}f'(\frac{x}{z})+\frac{{{x}^{2}}}{{{z}^{4}}}f''(\frac{x}{z})\mp \frac{1}{x}f'(\frac{z}{x})\mp \frac{1}{x}f'(\frac{z}{x})\mp \frac{z}{{{x}^{2}}}f''(\frac{z}{x})
忘记2011+\frac{2y}{{{z}^{3}}}f'(\frac{y}{z})+\frac{{{y}^{2}}}{{{z}^{4}}}f''(\frac{y}{z})\mp \frac{1}{y}f'(\frac{z}{y})\mp \frac{1}{y}f'(\frac{z}{y})\mp \frac{z} {{{y}^{2}}}f''(\frac{z}{y})
于是
{{x}^{2}}{{g}_{xx}}+{{y}^{2}}{{g}_{yy}}-{{z}^{2}}{{g}_{zz}}=-\frac{2x}{z}f(\frac{x}{z})-\frac{2y}{z}f(\frac{y}{z})
（2）设0<{{a}_{1}}<{{a}_{2}},0<{{b}_{1}}<{{b}_{2}},1<{{c}_{1}}<{{c}_{2}}.则
原式=-2\iiint_{\Omega }{[\frac{x}{z}f'(\frac{x}{z})+\frac{y}{z}f'(\frac{y}{z})dxdydz}
=-2\int\limits_{{{a}_{1}}}^{{{a}_{2}}}{du\int\limits_{{{b}_{1}}}^{{{b}_{2}}}{dv\int\limits_{{{c}_{1}}}^{{{c}_{2}}}{[\frac{1}{u}f'(\frac{1}{u})+\frac{1} {v}f'(\frac{1}{v})]\cdot \frac{2}{3}{{u}^{-2}}{{v}^{-2}}dw}}}
（其中令u=\frac{z}{x},v=\frac{z}{y},w={{z}^{3}}）
\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\int\limits_{{{a}_{1}}}^{{{a}_{2}}}{du\int\limits_{{{b}_{1}}}^{{{b}_{2}}}{[\frac{1}{{{u}^{2}}v}f'(\frac{1}{u})+\frac{1} {u{{v}^{2}}}f'(\frac{1}{v})}}]dv
=\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\int\limits_{{{a}_{1}}}^{{{a}_{2}}}{\{\frac{1}{{{u}^{2}}}f'(\frac{1}{u})In\frac{{{b}_{2}}}{{{b}_{1}}}-\frac{1}{u}[f(\frac{1} {{{b}_{2}}})-f(\frac{1}{{{b}_{1}}})]In\frac{{{a}_{2}}}{{{a}_{1}}}\}du}
=\frac{-2({{c}_{2}}-{{c}_{1}})}{3}\left\{ -In\frac{{{b}_{2}}}{{{b}_{1}}}[ \right.f(\frac{1}{{{a}_{2}}})-f(\frac{1}{{{a}_{1}}})]-[f(\frac{1}{{{b}_{2}}})-f(\frac{1} {{{b}_{1}}})]In\frac{{{a}_{2}}}{{{a}_{1}}}\}
=\frac{2({{c}_{2}}-{{c}_{1}})}{3}\left\{ In\frac{{{b}_{2}}}{{{b}_{1}}}[ \right.f(\frac{1}{{{a}_{2}}})+f(\frac{1}{{{a}_{1}}})]-[f(\frac{1}{{{b}_{2}}})-f(\frac{1} {{{b}_{1}}})]In\frac{{{a}_{2}}}{{{a}_{1}}}\}
8：证明：\iint_{D}{(x\frac{\partial u}{\partial x}}+y\frac{\partial u}{\partial y})dxdy=\int\limits_{0}^{1}{dr}\int_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}} {(}x{{u}_{x}}+y{{u}_{y}})dS
=\int\limits_{0}^{1}{r}\int_{{{x}^{2}}+{{y}^{2}}={{r}^{2}}}{\frac{\partial u}{\partial n}}dS=\int\limits_{0}^{1}{r}\iint_{{{x}^{2}}+{{y}^{2}}\le {{r}^{2}}}{\nabla udxdy}
=\int\limits_{0}^{1}{r}\iint_{{{x}^{2}}+{{y}^{2}}\le {{r}^{2}}}{\cos (\pi ({{x}^{2}}+{{y}^{2}}))dxdy}
=\int\limits_{0}^{1}{r}\iint_{{{x}^{2}}+{{y}^{2}}\le {{r}^{2}}}{\cos (\pi {{s}^{2}})\cdot 2\pi sdS}
=\int\limits_{0}^{1}{r\sin (\pi r)dr=\frac{1}{\pi }}
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