[尊重原有作者劳动成果]
一：
1：解：\[\because \underset{x\to 0}{\mathop{\lim }}\,\ln (1+x)=x\]
\[\therefore \underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt[n]{1+x}-1}{\ln (1+x)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt[n]{1+x}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{n}{ {(1+x)}^{\frac{1}{n}-1}}=\frac{1}{n}\]
2：解：$\int{\frac{x\ln (x+\sqrt{1+{ {x}^{2}}})}{ { {(1+{ {x}^{2}})}^{2}}}}dx=-\frac{1}{2}\int{\frac{-2x}{ { {(1+{ {x}^{2}})}^{2}}}\ln (x+\sqrt{1+{ {x}^{2}}})}dx=$
$-\frac{1}{2}[\frac{1}{(1+{ {x}^{2}})}\ln (x+\sqrt{1+{ {x}^{2}}})-\int{\frac{1}{(1+{ {x}^{2}})}\cdot \frac{\frac{x+\sqrt{1+{ {x}^{2}}}}{\sqrt{1+{ {x}^{2}}}}}{x+\sqrt{1+{ {x}^{2}}}}}dx=\int{\frac{dx}{ { {(1+{ {x}^{2}})}^{\frac{3}{2}}}}}$
而$\int{\frac{dx}{ { {(1+{ {x}^{2}})}^{\frac{3}{2}}}}}\overset{x=\tan \theta }{\mathop{=}}\,\int{\frac{d\tan \theta }{\frac{1}{ { {(\cos \theta )}^{3}}}}}=\int{\cos \theta d\theta }=\sin \theta +C=\frac{x}{\sqrt{1+{ {x}^{2}}}}+C$
于是$\int{\frac{x\ln (x+\sqrt{1+{ {x}^{2}}})}{ { {(1+{ {x}^{2}})}^{2}}}}dx=-\frac{\ln (x+\sqrt{1+{ {x}^{2}}})}{2(1+{ {x}^{2}})}+\frac{x}{2\sqrt{1+{ {x}^{2}}}}+C$（其中$C$为任意常数）
3：解：$\because \int_{0}^{\frac{\pi }{2}}{\sqrt{1-\sin 2x}}dx=\int_{0}^{\frac{\pi }{2}}{\sqrt{ { {(\sin x-\cos x)}^{2}}}}dx=\int_{0}^{\frac{\pi }{2}}{\left| \sin x-\cos x \right|}dx$
\[=\int_{0}^{\frac{\pi }{4}}{(\cos x-\sin x)dx+}\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}{(sinx-\cos x)dx=2(\sqrt{2}}-1)\]
4：解：$\because y=\arcsin x=x+\frac{1}{2}\cdot \frac{ { {x}^{3}}}{3}+\frac{1\times 3}{2\times 4}\cdot \frac{ { {x}^{5}}}{5}+\cdots +\frac{(2n-1)!!}{(2n)!!}\cdot \frac{ { {x}^{2n+1}}}{2n+1}+O({ {x}^{2n+1}})$
于是当$n=2k$时，${ {y}^{(n)}}(0)=0$
当$n=2k+1$时，${ {y}^{(2k+1)}}(x)=\frac{(2k-1)!!}{(2k)!!}\cdot \frac{(2k+1)!}{2k+1}+O(1)$
则\[{ {y}^{(2k+1)}}(0)=\frac{(2k-1)!!}{(2k)!!}\cdot \frac{(2k+1)!}{2k+1}={ {[(2k-1)!!]}^{2}}={ {[(n-2)!!]}^{2}}\]
5：解：$\because nx+k-1\sqrt{(nx+k)(nx+k-1)}nx+k$
\[\therefore \frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k-1}{n}}){ {S}_{n}}=\frac{1}{ { {n}^{2}}}\sum\limits_{k=1}^{n}{\sqrt{(nx+k)(nx+k-1)}}\frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k}{n}})\]
$\therefore \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k-1}{n}})\le \underset{n\to \infty }{\mathop{\lim }}\,{ {S}_{n}}\le \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k}{n}})$
而不等式左边
$\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{(x+\frac{k-1}{n}})\overset{i=k-1}{\mathop{=}}\,\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{1=0}^{n-1}{(x+\frac{i}{n}})=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}[\sum\limits_{1=1}^{n}{(x+\frac{i}{n}})+(x+0)-(x+1)]=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{1=1}^{n}{(x+\frac{i}{n}})=$由迫敛性知：$\underset{n\to \infty }{\mathop{\lim }}\,{ {S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{1=1}^{n}{(x+\frac{i}{n}})=\int_{0}^{1}{(x+t)dt=x+\frac{1}{2}}$
二：证明：由于${ {a}_{2}}={ {a}^{\frac{3}{4}}},{ {a}_{3}}={ {a}^{\frac{7}{8}}}$ $(a0)$
于是分三种情况：
1 当$a=1$时，此时${ {x}_{n}}=1$，则$\{ { {x}_{n}}\}$收敛且$\underset{n\to \infty }{\mathop{\lim }}\,{ {x}_{n}}=1$
2 当$0a1$时，下证$a{ {x}_{n}}\le \sqrt{a}$ （数学归纳法）
（1） 当$n=1$时，$a{ {a}_{1}}=\sqrt{a}$成立；
（2） 设$n=k$时，$a{ {x}_{k}}\le \sqrt{a}$，则$a{ {x}_{n+1}}=\sqrt{a{ {x}_{n}}}\le \sqrt{a\sqrt{a}}\le \sqrt{a}$
即当$n=k+1$时，也成立
于是对\[\forall n\in { {N}_{+}},a{ {x}_{n}}\le \sqrt{a}\]
则$\frac{ { {x}_{n+1}}}{ { {x}_{n}}}=\frac{\sqrt{a}}{\sqrt{ { {x}_{n}}}}1$
于是$\{ { {x}_{n}}\}$单调递减且${ {x}_{n}}a$
由单调有解原理知：$\{ { {x}_{n}}\}$收敛，设$\underset{n\to \infty }{\mathop{\lim }}\,{ {x}_{n}}=l$
由${ {x}_{n+1}}=\sqrt{a{ {x}_{n}}}$，两边取极限，于是有$\underset{n\to \infty }{\mathop{\lim }}\,{ {x}_{n}}=a$
3 当$a1$时，同理可证$\sqrt{a}\le { {x}_{n}}a$，得$\{ { {x}_{n}}\}$单调递增
同样由单调有解原理知：$\{ { {x}_{n}}\}$收敛并且$\underset{n\to \infty }{\mathop{\lim }}\,{ {x}_{n}}=a$
综上所述：对$\forall a0$，$\{ { {x}_{n}}\}$收敛且$\underset{n\to \infty }{\mathop{\lim }}\,{ {x}_{n}}=a$
三：证明：$\because \int_{0}^{+\infty }{\frac{dx}{ { {(1+x)}^{2}}(1+{ {x}^{\alpha }})}}=\int_{0}^{1}{\frac{dx}{ { {(1+x)}^{2}}(1+{ {x}^{\alpha }})}}+\int_{1}^{+\infty }{\frac{dx}{ { {(1+x)}^{2}}(1+{ {x}^{\alpha }})}}$
而\[\int_{1}^{+\infty }{\frac{dx}{ { {(1+x)}^{2}}(1+{ {x}^{\alpha }})}}\overset{t=\frac{1}{x}}{\mathop{=}}\,\int_{1}^{0}{\frac{-\frac{1}{ { {x}^{2}}}dx}{ { {(1+\frac{1}{x})}^{2}}(1+\frac{1}{ { {x}^{\alpha }}})}=}\int_{0}^{1}{\frac{ { {x}^{\alpha }}dx}{ { {(1+x)}^{2}}(1+{ {x}^{\alpha }})}}\]
于是
\[\int_{0}^{+\infty }{\frac{dx}{ { {(1+x)}^{2}}(1+{ {x}^{\alpha }})}}=\int_{0}^{1}{\frac{1+{ {x}^{\alpha }}}{ { {(1+x)}^{2}}(1+{ {x}^{\alpha }})}dx=\int_{0}^{1}{\frac{dx}{ { {(1+x)}^{2}}}}}=\frac{1}{2}(\]
反常积分\[\int_{0}^{+\infty }{\frac{dx}{ { {(1+x)}^{2}}(1+{ {x}^{\alpha }})}}\]与$\alpha $无关且值为$\frac{1}{2}$
四、证明：不妨设$F(x)=f(x+1)-f(x),x\in [0,1]$
于是$F(0)=f(1)-f(0)$
$F(1)=f(2)-f(1)=f(0)-f(1)$
于是$F(0)\cdot F(1)=-{ {[f(0)-f(1)]}^{2}}\le 0$
（1）若$F(0)=0$或，可得$f(0)=f(1)=f(2)$
则只需取$\xi =0$或$1$，即证
（2）若$F(0)\cdot F(1)0$，于是$F(0)$与$F(1)$异号，于是由介值定理知：
$\exists \xi \in [0,1]$，使得$f(\xi )=f(\xi +1)$
五、证明：$\because u=xy,v=x-y$
$\therefore \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial x}=y\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}$
$\frac{ { {\partial }^{2}}z}{\partial { {x}^{2}}}=y[\frac{\partial (\frac{\partial z}{\partial u})}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial (\frac{\partial z}{\partial u})}{\partial v}\cdot \frac{\partial v}{\partial x}]+[\frac{\partial (\frac{\partial z}{\partial v})}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial (\frac{\partial z}{\partial v})}{\partial v}\cdot \frac{\partial v}{\partial x}]$
$={ {y}^{2}}\frac{ { {\partial }^{2}}z}{\partial { {u}^{2}}}+2y\frac{ { {\partial }^{2}}z}{\partial u\partial v}+\frac{ { {\partial }^{2}}z}{\partial { {v}^{2}}}$
$\frac{ { {\partial }^{2}}z}{\partial x\partial y}=xy\frac{ { {\partial }^{2}}z}{\partial { {u}^{2}}}+(x-y)\frac{ { {\partial }^{2}}z}{\partial x\partial y}-\frac{ { {\partial }^{2}}z}{\partial { {v}^{2}}}+\frac{\partial z}{\partial u}$
同理可证：$\therefore \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial y}=x\frac{\partial z}{\partial u}-\frac{\partial z}{\partial v}$
$\frac{ { {\partial }^{2}}z}{\partial { {x}^{2}}}=={ {x}^{2}}\frac{ { {\partial }^{2}}z}{\partial { {u}^{2}}}-2x\frac{ { {\partial }^{2}}z}{\partial u\partial v}+\frac{ { {\partial }^{2}}z}{\partial { {v}^{2}}}$
带入已知化简得：\[\frac{ { {\partial }^{2}}z}{\partial { {u}^{2}}}+\frac{1}{ { {(x+y)}^{2}}}\frac{\partial z}{\partial u}=0\]
而${ {(x+y)}^{2}}={ {(x-y)}^{2}}+4xy={ {v}^{2}}+4u$
代入即得\[\frac{ { {\partial }^{2}}z}{\partial { {u}^{2}}}+\frac{1}{ { {v}^{2}}+4u}\frac{\partial z}{\partial u}=0\]
六：1：解：如图：
$A=\iint_{D}{\left| xy-\frac{1}{4} \right|}dxdy=A=\iint_{ { {D}_{1}}}{(xy-\frac{1}{4})}dxdy+\iint_{ { {D}_{2}}}{(-xy+\frac{1}{4})}dxdy$
$=\int_{\frac{1}{4}}^{1}{dx\int_{\frac{1}{4x}}^{1}{(xy-\frac{1}{4})dy+[}}\int_{0}^{\frac{1}{4}}{dx\int_{0}^{1}{(-xy+\frac{1}{4})dy+\int_{\frac{1}{4}}^{1}{dx\int_{0}^{\frac{1}{4x}}{(-xy+\frac{1}{4})dy]}}}}$
$=\frac{1}{8}\ln 2+\frac{3}{32}$
2：证明：$\because \left| \iint_{D}{(xy-\frac{1}{4})f(x,y)dxdy} \right|\le \iint_{D}{\left| xy-\frac{1}{4} \right|\left| f(x,y) \right|dxdy}$
由于$\left| xy-\frac{1}{4} \right|\ge 0,\left| f(x,y) \right|\ge 0$
于是$\exists ({ {x}^{*}},{ {y}^{*}})\in D$，使得
根据积分定理知：$\iint_{D}{\left| xy-\frac{1}{4} \right|\left| f(x,y) \right|dxdy}=\iint_{D}{\left| (xy-\frac{1}{4}) \right|dxdy}\cdot \left| f({ {x}^{*}},{ {y}^{*}}) \right|$
$=A\left| f({ {x}^{*}},{ {y}^{*}}) \right|$
而
\[\left| \iint_{D}{(xy-\frac{1}{4})f(x,y)dxdy} \right|=\left| \iint_{D}{xyf(x,y)dxdy-\frac{1}{4}\iint_{D}{f(x,y)dxdy}} \right|=1\]
于是$\exists ({ {x}^{*}},{ {y}^{*}})\in D$使得$\left| f({ {x}^{*}},{ {y}^{*}}) \right|\ge \frac{1}{A}$
七：解：设切点为$({ {x}_{0}},{ {y}_{0}},{ {z}_{0}})$，设$f(x,y,z)=\frac{ { {x}^{2}}}{ { {a}^{2}}}+\frac{ { {y}^{2}}}{ { {b}^{2}}}+\frac{ { {z}^{2}}}{ { {c}^{2}}}$
从而${ {f}_{x}}({ {x}_{0}},{ {y}_{0}},{ {z}_{0}})=\frac{2{ {x}_{0}}}{ { {a}^{2}}},{ {f}_{y}}({ {x}_{0}},{ {y}_{0}},{ {z}_{0}})=\frac{2{ {y}_{0}}}{ { {b}^{2}}},{ {f}_{z}}({ {x}_{0}},{ {y}_{0}},{ {z}_{0}})=\frac{2{ {z}_{0}}}{ { {c}^{2}}}$
从而$\pi $的表达式为$\frac{2{ {x}_{0}}}{ { {a}^{2}}}(x-{ {x}_{0}})+\frac{2{ {y}_{0}}}{ { {b}^{2}}}(y-{ {y}_{0}})+\frac{2{ {z}_{0}}}{ { {c}^{2}}}(z-{ {z}_{0}})=0$
且$\frac{x_{0}^{2}}{ { {a}^{2}}}+\frac{y_{0}^{2}}{ { {b}^{2}}}+\frac{z_{0}^{2}}{ { {c}^{2}}}=1$，代入化简得：$\frac{ { {x}_{0}}}{ { {a}^{2}}}x+\frac{ { {y}_{0}}}{ { {b}^{2}}}y+\frac{ { {z}_{0}}}{ { {c}^{2}}}z=1$
于是$\pi $在第一象限的部分与三个坐标的坐标分别为
$(\frac{ { {a}^{2}}}{ { {x}_{0}}},0,0),(0,\frac{ { {b}^{2}}}{ { {y}_{0}}},0),(0,0,\frac{ { {c}^{2}}}{ { {z}_{0}}})$，可知${ {x}_{0}},{ {y}_{0}},{ {z}_{0}}0$
于是$V=\frac{1}{6}\cdot \frac{ { {a}^{2}}}{ { {x}_{0}}}\cdot \frac{ { {b}^{2}}}{ { {y}_{0}}}\cdot \frac{ { {c}^{2}}}{ { {z}_{0}}}$，且$\frac{x_{0}^{2}}{ { {a}^{2}}}+\frac{y_{0}^{2}}{ { {b}^{2}}}+\frac{z_{0}^{2}}{ { {c}^{2}}}=1$
由广义均值不等式知：\[\frac{x_{0}^{2}}{ { {a}^{2}}}+\frac{y_{0}^{2}}{ { {b}^{2}}}+\frac{z_{0}^{2}}{ { {c}^{2}}}\ge 3\sqrt[3]{\frac{x_{0}^{2}}{ { {a}^{2}}}\cdot \frac{y_{0}^{2}}{ { {b}^{2}}}\cdot \frac{z_{0}^{2}}{ { {c}^{2}}}}\]
当且仅当${ {x}_{0}}=\frac{\sqrt{3}}{3}a,{ {y}_{0}}=\frac{\sqrt{3}}{3}b,{ {z}_{0}}=\frac{\sqrt{3}}{3}c$等号成立
于是当$\pi $的方程为$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\sqrt{3}$时，${ {V}_{\min }}=\frac{\sqrt{3}}{2}abc$
八、1：证明：$\because f(y)=\int_{0}^{+\infty }{x{ {e}^{-{ {x}^{2}}}}\cos xydx},-\infty y+\infty $
于是对$\forall { {y}_{0}}\in (-\infty ,+\infty )$，取$[a,b]\subset (-\infty ,+\infty )$，使得${ {y}_{0}}\in [a,b]$
由于\[\left| x{ {e}^{-{ {x}^{2}}}}\cos x{ {y}_{0}} \right|\le x{ {e}^{-{ {x}^{2}}}}\]，且\[\int_{0}^{+\infty }{x{ {e}^{-{ {x}^{2}}}}dx}=\frac{1}{2}\]收敛
从而$f(y)$在$[a,b]$上一致收敛，于是$f(y)$在$[a,b]$上连续
所以$f(y)$在点${ {y}_{0}}$连续，由${ {y}_{0}}$的任意性知，$f(y)$在点$(-\infty ,+\infty )$连续
记$F(x,y)=x{ {e}^{-{ {x}^{2}}}}\cos xy$，取$[c,d]\subset (-\infty ,+\infty )$，使得${ {y}_{0}}\in [c,d]$
则${ {F}_{y}}=-{ {x}^{2}}{ {e}^{-{ {x}^{2}}}}\sin xy$和$F(x,y)$在$[0.+\infty )\times (-\infty ,+\infty )$上连续
对$\forall A0$，由于$\int_{0}^{A}{\sin xydx}=\frac{-\operatorname{cosAy}}{y}$，而$\left| \frac{-\operatorname{cosAy}}{y} \right|\le \frac{1}{y}\le \frac{1}{c}$
即$\int_{0}^{A}{\sin xydx}$对$y$在$[c,d]$上一致有界
而当$x1$时，${ {x}^{2}}{ {e}^{-{ {x}^{2}}}}$是关于$x$的单调递减的函数，且$\underset{x\to \infty }{\mathop{\lim }}\,{ {x}^{2}}{ {e}^{-{ {x}^{2}}}}=0$
从而对一切$x$，有${ {x}^{2}}{ {e}^{-{ {x}^{2}}}}\to 0(x\to +\infty )$
从而由狄利克雷判别法知，$f(y)$有连续的导数
由于$F_{y}^{(2n)}={ {(-1)}^{n}}{ {x}^{2n+1}}{ {e}^{-{ {x}^{2}}}}\cos xy,F_{y}^{(2n+1)}={ {(-1)}^{n+1}}{ {x}^{2n+2}}{ {e}^{-{ {x}^{2}}}}\sin xy$
同理可证$f(y)$有$(2n)$和$(2n+1)$连续的导函数
于是$f(y)$有任意阶连续的导数
2：证明：由$F_{y}^{(2n)}={ {(-1)}^{n}}{ {x}^{2n+1}}{ {e}^{-{ {x}^{2}}}}\cos xy,F_{y}^{(2n+1)}={ {(-1)}^{n+1}}{ {x}^{2n+2}}{ {e}^{-{ {x}^{2}}}}\sin xy$
所以
$F_{y}^{(2n)}(x,0)={ {(-1)}^{n}}{ {x}^{2n+1}}{ {e}^{-{ {x}^{2}}}}\cos x0={ {(-1)}^{n}}{ {x}^{2n+1}}{ {e}^{-{ {x}^{2}}}},F_{y}^{(2n+1)}(x,0)={ {(-)}^{n+1}}{ {x}^{2n+2}}{ {e}^{-{ {x}^{2}}}}\sin x0=0$ 于是
\[{ {f}^{(n)}}(0)=\int_{0}^{+\infty }{ { {(-1)}^{n}}{ {x}^{2n+1}}{ {e}^{-{ {x}^{2}}}}dx}={ {(-1)}^{n}}\times (-\frac{1}{2})[{ {e}^{-{ {x}^{2}}}}\cdot { {x}^{2n}}|_{0}^{+\infty }-2n\cdot \int_{0}^{+\infty }{ { {e}^{-{ {x}^{2}}}}\cdot { {x}^{2n-1}}dx}]\] \[={ {(-1)}^{n+1}}\times \frac{1}{2}\times (2n)\times \int_{0}^{+\infty }{ { {e}^{-{ {x}^{2}}}}\cdot { {x}^{2n-1}}dx}\]
$={ {(-1)}^{n+2}}\times { {(\frac{1}{2})}^{2}}\times (2n)\times (2n-2)\times \int_{0}^{+\infty }{ { {e}^{-{ {x}^{2}}}}\cdot { {x}^{2n-3}}dx}$
$=\cdots $
$=-{ {(\frac{1}{2})}^{n}}\times (2n)!!\times \int_{0}^{+\infty }{ { {e}^{-{ {x}^{2}}}}\cdot xdx={ {(\frac{1}{2})}^{n+1}}\times (2n)!!}$
由泰勒展开式的定义知，$f(y)$的麦克劳林级数为
$f(y)=\sum\limits_{n=0}^{+\infty }{\frac{ { {(\frac{1}{2})}^{n+1}}\times (2n)!!}{(2n)!}}{ {y}^{2n}}=\sum\limits_{n=0}^{+\infty }{\frac{ { {y}^{2n}}}{ { {2}^{n+1}}(2n+1)!!}}$
九：法一：证明：由对称性知：$I=\iint\limits_{\sum }{({ {x}^{2}}}+{ {y}^{2}}+{ {z}^{2}}{ {)}^{-\frac{3}{2}}}{ {(\frac{ { {x}^{2}}}{ { {a}^{4}}}+\frac{ { {y}^{2}}}{ { {b}^{4}}}+\frac{ { {z}^{2}}}{ { {c}^{4}}})}^{-\frac{1}{2}}}dS$
$=8I=\iint\limits_{ { {S}_{1}}}{({ {x}^{2}}}+{ {y}^{2}}+{ {z}^{2}}{ {)}^{-\frac{3}{2}}}{ {(\frac{ { {x}^{2}}}{ { {a}^{4}}}+\frac{ { {y}^{2}}}{ { {b}^{4}}}+\frac{ { {z}^{2}}}{ { {c}^{4}}})}^{-\frac{1}{2}}}dS$
其中${ {S}_{1}}:z=c\sqrt{1-\frac{ { {x}^{2}}}{ { {a}^{2}}}-\frac{ { {y}^{2}}}{ { {b}^{2}}}},(x,y)\in { {D}_{1}}=\{\frac{ { {x}^{2}}}{ { {a}^{2}}}+\frac{ { {y}^{2}}}{ { {b}^{2}}}\le 1,x0,y0\}$
同时$\sqrt{1+z_{x}^{2}+z_{y}^{2}}=\sqrt{1+{ {[\frac{\frac{c}{ { {a}^{2}}}x}{\sqrt{1-\frac{ { {x}^{2}}}{ { {a}^{2}}}-\frac{ { {y}^{2}}}{ { {b}^{2}}}}}]}^{2}}+{ {[\frac{\frac{c}{ { {b}^{2}}}y}{\sqrt{1-\frac{ { {x}^{2}}}{ { {a}^{2}}}-\frac{ { {y}^{2}}}{ { {b}^{2}}}}}]}^{2}}}$
$=\frac{c}{z}\sqrt{\frac{ { {z}^{2}}}{ { {c}^{2}}}+\frac{ { {c}^{2}}{ {x}^{2}}}{ { {a}^{4}}}+\frac{ { {c}^{2}}{ {y}^{2}}}{ { {b}^{4}}}}=\frac{ { {c}^{2}}}{z}\sqrt{\frac{ { {z}^{2}}}{ { {c}^{4}}}+\frac{ { {x}^{2}}}{ { {a}^{4}}}+\frac{ { {y}^{2}}}{ { {b}^{4}}}}$
于是$I=8\iint\limits_{ { {S}_{1}}}{({ {x}^{2}}}+{ {y}^{2}}+{ {z}^{2}}{ {)}^{-\frac{3}{2}}}\cdot { {(\frac{ { {x}^{2}}}{ { {a}^{4}}}+\frac{ { {y}^{2}}}{ { {b}^{4}}}+\frac{ { {z}^{2}}}{ { {c}^{4}}})}^{-\frac{1}{2}}}\cdot \frac{ { {c}^{2}}}{z}\sqrt{\frac{ { {z}^{2}}}{ { {c}^{4}}}+\frac{ { {x}^{2}}}{ { {a}^{4}}}+\frac{ { {y}^{2}}}{ { {b}^{4}}}}dxdy$
$=8{ {c}^{2}}\iint_{ { {S}_{1}}}{\frac{dxdy}{z\cdot { {({ {x}^{2}}+{ {y}^{2}}+{ {z}^{2}})}^{\frac{3}{2}}}}}$
$=8c\iint_{ { {D}_{1}}}{\frac{dxdy}{\sqrt{1-\frac{ { {x}^{2}}}{ { {a}^{2}}}-\frac{ { {y}^{2}}}{ { {b}^{2}}}}\cdot { {[{ {x}^{2}}+{ {y}^{2}}+(1-\frac{ { {x}^{2}}}{ { {a}^{2}}}-\frac{ { {y}^{2}}}{ { {b}^{2}}})]}^{\frac{3}{2}}}}}$
其中 ${ {D}_{1}}=\{\frac{ { {x}^{2}}}{ { {a}^{2}}}+\frac{ { {y}^{2}}}{ { {b}^{2}}}\le 1,x0,y0\}$
于是设$x=ra\cos \theta ,y=rb\sin \theta ,\theta \in [0,\frac{\pi }{2}],r\in [0,1]$
原式$=8ab\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{rdr}{\sqrt{1-{ {r}^{2}}}{ {[{ {r}^{2}}{ {a}^{2}}{ {\cos }^{2}}\theta +{ {r}^{2}}{ {b}^{2}}{ {\sin }^{2}}\theta +{ {c}^{2}}(1-{ {r}^{2}})]}^{\frac{3}{2}}}}}$
$=4ab\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{d{ {r}^{2}}}{\sqrt{1-{ {r}^{2}}}{ {[{ {r}^{2}}{ {a}^{2}}{ {\cos }^{2}}\theta +{ {r}^{2}}{ {b}^{2}}{ {\sin }^{2}}\theta +{ {c}^{2}}(1-{ {r}^{2}})]}^{\frac{3}{2}}}}}$
\[\overset{t={ {r}^{2}}}{\mathop{=}}\,4ab\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{dt}{\sqrt{1-t}{ {[t{ {a}^{2}}{ {\cos }^{2}}\theta +t{ {b}^{2}}{ {\sin }^{2}}\theta +{ {c}^{2}}(1-t)]}^{\frac{3}{2}}}}}\]
\[=4ab\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{dt}{\sqrt{1-t}{ {[t[{ {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta -{ {c}^{2}}]+{ {c}^{2}}]}^{\frac{3}{2}}}}}\]
\[=\frac{4ab}{ { {c}^{3}}}\int_{0}^{\frac{\pi }{2}}{d\theta }\int_{0}^{1}{\frac{dt}{\sqrt{1-t}{ {[t[{ {(\frac{a}{c})}^{2}}{ {\cos }^{2}}\theta +{ {(\frac{b}{c})}^{2}}{ {\sin }^{2}}\theta -1]+1]}^{\frac{3}{2}}}}}\]
为此，设$S(d)=\int_{0}^{1}{\frac{dt}{\sqrt{1-t}{ {(1+dt)}^{\frac{3}{2}}}}}$，其中$d={ {(\frac{a}{c})}^{2}}{ {\cos }^{2}}\theta +{ {(\frac{b}{c})}^{2}}{ {\sin }^{2}}\theta -1$
令$h=\sqrt{1-t}$，则$S(d)=2\int_{0}^{1}{\frac{dh}{ { {(1+d-d{ {h}^{2}})}^{\frac{3}{2}}}}}$
易知：若$d=0$，则$S(0)=2$
若$d0$，则令\[h=\sqrt{\frac{1+d}{d}}\sin \varphi \]
于是\[S(d)=2\int_{0}^{\arcsin \sqrt{\frac{d}{1+d}}}{\frac{\sqrt{\frac{1+d}{d}}\cos \varphi }{ { {[1+d-(1+d){ {\sin }^{2}}\varphi ]}^{\frac{3}{2}}}}}d\varphi \]
=\[\frac{2}{\sqrt{d}(1+d)}\int_{0}^{\arcsin \sqrt{\frac{d}{1+d}}}{\frac{d\varphi }{ { {\cos }^{2}}\varphi }}\]
=$\frac{2}{\sqrt{d}(1+d)}\tan (arcsin\sqrt{\frac{d}{1+d}})$
于是令
$m=\arcsin \sqrt{\frac{d}{1+d}}$，则$\operatorname{sinm}=\sqrt{\frac{d}{1+d}}$
从而
\[\tan (arcsin\sqrt{\frac{d}{1+d}})=\tan m=\sqrt{d}\]
即 $S(d)=\frac{2}{1+d}$
同理，若$d0$
$S(d)=2\int_{0}^{1}{\frac{dh}{ { {(1+d+\left| d \right|{ {h}^{2}})}^{\frac{3}{2}}}}}$
于是令$k=\sqrt{\frac{1+d}{\left| d \right|}}\tan \varphi $
则\[S(d)=2\int_{0}^{\arctan \sqrt{\frac{\left| d \right|}{1+d}}}{\frac{\sqrt{\frac{1+d}{\left| d \right|}}\frac{1}{ { {\cos }^{2}}\varphi }}{ { {[1+d+(1+d){ {\tan }^{2}}\varphi ]}^{\frac{3}{2}}}}}d\varphi \]
=\[\frac{2}{\sqrt{\left| d \right|}(1+d)}\int_{0}^{\arctan \sqrt{\frac{\left| d \right|}{1+d}}}{\cos \varphi d\varphi }\]
$=\frac{2}{1+d}$
这里$d$不可能为$-1$，原因是$d={ {(\frac{a}{c})}^{2}}{ {\cos }^{2}}\theta +{ {(\frac{b}{c})}^{2}}{ {\sin }^{2}}\theta -1$
综上所述：$S(d)=\frac{2}{1+c}$
于是$S({ {(\frac{a}{c})}^{2}}{ {\cos }^{2}}\theta +{ {(\frac{b}{c})}^{2}}{ {\sin }^{2}}\theta -1)=\frac{2}{ { {(\frac{a}{c})}^{2}}{ {\cos }^{2}}\theta +{ {(\frac{b}{c})}^{2}}{ {\sin }^{2}}\theta }$
$=\frac{2{ {c}^{2}}}{ { {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta }$
于是$I=\frac{8ab}{c}\int_{0}^{\frac{\pi }{2}}{\frac{d\theta }{ { {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta }}$
$=\frac{8ab}{c}\int_{0}^{\frac{\pi }{2}}{\frac{1+{ {\tan }^{2}}\theta }{ { {a}^{2}}+{ {b}^{2}}{ {\tan }^{2}}\theta }}d\theta $
于是令$v=\tan \theta $
则$I=\frac{8ab}{c}\int_{0}^{+\infty }{\frac{1+{ {v}^{2}}}{ { {a}^{2}}+{ {b}^{2}}{ {v}^{2}}}\cdot \frac{1}{1+{ {v}^{2}}}}dv=\frac{4\pi }{c}$
法二：解：设
$\left\{\begin{array}{ll} x = a\sin \varphi \cos \theta , \\ y = b\sin \varphi \sin \theta ， \\ z = c\cos \varphi . \end{array} \right.$ $0 \le \varphi \le \pi ,0 \le \theta \le 2\pi$
则
${ {x}_{\varphi }}=a\cos \varphi \cos \theta ,{ {y}_{\varphi }}=b\cos \varphi \sin \theta ,{ {z}_{\varphi }}=-c\sin \varphi $
${ {x}_{\theta }}=-a\sin \varphi \sin \theta ,{ {y}_{\theta }}=b\sin \varphi \cos \theta ,{ {z}_{\theta }}=0$
从而
\[E={ {\cos }^{2}}\varphi ({ {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta )+{ {c}^{2}}{ {\sin }^{2}}\varphi \]
\[F={ {\sin }^{2}}\varphi ({ {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta )\]
\[G=({ {b}^{2}}-{ {a}^{2}})\sin \varphi \cos \varphi \sin \theta \cos \theta \]
$\sqrt{EF-{ {G}^{2}}}=\sin \varphi \sqrt{ { {a}^{2}}{ {b}^{2}}{ {\cos }^{2}}\varphi +{ {b}^{2}}{ {c}^{2}}{ {\sin }^{2}}\varphi { {\cos }^{2}}\theta +{ {a}^{2}}{ {c}^{2}}{ {\sin }^{2}}\varphi { {\sin }^{2}}\theta }$
$=abc\sin \varphi \sqrt{\frac{ { {x}^{2}}}{ { {a}^{4}}}+\frac{ { {y}^{2}}}{ { {b}^{4}}}+\frac{ { {z}^{2}}}{ { {c}^{4}}}}$
于是
$I=\iint\limits_{0\le \varphi \le \pi ,0\le \theta \le 2\pi }{abc[{ {a}^{2}}{ {\sin }^{2}}\varphi { {\cos }^{2}}}\theta +{ {b}^{2}}{ {\sin }^{2}}\varphi { {\sin }^{2}}\theta +{ {c}^{2}}{ {\cos }^{2}}\varphi { {]}^{-\frac{3}{2}}}\sin \varphi d\varphi d\theta $
于是设
$J=\int_{0}^{\pi }{ { {({ {a}^{2}}{ {\sin }^{2}}\varphi { {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\varphi { {\sin }^{2}}\theta +{ {c}^{2}}{ {\cos }^{2}}\varphi )}^{-\frac{3}{2}}}\sin \varphi d\varphi }$
\[\overset{t=\cos \varphi }{\mathop{=}}\,\int_{-1}^{1}{[({ {a}^{2}}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta )(1-{ {t}^{2}})+{ {c}^{2}}{ {t}^{2}}{ {]}^{-\frac{3}{2}}}dt\]
$=\frac{2}{ { {A}^{3}}}\int_{0}^{1}{[1-(\frac{ { {A}^{2}}-{ {c}^{2}}}{ { {A}^{2}}}}){ {t}^{2}}{ {]}^{-\frac{3}{2}}}dt$
其中$A=\sqrt{ { {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta }$
不失一般性，不妨设$a\ge b\ge c$
于是
${ {A}^{2}}={ {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta ={ {b}^{2}}+({ {a}^{2}}-{ {b}^{2}}){ {\cos }^{2}}\theta \ge { {b}^{2}}\ge { {c}^{2}}$
当且仅当$a=b=c$取等号
（1）若$A=c$，则$a=b=c$，由此可得$I=4\pi $
（2）若$Ac$，则
$J=\frac{2}{ { {A}^{2}}\sqrt{ { {A}^{2}}-{ {c}^{2}}}}\int_{0}^{\frac{\sqrt{ { {A}^{2}}-{ {c}^{2}}}}{A}}{ { {(1-{ {s}^{2}})}^{-\frac{3}{2}}}}ds$
$=\frac{2}{ { {A}^{2}}\sqrt{ { {A}^{2}}-{ {c}^{2}}}}\frac{s}{\sqrt{1-{ {s}^{2}}}}|_{0}^{\frac{\sqrt{ { {A}^{2}}-{ {c}^{2}}}}{A}}$
$=\frac{2}{ { {A}^{2}}c}$
于是
$I=2ab\int_{0}^{2\pi }{\frac{1}{ { {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta }}d\theta $
$=8ab\int_{0}^{\frac{\pi }{2}}{\frac{1}{ { {a}^{2}}{ {\cos }^{2}}\theta +{ {b}^{2}}{ {\sin }^{2}}\theta }}d\theta $
$\overset{y=\tan \theta }{\mathop{=}}\,8ab\int_{0}^{+\infty }{\frac{dt}{({ {a}^{2}}-{ {b}^{2}})+{ {b}^{2}}(1+{ {t}^{2}})}}$
$=8ab\int_{0}^{+\infty }{\frac{dt}{ { {a}^{2}}+{ {b}^{2}}{ {t}^{2}}}}$
$=4\pi $
由（1）（2）可知，$I=4\pi $