2013武大数分压轴题

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

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

 

故

\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*}

而

\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

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

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另外有更好的方法：(Hansschwarzkopf)