曲线积分的计算
张元博问了这么一道题：计算曲线积分\[\oint_C {\left( {{{\left( {x + 1} \right)}^2} + {{\left( {y - 2} \right)}^2}} \right)ds} ,\]其中$C$表示曲面$x^2+y^2+z^2=1$与$x+y+z=1$的交线.
解:我是利用常规的三角换元解决的.联立方程有
\[\left\{ \begin{array}{l}{x^2} + {y^2} + {z^2} = 1\\x + y + z = 1\end{array} \right. \Rightarrow {\left( {x - \frac{1}{2} + \frac{y}{2}} \right)^2} + \frac{3}{4}{\left( {y - \frac{1}{3}} \right)^2} = \frac{1}{3}.\]
令
\[\left\{ \begin{array}{l}x - \frac{1}{2} + \frac{y}{2} = \frac{1}{{\sqrt 3 }}\cos \theta \\\frac{{\sqrt 3 }}{2}\left( {y - \frac{1}{3}} \right) = \frac{1}{{\sqrt 3 }}\sin \theta \end{array} \right. \Rightarrow \left\{ \begin{array}{l}x = \frac{1}{3} - \frac{1}{3}\sin \theta + \frac{1}{{\sqrt 3 }}\cos \theta \\y = \frac{1}{3} + \frac{2}{3}\sin \theta \\z = 1 - x - y = \frac{1}{3} - \frac{1}{3}\sin \theta - \frac{1}{{\sqrt 3 }}\cos \theta \end{array} \right.\]
则\[ds = \sqrt {{{\left( {x'\left( \theta \right)} \right)}^2} + {{\left( {y'\left( \theta \right)} \right)}^2} + {{\left( {z'\left( \theta \right)} \right)}^2}} = \frac{{\sqrt 6 }}{3}.\]
因此
\begin{align*}&\oint_C {\left( {{{\left( {x + 1} \right)}^2} + {{\left( {y - 2} \right)}^2}} \right)ds} \\= &\frac{{\sqrt 6 }}{3}\int_0^{2\pi } {\left( {\frac{5}{9}{{\sin }^2}\theta + \frac{1}{3}{{\cos }^2}\theta - \frac{{28}}{9}\sin \theta + \frac{8}{{3\sqrt 3 }}\cos \theta - \frac{2}{{3\sqrt 3 }}\sin \theta \cos \theta + \frac{{41}}{9}} \right)d\theta } \\= &\frac{{\sqrt 6 }}{3} \times 5 = \frac{{5\sqrt 6 }}{3}.\end{align*}
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