[再寄小读者之数学篇](2014-04-01 from [email protected] 曲线积分)

求 $\int_\vGa y^2\rd s$, 其中 $\vGa$ 由 $\dps{\sedd{\ba{rl} x^2+y^2+z^2&=a^2\\ x+z&=a \ea}}$ 决定.

解答: $\vGa$: $$\bex \sedd{\ba{rl} \sex{x-\cfrac{a}{2}}^2+y^2+\sex{z-\cfrac{a}{2}}^2&=\cfrac{a^2}{2}\\ \sex{x-\cfrac{a}{2}}+\sex{y-\cfrac{a}{2}}&=0 \ea}. \eex$$ 作变换 $$\bex u=x-\cfrac{a}{2},\quad v=y,\quad w=z-\cfrac{a}{2}, \eex$$ 则 $$\beex \bea \int_\vGa y^2\rd s &=\int_l v^2\rd s\quad\sex{l:\ \sedd{\ba{rl} u^2+v^2+w^2&=\cfrac{a^2}{2}\\ u+w=0 \ea}}\\ &=\int_0^{2\pi} \cfrac{a^2}{2}\sin^2\tt \sqrt{\sex{\cfrac{\rd u}{\rd \tt}}^2 +\sex{\cfrac{\rd v}{\rd t}}^2 +\sex{\cfrac{\rd w}{\rd t}}^2}\rd \tt\\ &\quad\sex{l:\ \sedd{\ba{rl} u=\cfrac{a}{2}\cos\tt\\ v=\cfrac{a}{\sqrt{2}}\sin\tt\\ w=-\cfrac{a}{2}\cos\tt \ea}, 0\leq \tt\leq 2\pi}\\ &=\int_0^{2\pi} \cfrac{a^2}{2}\sin^2\tt \cdot \cfrac{a}{\sqrt{2}}\rd \tt\\ &=\cfrac{a^3\pi}{2\sqrt{2}}. \eea \eeex$$