下册 9.1 第一型曲线积分与第一型曲面积分 第2题

\section*{解题过程：} 平面 $x+y+z=0$ 与球面 $x^{2}+y^{2}+z^{2}=a^{2}$ 的交线 $L$ 是该球面上的极大圆．由对称性有 $$ \begin{aligned} & \int_{L} x^{2} \mathrm{~d} s=\int_{L} y^{2} \mathrm{~d} s=\int_{L} z^{2} \mathrm{~d} s=\frac{1}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s \\ & \begin{aligned} \int_{L}\left(x^{2}+y^{2}\right) \mathrm{d} s & =\int_{L}\left(y^{2}+z^{2}\right) \mathrm{d} s=\int_{L}\left(z^{2}+x^{2}\right) \mathrm{d} s \\ & =\frac{2}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s=\frac{2}{3} \int_{L} a^{2} \mathrm{~d} s=\frac{4}{3} \pi a^{3} \end{aligned} \\ & \int_{L} x \mathrm{~d} s=\int_{L} y \mathrm{~d} s \\ & \text { 盾 } z \mathrm{~d} s=\frac{1}{3} \int_{L}(x+y+z) \mathrm{d} s=0 \end{aligned} $$ 于是（1） $\displaystyle \int_{L} x^{2} \mathrm{~d} s=\int_{L} y^{2} \mathrm{~d} s=\frac{1}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s=\frac{1}{3} \int_{L} a^{2} \mathrm{~d} s=\frac{2 \pi}{3} a^{3}$ ． 特别地，当 $a=1$ 时， $\displaystyle \int_{L} x^{2} \mathrm{~d} s=\frac{1}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s=\frac{1}{3} \int_{L} \mathrm{~d} s=\frac{2 \pi}{3}$ ． （2）如图 9.3 所示， $\displaystyle \oint_{\Gamma}\left(x^{2}+2 z\right) \mathrm{d} s=\frac{1}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s+\frac{2}{3} \int_{L}(x+y+z) \mathrm{d} s=\frac{1}{3} \int_{L} a^{2} \mathrm{~d} s=\frac{2 \pi a^{3}}{3}$ ． $\displaystyle \oint_{L}\left(z^{2}+x\right) \mathrm{d} s=\frac{1}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s+\frac{1}{3} \int_{L}(x+y+z) \mathrm{d} s=\frac{1}{3} \int_{L} a^{2} \mathrm{~d} s=\frac{2 \pi a^{3}}{3}$ ． $\displaystyle \oint_{L}\left(x^{2}+2 y+z\right) \mathrm{d} s=\frac{1}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s+\int_{L}(x+y+z) \mathrm{d} s=\frac{1}{3} \int_{L} a^{2} \mathrm{~d} s=\frac{2 \pi a^{3}}{3}$ ． \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-231.jpg?height=1189&width=1230&top_left_y=6865&top_left_x=4372} \captionsetup{labelformat=empty} \caption{图 9.3} \end{figure} （3） $\displaystyle \int_{L}\left(x^{2}+y^{2}\right) \mathrm{d} s=\frac{2}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s=\frac{2}{3} \int_{L} a^{2} \mathrm{~d} s=\frac{4}{3} \pi a^{3}$ ． （4） $\displaystyle \int_{L}\left(x^{2}+y^{2}+y\right) \mathrm{d} s=\frac{2}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s+\frac{1}{3} \int_{L}(x+y+z) \mathrm{d} s=\frac{2}{3} \int_{L} a^{2} \mathrm{~d} s=\frac{4}{3} a^{3} \pi \xlongequal{a=1} \frac{4}{3} \pi$ ． （5） $\displaystyle \int_{L}\left(x^{2}+y^{2}+2 z\right) \mathrm{d} s=\frac{2}{3} \int_{L}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} s+\frac{2}{3} \int_{L}(x+y+z) \mathrm{d} s=\frac{2}{3} \int_{L} a^{2} \mathrm{~d} s=\frac{4}{3} a^{3} \pi$ ． （6）在交线上，$(x-1)^{2}+(y-1)^{2}+(z-1)^{2}=x^{2}+y^{2}+z^{2}-2(x+y+z)+3=4$ 。所以 $$ \int_{L}\left[(x-1)^{2}+(y-1)^{2}+(z-1)^{2}\right] \mathrm{d} s=\int_{L} 4 \mathrm{~d} s=4 \cdot 2 \pi=8 \pi $$ （7）因为在 $L$ 上，$\displaystyle x y+y z+z x=\frac{1}{2}\left[(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)\right]$ ，所以 $$ \begin{aligned} & \int_{L}(x y+y z+z x) \mathrm{d} s=\frac{1}{2} \int_{L}\left[(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)\right] \mathrm{d} s=-\frac{a^{2}}{2} \int_{L} \mathrm{~d} s=-\pi a^{3} . \\ & \int_{L} x y \mathrm{~d} s=\frac{1}{3} \int_{L}(x y+y z+z x) \mathrm{d} s=-\frac{1}{3} \pi a^{3} . \end{aligned} $$ 注：下面介绍直接求 $\int_{L} x^{2} \mathrm{~d} s$ 的解法，其中 $L$ 为曲面 $x^{2}+y^{2}+z^{2}=a^{2}$ 与 $x+y+z=0$ 的交线． 把 $z=-(x+y)$ 代人 $x^{2}+y^{2}+z^{2}=a^{2}$ 得 $\displaystyle x^{2}+y^{2}+x y=\frac{1}{2} a^{2}$ ．从而 $\displaystyle \left(\frac{1}{2} x+y\right)^{2}+\frac{3}{4} x^{2}=\frac{1}{2} a^{2}$ ．利用椭 圆的参数方程得 $L$ 的参数方程为 $$ L: x=\sqrt{\frac{2}{3}} a \cos \theta, y=\frac{a}{\sqrt{2}} \sin \theta-\frac{a}{\sqrt{6}} \cos \theta, z=-\frac{a}{\sqrt{6}} \sin \theta-\frac{a}{\sqrt{6}} \cos \theta, 0 \leqslant \theta \leqslant 2 \pi $$ 由于 $\mathrm{d} s=\sqrt{x^{\prime 2}+y^{\prime 2}+z^{\prime 2}} \mathrm{~d} \theta=a \mathrm{~d} \theta$ ，则 $\displaystyle \int_{L} x^{2} \mathrm{~d} s=\int_{0}^{2 \pi} \frac{2}{3} a^{3} \cos ^{2} \theta \mathrm{~d} \theta=\frac{2}{3} a^{3} \pi$ ．