下册 9.1 第一型曲线积分与第一型曲面积分 第8题

\section*{解题过程：} （1）如图 9.10 所示，$\displaystyle z=\sqrt{a^{2}-x^{2}-y^{2}}, \sqrt{1+z_{x}{ }^{2}+z_{y}{ }^{2}}=\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}}$ 。从而 $$ \begin{aligned} \iint_{S}(x+y+z) \mathrm{d} S & =\int_{-a}^{a} \mathrm{~d} x \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}} \frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}}\left(x+y+\sqrt{a^{2}-x^{2}-y^{2}}\right) \mathrm{d} y \\ & =\int_{-a}^{a}\left(\pi a x+2 a \sqrt{a^{2}-x^{2}}\right) \mathrm{d} x=\pi a^{3} . \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-235.jpg?height=906&width=1037&top_left_y=6464&top_left_x=1243} \captionsetup{labelformat=empty} \caption{图 9.10} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-235.jpg?height=885&width=1037&top_left_y=6485&top_left_x=3626} \captionsetup{labelformat=empty} \caption{图9．11} \end{figure} （2）曲面 $S$ 由两部分 $S_{1}, S_{2}$ 组成，其中 $S_{1}: x=a+\sqrt{a^{2}-\left(y^{2}+z^{2}\right)}, S_{2}: x=a-\sqrt{a^{2}-\left(y^{2}+z^{2}\right)}$ ．它们在 $z O y$ 平面上的投影区域都是 $D_{z y}: z^{2}+y^{2} \leqslant a^{2}$ ． $$ 1+x_{y}^{2}+x_{z}^{2}=1+\frac{y^{2}}{a^{2}-\left(y^{2}+z^{2}\right)}+\frac{z^{2}}{a^{2}-\left(y^{2}+z^{2}\right)}=\frac{a^{2}}{a^{2}-\left(y^{2}+z^{2}\right)} . $$ 于是 $\iint_{S}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} S$ $$ \begin{aligned} & =\iint_{S} 2 a x \mathrm{~d} S=2 a \iint_{S_{1}} x \mathrm{~d} S+2 a \iint_{S_{2}} x \mathrm{~d} S \\ & =2 a \iint_{D_{n z}}\left(a+\sqrt{a^{2}-\left(y^{2}+z^{2}\right)}\right) \sqrt{\frac{a^{2}}{a^{2}-\left(y^{2}+z^{2}\right)}} \mathrm{d} y \mathrm{~d} z+2 a \iint_{D_{n z}}\left(a-\sqrt{a^{2}-\left(y^{2}+z^{2}\right)}\right) \sqrt{\frac{a^{2}}{a^{2}-\left(y^{2}+z^{2}\right)}} \mathrm{d} y \mathrm{~d} z \\ & =4 a^{3} \iint_{D_{n z}} \frac{1}{\sqrt{a^{2}-\left(y^{2}+z^{2}\right)}} \mathrm{d} y \mathrm{~d} z=4 a^{3} \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{a} \frac{r}{\sqrt{a^{2}-r^{2}}} \mathrm{~d} r=-\left.8 \pi a^{3} \sqrt{a^{2}-r^{2}}\right|_{0} ^{a}=8 \pi a^{4} . \end{aligned} $$ （3）如图 9.12 所示，所截得的曲面 $S$ 为 $z=\sqrt{x^{2}+y^{2}}$ ， $$ \begin{gathered} D=\left\{(x, y): x^{2}+y^{2} \leqslant a x\right\}=\left\{(x, y):\left(x-\frac{a}{2}\right)^{2}+y^{2} \leqslant\left(\frac{a}{2}\right)^{2}\right\} \\ \mathrm{d} S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} \mathrm{~d} x \mathrm{~d} y=\sqrt{2} \mathrm{~d} x \mathrm{~d} y \end{gathered} $$ 于是 $$ \iint_{S} x \mathrm{~d} S=\iint_{D} x \sqrt{2} \mathrm{~d} x \mathrm{~d} y $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-236.jpg?height=1092&width=1141&top_left_y=2935&top_left_x=4323} \captionsetup{labelformat=empty} \caption{图 9.12} \end{figure} 作极坐标变换 $\displaystyle x=\frac{a}{2}+r \cos \theta, y=r \sin \theta$ ，则 $\displaystyle 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant \frac{1}{2} a, \mathrm{~d} x \mathrm{~d} y=r \mathrm{~d} r \mathrm{~d} \theta$ 。于是 $$ \iint_{S} x \mathrm{~d} S=\iint_{D} x \sqrt{2} \mathrm{~d} x \mathrm{~d} y=\sqrt{2} \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{1}{2} a}\left(\frac{a}{2}+r \cos \theta\right) \cdot r \mathrm{~d} r=\frac{\sqrt{2}}{2} 2 \pi \cdot a \cdot \frac{1}{2} \cdot\left(\frac{1}{2} a\right)^{2}=\frac{\sqrt{2}}{8} \pi a^{3} $$ （4）如图 9.12 所示，曲面 $S$ 的方程为 $z=\sqrt{x^{2}+y^{2}} . S$ 在 $x O y$ 平面上的投影区域为 $D$ ： $x^{2}+y^{2} \leqslant 2 x . \mathrm{d} S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} \mathrm{~d} x \mathrm{~d} y=\sqrt{2} \mathrm{~d} x \mathrm{~d} y$ 。于是 $$ \iint_{S} z \mathrm{~d} s=\iint_{D} \sqrt{x^{2}+y^{2}} \sqrt{2} \mathrm{~d} x \mathrm{~d} y=\sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{2 \cos \theta} r^{2} \mathrm{~d} r=\frac{16}{3} \sqrt{2} \int_{0}^{\frac{\pi}{2}} \cos ^{3} \theta \mathrm{~d} \theta=\frac{32}{9} \sqrt{2} $$ （5）如图 9.13 所示，曲面 $S$ 关于 $y O z, z O x$ 两坐标平面对称．由对称性 $$ \iint_{S} x \mathrm{~d} S=\iint_{D} x \frac{a}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x \mathrm{~d} y=0 $$ 在对称点上被积函数 $f(x, y, z)=z$ 大小相等，符号相同．因此积分 $\iint_{S} z \mathrm{~d} S$ 等于 $S$ 在第一卦限内的部分 $S_{1}$ 上积分的 4 倍。 由 $x^{2}+z^{2}=2 a z, z=\sqrt{x^{2}+y^{2}}$ ，消去 $z$ ，可知 $S_{1}$ 在 $x O y$ 平面上的投影区域为第一象限内 $x, y$ 轴与曲线 $2 x^{2}+y^{2}=2 a \sqrt{x^{2}+y^{2}}$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-236.jpg?height=1134&width=1652&top_left_y=6340&top_left_x=3867} \captionsetup{labelformat=empty} \caption{图 9.13} \end{figure}所围的区域 $D$ ．所截得的曲面 $S$ 为 $z=a+\sqrt{a^{2}-x^{2}}$ ． $$ \mathrm{d} S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} \mathrm{~d} x \mathrm{~d} y=\frac{a}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x \mathrm{~d} y $$ 于是 $$ \begin{aligned} & \iint_{S}(x+z) \mathrm{d} S=\iint_{S_{1}} z \mathrm{~d} S \\ & \iint_{S_{1}} z \mathrm{~d} S=\iint_{D}\left(a+\sqrt{a^{2}-x^{2}}\right) \frac{a}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x \mathrm{~d} y=a^{2} \iint_{D} \frac{1}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x \mathrm{~d} y+a \iint_{D} \mathrm{~d} x \mathrm{~d} y \end{aligned} $$ 作极坐标变换 $x=\operatorname{arcos} \theta, y=\operatorname{ar} \sin \theta$ ，则 $\displaystyle 0 \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant \frac{2}{1+\cos ^{2} \theta}, J=a^{2} r$ ． $$ \begin{gathered} \iint_{D} \mathrm{~d} x \mathrm{~d} y=a^{2} \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{2}{1+\cos ^{2} \theta}} r \mathrm{~d} r=\frac{1}{2} \cdot 4 a^{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\left(1+\cos ^{2} \theta\right)^{2}} \mathrm{~d} \theta=2 a^{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\left(1+\cos ^{2} \theta\right)^{2}} \mathrm{~d} \theta \\ \iint_{D} \frac{1}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x \mathrm{~d} y=a \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{2}{1+\cos ^{2} \theta}} \frac{1}{\sqrt{1-r^{2} \cos ^{2} \theta} r \mathrm{~d} r=\left.a \int_{0}^{\frac{\pi}{2}}\left(-\frac{1}{\cos ^{2} \theta} \sqrt{1-r^{2} \cos ^{2} \theta}\right)\right|_{0} ^{\frac{2}{1+\cos ^{2} \theta}} \mathrm{~d} \theta} \\ =a \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos ^{2} \theta}\left(1-\frac{1-\cos ^{2} \theta}{1+\cos ^{2} \theta}\right) \mathrm{d} \theta=2 a \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos ^{2} \theta} \mathrm{~d} \theta \end{gathered} $$ 令 $t=\tan \theta$ ，则 $$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos ^{2} \theta} \mathrm{~d} \theta & =\int_{0}^{+\infty} \frac{1}{2+t^{2}} \mathrm{~d} t=\left.\frac{1}{\sqrt{2}} \arctan \left(\frac{t}{\sqrt{2}}\right)\right|_{0} ^{+\infty}=\frac{\pi}{2 \sqrt{2}} \\ \int_{0}^{\frac{\pi}{2}} \frac{1}{\left(1+\cos ^{2} \theta\right)^{2}} \mathrm{~d} \theta & =\int_{0}^{+\infty} \frac{1+t^{2}}{\left(2+t^{2}\right)^{2}} \mathrm{~d} t=\int_{0}^{+\infty}\left[\frac{1}{2+t^{2}}-\frac{1}{\left(2+t^{2}\right)^{2}}\right] \mathrm{d} t \\ & =\frac{\pi}{2 \sqrt{2}}-\left.\frac{1}{4 \sqrt{2}}\left(\arctan \frac{t}{\sqrt{2}}+\frac{\sqrt{2} t}{2+t^{2}}\right)\right|_{0} ^{+\infty}=\frac{\pi}{2 \sqrt{2}}-\frac{1}{4 \sqrt{2}} \frac{\pi}{2}=\frac{3 \pi}{8 \sqrt{2}} \end{aligned} $$ 于是 $$ \iint_{S_{1}} z \mathrm{~d} S=a^{2} \cdot 2 a \cdot \frac{\pi}{2 \sqrt{2}}+a \cdot 2 a^{2} \frac{3 \pi}{8 \sqrt{2}}=2 a^{3} \frac{7 \pi}{8 \sqrt{2}} $$ 从而 $$ \iint_{S}(x+z) \mathrm{d} S=4 \iint_{S_{1}} z \mathrm{~d} S=8 a^{3} \frac{7 \pi}{8 \sqrt{2}}=\frac{7 \sqrt{2}}{2} \pi a^{3} . $$ （6）如图 9.14 所示，曲面 $S$ 的方程为 $z=\sqrt{x^{2}+y^{2}} . S$ 在 $x O y$ 平面上的投影区域为 $D: x^{2}+y^{2} \leqslant 4$ ． $$ \mathrm{d} S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} \mathrm{~d} x \mathrm{~d} y=\sqrt{2} \mathrm{~d} x \mathrm{~d} y $$ 于是 $$ \iint_{S} z \mathrm{~d} S=\iint_{D} \sqrt{x^{2}+y^{2}} \sqrt{2} \mathrm{~d} \sigma=\sqrt{2} \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{2} r^{2} \mathrm{~d} r=\frac{16 \sqrt{2}}{3} \pi $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-237.jpg?height=1306&width=1100&top_left_y=5926&top_left_x=4323} \captionsetup{labelformat=empty} \caption{图9．14} \end{figure}