下册 9.1 第一型曲线积分与第一型曲面积分 第10题

\section*{解题过程：} （1）曲面 $\displaystyle z=\sqrt{a^{2 \cdot}-x^{2}-y^{2}},(x, y) \in D=\left\{(x, y) \mid x^{2}+y^{2} \leqslant a^{2}\right\} \cdot \sqrt{1+z_{x}^{2}+z_{y}^{2}}=\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}}$ 。于是 $$ \begin{aligned} \iint_{S}\left(2 x^{2}+z^{2}\right) \mathrm{d} S & =\iint_{D}\left(2 x^{2}+a^{2}-x^{2}-y^{2}\right) \frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}} \mathrm{~d} x \mathrm{~d} y \\ & =\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{a}\left(a^{2}+r^{2} \cos ^{2} \theta-r^{2} \sin ^{2} \theta\right) \frac{a}{\sqrt{a^{2}-r^{2}}} r \mathrm{~d} r \\ & =\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{a}\left(a^{2}+r^{2} \cos 2 \theta\right) \frac{a}{\sqrt{a^{2}-r^{2}}} r \mathrm{~d} r=-\left.2 \pi a^{3} \sqrt{a^{2}-r^{2}}\right|_{0} ^{a}=2 \pi a^{4} \end{aligned} $$ （2）如图9．16所示，因为在 $S$ 上，$x^{2}+y^{2}+z^{2}=1$ ，所以 $\iint_{S}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} S=\iint_{S} \mathrm{~d} S=2 \pi$ ． \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-239.jpg?height=775&width=975&top_left_y=5380&top_left_x=1229} \captionsetup{labelformat=empty} \caption{图 9.16} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-239.jpg?height=1071&width=1002&top_left_y=5097&top_left_x=3605} \captionsetup{labelformat=empty} \caption{图 9.17} \end{figure} （3）如图 9.17 所示，曲面 $z=\sqrt{1-x^{2}-y^{2}},(x, y) \in D, D: x^{2}+y^{2} \leqslant 1, x \geqslant 0, y \geqslant 0$ 。于是 $$ \begin{aligned} \iint_{S}\left(x^{2}+y^{2}\right) z^{3} \mathrm{~d} S & =\iint_{D}\left(x^{2}+y^{2}\right)\left(\sqrt{1-x^{2}-y^{2}}\right)^{3} \frac{1}{\sqrt{1-x^{2}-y^{2}}} \mathrm{~d} x \mathrm{~d} y \\ & =\iint_{D}\left(x^{2}+y^{2}\right)\left(1-x^{2}-y^{2}\right) \mathrm{d} x \mathrm{~d} y=\int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{1} r^{2}\left(1-r^{2}\right) r \mathrm{~d} r \\ & =\frac{\pi}{2} \int_{0}^{1} r^{2}\left(1-r^{2}\right) r \mathrm{~d} r=\left.\frac{\pi}{2}\left(\frac{1}{4} r^{4}-\frac{1}{6} r^{6}\right)\right|_{0} ^{1}=\frac{1}{24} \pi \end{aligned} $$ （4）如图 9.18 所示，曲面 $S$ 由上下两部分 $S_{1}, S_{2}$ 组成 $$ \begin{gathered} S_{1}: z=\sqrt{1-x^{2}-y^{2}}, S_{2}: z=-\sqrt{1-x^{2}-y^{2}},(x, y) \in D, D: x^{2}+y^{2} \leqslant 1 . \\ \frac{\partial z}{\partial x}=\mp \frac{x}{\sqrt{1-x^{2}-y^{2}}}, \frac{\partial z}{\partial y}=\mp \frac{y}{\sqrt{1-x^{2}-y^{2}}}, \\ \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}}=\frac{1}{\sqrt{1-x^{2}-y^{2}}} . \end{gathered} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-240.jpg?height=1189&width=1238&top_left_y=863&top_left_x=4254} \captionsetup{labelformat=empty} \caption{图 9.18} \end{figure} 于是 $$ \begin{aligned} \iint_{S}\left(x^{2}+y^{2}\right)^{\beta} \mathrm{d} s & =2 \iint_{D}\left(x^{2}+y^{2}\right)^{\beta} \frac{1}{\sqrt{1-x^{2}-y^{2}}} \mathrm{~d} x \mathrm{~d} y=2 \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} r^{2 \beta} \frac{1}{\sqrt{1-r^{2}}} r \mathrm{~d} r \\ & =4 \pi \int_{0}^{1} r^{2 \beta} \frac{1}{\sqrt{1-r^{2}}} r \mathrm{~d} r=2 \pi \int_{0}^{1} r^{\beta}(1-r)^{-\frac{1}{2}} \mathrm{~d} r=2 \pi B\left(\frac{1}{2}, \beta+1\right)=2 \pi \frac{\sqrt{\pi} \Gamma(\beta+1)}{\Gamma\left(\beta+\frac{3}{2}\right)} \end{aligned} $$