下册 8.2 三重积分 第15题

\section*{解题过程：} （1）如图8．156所示，交换积分顺序得 $$ \begin{aligned} I & =\int_{0}^{1} \mathrm{~d} x \int_{x}^{1} \mathrm{~d} y \int_{y}^{1} y \sqrt{1+z^{4}} \mathrm{~d} z=\int_{0}^{1} \mathrm{~d} x \int_{x}^{1} \mathrm{~d} z \int_{x}^{z} y \sqrt{1+z^{4}} \mathrm{~d} y=\left.\int_{0}^{1} \mathrm{~d} x \int_{x}^{1} \frac{y^{2}}{2} \sqrt{1+z^{4}} \mathrm{~d} z\right|_{y=x} ^{y=z} \\ & =\int_{0}^{1} \mathrm{~d} x \int_{x}^{1} \frac{1}{2}\left(z^{2}-x^{2}\right) \sqrt{1+z^{4}} \mathrm{~d} z=\int_{0}^{1} \mathrm{~d} z \int_{0}^{z} \frac{1}{2}\left(z^{2}-x^{2}\right) \sqrt{1+z^{4}} \mathrm{~d} x=\int_{0}^{1} \frac{1}{2} \sqrt{1+z^{4}} \mathrm{~d} z \int_{0}^{z}\left(z^{2}-x^{2}\right) \mathrm{d} x \\ & =\int_{0}^{1} \frac{1}{3} z^{3} \sqrt{1+z^{4}} \mathrm{~d} z=\frac{1}{18}(2 \sqrt{2}-1) \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-210.jpg?height=1424&width=1272&top_left_y=925&top_left_x=1091} \captionsetup{labelformat=empty} \caption{图8．156} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-210.jpg?height=1527&width=1272&top_left_y=801&top_left_x=3280} \captionsetup{labelformat=empty} \caption{图8．157} \end{figure} （2）如图 8．157，记 $\Omega: x^{2}+y^{2}+(z-1)^{2} \leqslant 1, y \geqslant 0, z \geqslant 1, D(z): x^{2}+y^{2} \leqslant 1-(z-1)^{2}, y \geqslant 0$ ，则 $$ \begin{aligned} I & =\int_{-1}^{1} \mathrm{~d} x \int_{0}^{\sqrt{1-x^{2}}} \mathrm{~d} y \int_{1}^{1+\sqrt{1-x^{2}-y^{2}}} \frac{\mathrm{~d} z}{x^{2}+y^{2}+z^{2}}=\iiint_{\Omega} \frac{\mathrm{d} z}{x^{2}+y^{2}+z^{2}}=\int_{1}^{2} \mathrm{~d} z \iint_{D(z)} \frac{1}{x^{2}+y^{2}+z^{2}} \\ & =\int_{1}^{2} \mathrm{~d} z \int_{0}^{\sqrt{1-(z-1)^{2}}} \mathrm{~d} r \int_{0}^{\pi} \frac{r \mathrm{~d} \theta}{r^{2}+z^{2}}=\pi \int_{1}^{2} \mathrm{~d} z \int_{0}^{\sqrt{1-(z-1)^{2}}} \frac{r}{r^{2}+z^{2}} \mathrm{~d} r=\left.\frac{\pi}{2} \int_{1}^{2} \ln \left(r^{2}+z^{2}\right)\right|_{0} ^{\sqrt{1-(z-1)^{2}}} \mathrm{~d} z \\ & =\frac{\pi}{2} \int_{1}^{2}(\ln 2-\ln z) \mathrm{d} z=\frac{\pi}{2}(1-\ln 2) \end{aligned} $$ （3）方法 1：如图 8．158，应用柱坐标变换，积分区域变为 $$ V^{\prime}=\left\{(r, \theta, z) \mid 0 \leqslant r \leqslant 1,0 \leqslant \theta \leqslant \frac{\pi}{2}, r \leqslant z \leqslant \sqrt{2-r^{2}}\right\} . $$ 从而 $$ \begin{aligned} \int_{0}^{1} \mathrm{~d} x \int_{0}^{\sqrt{1-x^{2}}} \mathrm{~d} y \int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^{2}-y^{2}}} z^{2} \mathrm{~d} z & =\int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{1} r \mathrm{~d} r \int_{r}^{\sqrt{2-r^{2}}} z^{2} \mathrm{~d} z \\ & =\frac{\pi}{2} \cdot \frac{1}{3} \int_{0}^{1}\left[\left(2-r^{2}\right)^{\frac{3}{2}}-r^{3}\right] r \mathrm{~d} r=\frac{\pi}{15}(2 \sqrt{2}-1) \end{aligned} $$ 方法 2：应用球面坐标，积分区域变为 $\displaystyle V^{\prime}=\left\{(r, \varphi, \theta): 0 \leqslant r \leqslant \sqrt{2}, 0 \leqslant \varphi \leqslant \frac{\pi}{4}, 0 \leqslant \theta \leqslant \frac{\pi}{2}\right\}$ ．于是 $$ \begin{aligned} \int_{0}^{1} \mathrm{~d} x \int_{0}^{\sqrt{1-x^{2}}} \mathrm{~d} y \int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^{2}-y^{2}}} z^{2} \mathrm{~d} z & =\int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{4}} \sin \varphi \mathrm{~d} \varphi \int_{0}^{\sqrt{2}} r^{2} \cos ^{2} \varphi r^{2} \mathrm{~d} r \\ & =\frac{2 \sqrt{2}}{5} \pi \int_{0}^{\frac{\pi}{4}} \sin \varphi \cos ^{2} \varphi \mathrm{~d} \varphi=\frac{2 \sqrt{2}-1}{15} \pi \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-210.jpg?height=1237&width=1238&top_left_y=6741&top_left_x=966} \captionsetup{labelformat=empty} \caption{图8．158} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-210.jpg?height=1334&width=1320&top_left_y=6637&top_left_x=3391} \captionsetup{labelformat=empty} \caption{图8．159} \end{figure} （4）如图8．159所示，交换积分次序。先交换 $y, z$ 有 $$ \begin{aligned} I & =\int_{0}^{1} \mathrm{~d} x \int_{0}^{1-x} \mathrm{~d} y \int_{0}^{\frac{y}{2}} \frac{\cos z}{(2 z-1)^{2}} \mathrm{~d} z=\int_{0}^{1} \mathrm{~d} x \int_{0}^{\frac{1-x}{2}} \mathrm{~d} z \int_{2 z}^{1-x} \frac{\cos z}{(2 z-1)^{2}} \mathrm{~d} y \\ & =\int_{0}^{1} \mathrm{~d} x \int_{0}^{\frac{1-x}{2}} \frac{\cos z}{(2 z-1)^{2}}(1-x-2 z) \mathrm{d} z=-\int_{0}^{1} \mathrm{~d} x \int_{0}^{\frac{1-x}{2}}\left(\frac{x \cos z}{(2 z-1)^{2}}+\frac{\cos z}{2 z-1}\right) \mathrm{d} z \end{aligned} $$ 再交换 $x, z$ 得 $$ I=-\int_{0}^{1} \mathrm{~d} x \int_{0}^{\frac{1-x}{2}}\left(\frac{x \cos z}{(2 z-1)^{2}}+\frac{\cos z}{2 z-1}\right) \mathrm{d} z=-\int_{0}^{\frac{1}{2}} \mathrm{~d} z \int_{0}^{1-2 z}\left(\frac{x \cos z}{(2 z-1)^{2}}+\frac{\cos z}{2 z-1}\right) \mathrm{d} x=\frac{1}{2} \int_{0}^{\frac{1}{2}} \cos z \mathrm{~d} z=\frac{1}{2} \sin \frac{1}{2} . $$