下册 8.2 三重积分 第6题

\section*{解题过程：} （1）如图8．139所示，作柱面坐标变换 $x=r \cos \theta, y=r \sin \theta, z=z$ ，则 $V$ 可表示为 $0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant 2, r^{2} \leqslant z \leqslant 4$ 。于是 $$ \iiint_{V}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{2} r^{3} \mathrm{~d} r \int_{r^{2}}^{4} \mathrm{~d} z=\frac{32 \pi}{3} . $$ （2）如图8．139所示，作柱面坐标变换 $x=r \cos \theta, y=r \sin \theta, z=z$ ，则 $V$ 可表示为 $\displaystyle 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant 2 \sqrt{2}, \frac{r^{2}}{2} \leqslant z \leqslant 4$ ．于是 \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-197.jpg?height=1182&width=982&top_left_y=5373&top_left_x=4641} \captionsetup{labelformat=empty} \caption{图8．139} \end{figure} $\displaystyle \iiint_{V}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} V=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{2 \sqrt{2}} r \mathrm{~d} r \int_{\frac{r^{2}}{2}}^{4}\left(r^{2}+z^{2}\right) \mathrm{d} z=2 \pi\left(\frac{320}{3}-\frac{128 \sqrt{2}}{21}\right)$. （3）方法1： $\displaystyle \iint_{V} \sqrt{x^{2}+y^{2}} \mathrm{~d} V=\int_{0}^{1} \mathrm{~d} z \iint_{x^{2}+y^{2}<1} \sqrt{x^{2}+y^{2}} \mathrm{~d} x \mathrm{~d} y=\int_{0}^{1} \mathrm{~d} z \int_{0}^{1} r 2 \pi r=\frac{\pi}{6}$ ． 方法 2：如图 8．140，在柱面坐标下，积分区域 $V$ 可表示为 $0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant 1, r \leqslant z \leqslant 1$ ．于是 $$ \iiint_{V} \sqrt{x^{2}+y^{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} r \cdot r \mathrm{~d} r \int_{r}^{1} \mathrm{~d} z=2 \pi \int_{0}^{1}\left(r^{2}-r^{3}\right) \mathrm{d} r=\frac{\pi}{6} . $$ （4）如图8．140所示，在柱面坐标下，积分区域 $V$ 可表示为 $$ 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant 1, r \leqslant z \leqslant 1 . $$ 于是 $$ \begin{gathered} \iiint_{V}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} r^{2} \cdot r \mathrm{~d} r \int_{r}^{1} \mathrm{~d} z=2 \pi \int_{0}^{1}\left(r^{3}-r^{4}\right) \mathrm{d} r=\frac{\pi}{10} \\ \iiint_{V} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{1} z \mathrm{~d} z \iint_{x^{2}+y^{2}