下册 8.2 三重积分 第4题

\section*{解题过程：} （1）如图8．131 所示，在球面坐标下，积分区域 $V$ 可表示为 $\displaystyle 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant \varphi \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant \cos \varphi$ 。于是 $$ \iiint_{V} \sqrt{x^{2}+y^{2}+z^{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \sin \varphi \mathrm{~d} \varphi \int_{0}^{\cos \varphi} r^{3} \mathrm{~d} r=\frac{\pi}{10} . $$ （2）如图8．131 所示，在球面坐标下，积分区域 $V$ 可表示为 $\displaystyle 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant \varphi \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant 2 \cos \varphi$ 。于是 \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-193.jpg?height=1237&width=1113&top_left_y=925&top_left_x=4420} \captionsetup{labelformat=empty} \caption{图 8.131} \end{figure} $$ \iiint_{x^{2}+y^{2}+z^{2}<2 z}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{5}{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \sin \varphi \mathrm{~d} \varphi \int_{0}^{2 \cos \varphi} r^{7} \mathrm{~d} r=\frac{64 \pi}{9} . $$ （3）如图 8．131．所示，在球面坐标下，积分区域 $V$ 可表示为 $\displaystyle 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant \varphi \leqslant \frac{\pi}{2}$ ， $0 \leqslant r \leqslant 2 \cos \varphi$ ．于是 $$ \iiint_{V} \frac{\mathrm{~d} V}{\sqrt{x^{2}+y^{2}+z^{2}}}=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \sin \varphi \mathrm{~d} \varphi \int_{0}^{2 \cos \varphi} r \mathrm{~d} r=\frac{4 \pi}{3} . $$ （4）由 $\partial V=\left\{(x, y, z) \mid(x-a)^{2}+(y-a)^{2}+(z-a)^{2}=a^{2}\right\}$ 知，$V$ 是以 $(a, a, a)$ 为心，$a$ 为半径的球 （见图8．132）。 由泰勒公式有 $x^{3}=(x-a)^{3}+3 a(x-a)^{2}+3 a^{2}(x-a)+a^{3}$ ．所以 $$ \begin{aligned} & \iiint_{V}\left(x^{3}+y^{3}+z^{3}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & =3 \iiint_{V} x^{3} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=3 \iiint_{V}\left[(x-a)^{3}+3 a(x-a)^{2}+3 a^{2}(x-a)+a^{3}\right] \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & =3 \iiint_{V}\left[3 a(x-a)^{2}+a^{3}\right] \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=9 a \cdot \frac{1}{3} \iiint_{V}\left[(x-a)^{2}+(y-a)^{2}+(z-a)^{2}\right] \mathrm{d} x \mathrm{~d} y \mathrm{~d} z+3 \iiint_{V} a^{3} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \\ & =3 a \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\pi} \sin \varphi \mathrm{d} \varphi \int_{0}^{a} r^{4} \mathrm{~d} r+3 a^{3} \cdot \frac{4}{3} \pi a^{3}=\frac{32}{5} \pi a^{6} \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-193.jpg?height=920&width=1003&top_left_y=6133&top_left_x=1215} \captionsetup{labelformat=empty} \caption{图8．132} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-193.jpg?height=1244&width=1016&top_left_y=5815&top_left_x=3633} \captionsetup{labelformat=empty} \caption{图8．133} \end{figure} （5）分别记 $V_{1}, V_{2}$ 为球面 $x^{2}+y^{2}+z^{2}=2 a z$ 及 $x^{2}+y^{2}+z^{2}=a z$ 所包围的区域（见图8．133）． $$ \begin{aligned} \iiint_{D} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z & =\iiint_{V_{1}} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z-\iiint_{V_{2}} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \\ & =\left(a \iiint_{V_{1}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z+\iiint_{V_{1}}(z-a) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z\right)-\left(\frac{a}{2} \iiint_{V_{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z+\iiint_{V_{2}}\left(z-\frac{a}{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z\right) \end{aligned} $$ $$ =a \iiint_{V_{1}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z-\frac{a}{2} \iiint_{V_{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{4 \pi}{3}\left[a^{4}-\left(\frac{a}{2}\right)^{4}\right]=\frac{5 \pi}{4} a^{4} $$