下册 8.2 三重积分 第10题

解题过程： （1）旋转后形成的立体 $V$ 为曲面：$x^{2}+y^{2}+z^{2}=1$ ，锥体 $\displaystyle \sqrt{x^{2}+y^{2}} \leqslant \frac{z+1}{2}$ 及平面：$z=0$ 所围成． \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-204.jpg?height=1375&width=3144&top_left_y=787&top_left_x=1339} \captionsetup{labelformat=empty} \caption{图8．150} \end{figure} 方法 1：将 $V$ 看成两部分，如图 8．150（2）所示：$V=V_{1}-V_{2}$ ，其中 $$ \begin{aligned} & V_{1}=\left\{(x, y, z) \mid x^{2}+y^{2}+z^{2} \leqslant 1, \sqrt{x^{2}+y^{2}} \leqslant \frac{z+1}{2}\right\}, \\ & V_{2}=\left\{(x, y, z) \mid z \leqslant 0, \sqrt{x^{2}+y^{2}} \leqslant \frac{z+1}{2}\right\} . \end{aligned} $$ 记 $\displaystyle D_{x y}=\left\{(x, y) \left\lvert\, x^{2}+y^{2} \leqslant\left(\frac{4}{5}\right)^{2}\right.\right\}$ ，则 $$ \begin{aligned} \iiint_{V_{1}} \frac{2 z}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} V & =\iint_{D_{v y}} \frac{1}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} x \mathrm{~d} y \int_{2 \sqrt{x^{2}+y^{2}-1}}^{\sqrt{1-x^{2}-y^{2}}} 2 z \mathrm{~d} z=\iint_{D_{v y}} \frac{1}{\sqrt{x^{2}+y^{2}}}\left[1-x^{2}-y^{2}-\left(2 \sqrt{x^{2}+y^{2}}-1\right)^{2}\right] \mathrm{d} x \mathrm{~d} y \\ & =\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{4}{5}}\left[1-r^{2}-(2 r-1)^{2}\right] r \mathrm{~d} r=\frac{4 \pi}{3}\left(\frac{4}{5}\right)^{2} \end{aligned} $$ $$ \iiint_{V_{2}} \frac{2 z}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} V=\int_{-1}^{0} 2 z \mathrm{~d} z \iint_{D(z)} \frac{1}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} x \mathrm{~d} y=\int_{-1}^{0} 2 z \cdot\left(2 \pi \frac{z+1}{2}\right) \mathrm{d} z=2 \pi \int_{-1}^{0} z \cdot(z+1) \mathrm{d} z=-\frac{1}{3} \pi . $$ 所以 $\displaystyle \iiint_{V} \frac{2 z}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{4 \pi}{3}\left(\frac{4}{5}\right)^{2}-\left(-\frac{1}{3} \pi\right)=\frac{89 \pi}{75}$ ． 方法 2：将 $V$ 看成两部分，如图 8．150（2）所示：$V=V_{1}+V_{2}$ ，其中 $$ \begin{gathered} V_{1}=\left\{(x, y, z) \mid x^{2}+y^{2}+z^{2} \leqslant 1, \sqrt{x^{2}+y^{2}} \leqslant z \tan 2 \alpha\right\}, \\ V_{2}=\left\{(x, y, z) \mid, \sqrt{x^{2}+y^{2}} \geqslant z \tan 2 \alpha, \sqrt{x^{2}+y^{2}} \leqslant \frac{z+1}{2}\right\}, \tan \alpha=\frac{1}{2} . \\ \iiint_{V} \frac{2 z}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} V=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{2 \alpha} \mathrm{~d} \varphi \int_{0}^{1} \frac{2 r \cos \varphi}{r \sin \varphi} r^{2} \sin \varphi \mathrm{~d} r+\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{2 \alpha}^{\frac{\pi}{2}} \mathrm{~d} \varphi r \mathrm{~d} r \int_{0}^{\frac{1}{2 \sin \varphi-\cos \varphi}} \frac{2 r \cos \varphi}{r \sin \varphi} r^{2} \sin \varphi \mathrm{~d} r \\ =2(2 \pi) \int_{0}^{2 \alpha} \cos \varphi \mathrm{~d} \varphi \int_{0}^{1} r^{2} \mathrm{~d} r+2(2 \pi) \int_{2 \alpha}^{\frac{\pi}{2}} \cos \varphi \mathrm{~d} \varphi \int_{0}^{\frac{1}{2 \sin \varphi-\cos \varphi}} r^{2} \mathrm{~d} r=\frac{89 \pi}{75} . \end{gathered} $$ （2）如图8．151（1），（2）所示，旋转曲面的方程为 $x^{2}+y^{2}=2 z, 2 \leqslant z \leqslant 8$ 。将 $\Omega$ 向 $z$ 轴投影得投影区间 $[2,8]$ ．由于 $\Omega$ 由 $z=2, z=8$ 及曲面 $x^{2}+y^{2}=2 z$ 围成，所以 $D(z)$ 为 $x^{2}+y^{2} \leqslant 2 z$ ．于是 $$ \iiint_{\Omega}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} x=\int_{2}^{8} \mathrm{~d} z \iint_{D(z)}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y=\int_{2}^{8} \mathrm{~d} z \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\sqrt{2 z}} r^{2} \cdot r \mathrm{~d} y=2 \pi \int_{2}^{8} \frac{4 z^{2}}{4} \mathrm{~d} z=336 \pi $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-205.jpg?height=1210&width=1023&top_left_y=1574&top_left_x=849} \captionsetup{labelformat=empty} \caption{图8．151} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-205.jpg?height=1479&width=920&top_left_y=1298&top_left_x=2493} \captionsetup{labelformat=empty} \caption{图8．151} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-205.jpg?height=1175&width=865&top_left_y=1326&top_left_x=4171} \captionsetup{labelformat=empty} \caption{（ 3 ）} \end{figure} （3）如图8．151（1），（3）所示，$\Omega$ 由旋转抛物面 $y^{2}+x^{2}=2 z$ 与平面 $z=4$ 围成．将 $\Omega$ 投影到 $z$ 轴得投影区间 $[0,4]$ ，此时可得 $D(z): x^{2}+y^{2} \leqslant 2 z$ ．于是 $$ \iiint_{\Omega}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{4} \mathrm{~d} z \iint_{D(z)}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y=\int_{0}^{4} \mathrm{~d} z \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\sqrt{2 z}}\left(r^{2}+z^{2}\right) r \mathrm{~d} r=\frac{512}{3} \pi $$