下册 8.2 三重积分 第18题

解题过程： （1）如图8．172 所示，由 $\displaystyle \frac{x^{2}}{12}+\frac{y^{2}}{36} \leqslant 2 z, \frac{x^{2}}{6}+\frac{y^{2}}{12} \leqslant 2(2-z)$ 得投影区域为 $\displaystyle D: \frac{x^{2}}{16}+\frac{y^{2}}{36} \leqslant 1$ ．利用极坐标变换得 $$ \begin{aligned} V= & \iint_{D}\left(2-\frac{x^{2}}{12}-\frac{y^{2}}{24}\right) \mathrm{d} x \mathrm{~d} y-\iint_{D}\left(\frac{x^{2}}{24}+\frac{y^{2}}{72}\right) \mathrm{d} x \mathrm{~d} y \\ = & 48 \pi-\frac{1}{12} \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} 16 r^{3} \cos ^{2} \theta \mathrm{~d} r-\frac{1}{24} \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} 36 r^{3} \sin ^{2} \theta \mathrm{~d} r \\ & -\frac{1}{24} \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} 16 r^{3} \cos ^{2} \theta \mathrm{~d} r-\frac{1}{72} \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} 36 r^{3} \sin ^{2} \theta \mathrm{~d} r \\ = & 48 \pi-\frac{\pi}{3}-\frac{3 \pi}{8}-\frac{\pi}{6}-\frac{\pi}{8}=47 \pi \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-216.jpg?height=1085&width=1811&top_left_y=6064&top_left_x=911} \captionsetup{labelformat=empty} \caption{图8．172} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-216.jpg?height=1161&width=1168&top_left_y=5988&top_left_x=3771} \captionsetup{labelformat=empty} \caption{图8．173} \end{figure} （2）如图8．173所示，设立方体在 $x y$ 平面的投影区域为 $\displaystyle D=\left\{(x, y) \left\lvert\, \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant \frac{1}{2}\right.\right\}$ ．于是 $$ V=\iint_{D}\left(c \sqrt{1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}}-c \sqrt{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}}\right) \mathrm{d} x \mathrm{~d} y $$ 令 $x=a r \cos \theta, y=a r \sin \theta$ ，则 $\displaystyle J=a b r, D^{\prime}: r^{2} \leqslant \frac{1}{2}$ ．则 $$ V=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{1}{\sqrt{2}}}\left(c \sqrt{1-r^{2}}-r^{2}\right) a b r \mathrm{~d} r=2 \pi a b c \int_{0}^{\frac{1}{\sqrt{2}}}\left(r \sqrt{1-r^{2}}-r^{2}\right) \mathrm{d} r=\frac{2-\sqrt{2}}{3} \pi a b c . $$ （3）方法 1：如图8．174所示，用平面 $z=z$ 去截椭球，截面为椭圆．椭圆面积为 $\displaystyle \pi a b\left(1-\frac{z^{2}}{c^{2}}\right)$ ． 于是 $$ V=\int_{-c}^{c} \pi a b\left(1-\frac{z^{2}}{c^{2}}\right) \mathrm{d} z=\left.\pi a b\left(z-\frac{1}{3} \frac{z^{3}}{c^{2}}\right)\right|_{-c} ^{c}=\pi a b \cdot \frac{4}{3} c=\frac{4}{3} \pi a b c . $$ 方法 2：$\displaystyle z=c \sqrt{1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}}, D=\left\{(x, y): \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant 1\right\}$ ．则 $$ \begin{aligned} V & =2 \iint_{D} c \sqrt{1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}} \mathrm{~d} x \mathrm{~d} y=2 c \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} \sqrt{1-r^{2}} a b r \mathrm{~d} r \\ & =\left.4 \pi a b c\left[-\frac{1}{3}\left(1-r^{2}\right)^{\frac{3}{2}}\right]\right|_{0} ^{1}=4 \pi a b c \cdot \frac{1}{3}=\frac{4}{3} \pi a b c . \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-217.jpg?height=933&width=1196&top_left_y=2058&top_left_x=4351} \captionsetup{labelformat=empty} \caption{图 8.174} \end{figure} 方法 3：所得立体的体积 $$ V=\iiint_{V} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\pi} \mathrm{d} \varphi \int_{0}^{-1} a b c r^{2} \sin \varphi \mathrm{~d} r=2 \pi a b c \cdot \frac{1}{3} \cdot 2=\frac{4}{3} \pi a b c . $$ （4）由曲面方程可知，所求儿何体位于 $x$ 轴的正半轴，且关于两个坐标面对称，故可利用对称性． 作广义球面坐标 $x=a r \sin \varphi \cos \theta, y=b r \sin \varphi \sin \theta, z=c r \cos \varphi$ ，则 $|J|=a b c r^{2} \sin \varphi$ ，曲面方程为 $$ r^{4}=\frac{1}{h} \operatorname{ar} \sin \varphi \cos \theta \text {, 即 } r=\sqrt[3]{\frac{a \sin \varphi \cos \theta}{h}} . $$ 若 $h>0$ ，则立体落在 $x>0$ 的四个挂限内，且在每个挂限内的体积均等于第一挂限内的体积，从而所求立体的体积是第一挂限内那部分体积的四倍．于是 $$ V=\iiint_{V} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=4 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{\sqrt{\frac{a \sin \varphi \cos \theta}{h}}} a b c r^{2} \sin \varphi \mathrm{~d} r=\frac{\pi a^{2} b c}{3 h} . $$ 特别地，当 $a=b=c=h$ 时，$\displaystyle V=\iiint_{V} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=4 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{a \sqrt{\cos \varphi}} r^{2} \sin \varphi \mathrm{~d} r=\frac{1}{3} \pi a^{3}$ ． （5）令 $x=a r \sin \varphi \cos \theta, y=b r \sin \varphi \sin \theta, z=c r \cos \varphi$ ，并利用对称性得 $$ V=8 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{\sin \varphi} a b c r^{2} \sin \varphi \mathrm{~d} r=8 \frac{\pi}{2} \frac{a b c}{3} \int_{0}^{\frac{\pi}{2}} \sin ^{4} \varphi \mathrm{~d} \varphi=\frac{4 \pi a b c}{3} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi^{2}}{4} a b c . $$ （6）由曲面方程知所围立体只能位于满足 $x y z \geqslant 0$ 的四个卦限，且体积为第一卦限立体体积 $V_{1}$ 的 4 倍，即 $V=4 V_{1}=\iiint_{V_{1}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z$ ． 在广义球面坐标变换下，曲面方程为 $\displaystyle r=\left(\sin ^{2} \varphi \cos \varphi \sin \theta \cos \theta\right)^{\frac{1}{3}}$ ，其中 $$ 0 \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant \varphi \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant\left(\sin ^{2} \varphi \cos \varphi \sin \theta \cos \theta\right)^{\frac{1}{3}}, J=a b c r^{2} \sin \varphi $$ 于是 $$ \begin{aligned} V & =4 \iiint_{V_{1}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=4 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{\left(\sin ^{2} \varphi \cos \varphi \sin \theta \cos \theta\right)^{\frac{1}{3}}} a b c r^{2} \sin \varphi \mathrm{~d} r \\ & =\frac{4 a b c}{3} \int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \sin ^{3} \varphi \cos \varphi \mathrm{~d} \varphi=\frac{a b c}{6} \end{aligned} $$ 特别地，当 $a=b=c$ 时，$\displaystyle V=\iiint_{V} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{1}{6} a^{3}$ ． （7）令 $x=a r \sin \varphi \cos ^{2} \theta, y=b r \sin \varphi \sin ^{2} \theta, z=c r \cos \varphi$ ，则 于是 $$ \begin{aligned} 0 \leqslant & \theta \leqslant \frac{\pi}{2}, 0 \leqslant \varphi \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant 1, J=2 a b c r^{2} \sin \varphi \sin \theta \cos \theta \\ V & =\iiint_{V} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{1} 2 a b c r^{2} \sin \varphi \sin \theta \cos \theta \mathrm{~d} r \\ & =2 a b c \int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \sin \varphi \mathrm{~d} \varphi \int_{0}^{1} r^{2} \mathrm{~d} r=\frac{1}{3} a b c \end{aligned} $$ （8）由曲面方程知，所围立体只能位于满足 $y \geqslant 0$ 的四个卦限，且体积为第一卦限立体 $V_{1}$ 体积的 4 倍，即 $V=4 V_{1}=4 \iiint_{V_{1}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z$ ． 在球面坐标变换下，曲面方程为 $\displaystyle r=\left(\frac{\sin \varphi \cos \theta}{\sin ^{4} \varphi+\cos ^{4} \varphi}\right)^{\frac{1}{3}}$ ．于是 $$ \begin{aligned} V & =4 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{\left(\frac{\sin \varphi \cos \theta}{\sin ^{4} \varphi+\cos ^{4} \varphi}\right)^{\frac{1}{3}} r^{2} \sin \varphi \mathrm{~d} r} \\ & =\frac{4}{3} \int_{0}^{\frac{\pi}{2}} \sin \theta \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \varphi}{\sin ^{4} \varphi+\cos ^{4} \varphi} \mathrm{~d} \varphi=\frac{4}{3} \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \varphi}{\sin ^{4} \varphi+\cos ^{4} \varphi} \mathrm{~d} \varphi \end{aligned} $$ 令 $\displaystyle \varphi=\frac{\pi}{2}-t$ ，则 $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \varphi}{\sin ^{4} \varphi+\cos ^{4} \varphi} \mathrm{~d} \varphi=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} t}{\sin ^{4} t+\cos ^{4} t} \mathrm{~d} t$ ．于是 $$ V=\frac{2}{3} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin ^{4} \varphi+\cos ^{4} \varphi} \mathrm{~d} \varphi $$ 令 $u=\tan \varphi$ ，则 $\displaystyle V=\frac{2}{3} \int_{0}^{+\infty} \frac{1+t^{2}}{1+t^{4}} \mathrm{~d} t=\frac{\sqrt{2}}{3} \pi$ ． （9）在球面坐标变换 $x=r \sin \varphi \cos \theta, y=r \sin \varphi \sin \theta, z=r \cos \varphi$ 下，曲面方程为 $r=a \sqrt{\cos 2 \varphi}$ ． 记 $\displaystyle V_{1}: 0 \leqslant \theta \leqslant \frac{\pi}{2}, \frac{\pi}{4} \leqslant \varphi \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant a \sqrt{\cos 2 \varphi}$ ，则所求体积为立体体积 $V_{1}$ 的 8 倍，即 $$ V=8 V_{1}=8 \iiint_{V_{1}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z $$ 于是 $$ V=8 \iiint_{V_{1}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=8 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{a \sqrt{\cos 2 \varphi}} r^{2} \sin \varphi \mathrm{~d} r=\frac{4 a^{3} \pi}{3} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\cos 2 \varphi)^{\frac{3}{2}} \sin \varphi \mathrm{~d} \varphi $$ $$ \begin{aligned} & =\frac{4 a^{3} \pi}{3} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(1-2 \cos ^{2} \varphi\right)^{\frac{3}{2}} \sin \varphi \mathrm{~d} \varphi \stackrel{t=\cos \varphi}{4} \frac{4 a^{3} \pi}{3} \int_{0}^{\frac{1}{\sqrt{2}}}\left(1-2 t^{2}\right)^{\frac{3}{2}} \mathrm{~d} t \\ & \stackrel{t=\frac{1}{\sqrt{2}} \sin x}{4} \frac{4 a^{3} \pi}{3} \cdot \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \cos ^{4} x \mathrm{~d} x=\frac{4 a^{3} \pi}{3 \sqrt{2}} \cdot \frac{1}{2} B\left(\frac{1}{2}, \frac{5}{2}\right)=\frac{\pi^{2} a^{3}}{4 \sqrt{2}} . \end{aligned} $$