下册 8.2 三重积分 第11题

\section*{解题过程：} （1）令 $\displaystyle u=\frac{z}{y^{2}}, v=\frac{z}{x}, w=z$ ，即 $\displaystyle x=v^{-1} w, y=\sqrt{\frac{w}{u}}, z=w$ ，则 $V$ 变成 $\displaystyle \Delta=\{(u, v, w) \mid a \leqslant u \leqslant b, \alpha \leqslant v \leqslant \beta, 0 \leqslant w \leqslant h\}, J=\frac{\partial(x, y, z)}{\partial(u, v, w)}=-\frac{1}{2} u^{-\frac{3}{2}} v^{-2} w^{\frac{3}{2}}$ ． 于是 $$ \begin{aligned} \iiint_{V} x^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z & =\iiint_{\Delta} v^{-2} w^{2} \cdot \frac{1}{2} u^{-\frac{3}{2}} v^{-2} w^{\frac{3}{2}} \mathrm{~d} u \mathrm{~d} v \mathrm{~d} w=\frac{1}{2} \int_{a}^{b} u^{-\frac{3}{2}} \mathrm{~d} u \int_{a}^{\beta} v^{-4} \mathrm{~d} v \int_{0}^{h} w^{\frac{7}{2}} \mathrm{~d} w \\ & =\frac{2}{3}\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{a}}\right)\left(\frac{1}{\beta^{3}}-\frac{1}{\alpha^{3}}\right) h^{\frac{9}{2}} \end{aligned} $$ （2）作变换 $\displaystyle u=\frac{z}{x^{2}+y^{2}}, v=x y, w=\frac{y}{x}$ ，即 $\displaystyle x=\sqrt{\frac{v}{w}}, y=\sqrt{v w}, z=u\left(\frac{v}{w}+v w\right)$ ，则 $\Omega$ 变成 $\displaystyle V:\{(u, v, w) \mid p \leqslant u \leqslant q, a \leqslant v \leqslant b, \alpha \leqslant w \leqslant \beta\}, J=\frac{\partial(x, y, z)}{\partial(u, v, w)}=\frac{v}{2 w}\left(w^{-1}+w\right)$ ． 于是 $$ \begin{aligned} \iiint_{\Omega} x y z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z & =\iiint_{V} u v^{2}\left(w+w^{-1}\right) \cdot \frac{v}{2 w}\left(w+w^{-1}\right) \mathrm{d} u \mathrm{~d} v \mathrm{~d} w=\frac{1}{2} \int_{p}^{q} u \mathrm{~d} u \int_{a}^{b} v^{3} \mathrm{~d} v \int_{\alpha}^{\beta} \frac{\left(w+w^{-1}\right)^{2}}{w} \mathrm{~d} w \\ & =\frac{1}{16}\left(q^{2}-p^{2}\right)\left(b^{4}-a^{4}\right)\left(\frac{\beta^{2}-\alpha^{2}}{2}+2 \ln \frac{\beta}{\alpha}-\frac{1}{2}\left(\frac{1}{\beta^{2}}-\frac{1}{\alpha^{2}}\right)\right) \end{aligned} $$ （3）令 $u=x+y-z, v=x-y+z, w=y+z-x$ ，则 $\displaystyle \frac{\partial(u, v, w)}{\partial(x, y, z)}=-4, J=-\frac{1}{4}$ ． $$ \Omega \text { 变成 } \Delta=\{(u, v, w) \mid 0 \leqslant u \leqslant 10,0 \leqslant v \leqslant 10,0 \leqslant w \leqslant 10\} \text {. } $$ 于是 $$ \iiint_{\Omega}(x+y-z)(x-y+z)(y+z-x) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\frac{1}{4} \iiint_{\Delta} u v w \mathrm{~d} u \mathrm{~d} v \mathrm{~d} w=\frac{1}{4} \int_{0}^{10} u \mathrm{~d} u \int_{0}^{10} v \mathrm{~d} v \int_{0}^{10} w \mathrm{~d} w=31250 $$