[家里蹲大学数学杂志]第241期利用正交变换和对称性求解三重积分

求 $$\bex I=\iiint_V|x+y+2z|\cdot |4x+4y-z|\rd x\rd y\rd z, \eex$$ 其中 $V$ 是区域 $\dps{x^2+y^2+\frac{z^2}{4}\leq 1}$. 

解答: 作变换 $$\bex x=u,\quad y=v,\quad \frac{z}{2}=w, \eex$$ 则 $$\beex \bea I&=\iiint_{u^2+v^2+w^2\leq 1} |u+v+4w|\cdot |4u+4v-2w|\cdot 2\rd u\rd v\rd w\\ &=4\iiint_{u^2+v^2+w^2\leq 1}|u+v+4w|\cdot |2u+2v-w|\rd u\rd v\rd w. \eea \eeex$$ 再作变换 $$\bex \tilde u=\frac{u+v+4w}{3\sqrt{2}},\quad \tilde v=\frac{2u+2v-w}{3},\quad \tilde w=\frac{-u+v}{\sqrt{2}}, \eex$$ 则 $$\beex \bea I&=4\iiint_{\tilde u^2+\tilde v^2+\tilde w^2\leq 1} |3\sqrt{2}\tilde u|\cdot |3\tilde v|\rd \tilde u\rd \tilde v\rd \tilde w\\ &=36\sqrt{2}\iiint_{x^2+y^2+z^2\leq 1} |xy|\rd x\rd y\rd z\\ &=144\sqrt{2}\iiint_{x^2+y^2+z^2\leq 1\atop x\geq 0,y\geq 0} xy\rd x\rd y\rd z\\ &=144\sqrt{2}\int_{-1}^1 \rd z \iint_{x^2+y^2\leq 1-z^2\atop x\geq0,y\geq 0}xy\rd x\rd y\\ &=144\sqrt{2}\int_{-1}^1 \rd z \int_0^{\sqrt{1-z^2}}\rd r \int_0^\frac{\pi}{2} r\cos \theta\cdot r\sin\theta\cdot r\rd \theta\\ &=\frac{96\sqrt{2}}{5}. \eea \eeex$$ 

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