求 \bexI=∭V|x+y+2z|⋅|4x+4y−z|\rdx\rdy\rdz,\eex
其中 V 是区域 \dpsx2+y2+z24≤1.
解答: 作变换 \bexx=u,y=v,z2=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˜u=u+v+4w3√2,˜v=2u+2v−w3,˜w=−u+v√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