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求由曲面z=x2+2y2及z=6-2x2-y2所围成的立体的体积
人气:460 ℃ 时间:2020-06-03 06:30:35
解答
D:x2+2y2=6-2x2-y2整理:x2+y2=<26-2x2-y2>x2+2y2在D上对6-2x2-y2-(x2+2y2)积分令x=rsinα,y=rcosαds=rdαdr[6-2x2-y2-(x2+2y2]rdαdr[6-3r2]rdαdrr2:0->2,α:0->2π体积为:2π;œ‹D:x2...
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