求由曲面z=x2+2y2及z=6-2x2-y2所围成的立体的体积

2025-04-17 09:58:09
推荐回答(1个)
回答1:

D:x2+2y2=6-2x2-y2
整理:x2+y2=<2
6-2x2-y2>x2+2y2
在D上对6-2x2-y2-(x2+2y2)积分
令x=rsinα,y=rcosα
ds=rdαdr
[6-2x2-y2-(x2+2y2]rdαdr
[6-3r2]rdαdr
r2:0->2,α:0->2π
体积为:2π;œ‹D:x2+2y2=6-2x2-y2
整理:x2+y2=<2
6-2x2-y2>x2+2y2
在D上对6-2x2-y2-(x2+2y2)积分
令x=rsinα,y=rcosα
ds=rdαdr
[6-2x2-y2-(x2+2y2]rdαdr
[6-3r2]rdαdr
r2:0->2,α:0->2π
体积为:2π;œ‹啊这个啊呵呵不阵地