求平面x=0，y=0，x+y=1所围成的柱体，被平面z=0及平面x²+y²=6-z截得的立体的体积 求由平面x=0,y=0,x+y=1所围成柱体
\u8ba1\u7b97\u7531\u56db\u4e2a\u5e73\u9762x=0,y=0,x=1,y=1\u6240\u56f4\u6210\u7684\u67f1\u9762\u88ab\u5e73\u9762z=0\u53ca2x+3y+z=6\u622a\u5f97\u7684\u7acb\u4f53
\u6c42\u56db\u4e2a\u5e73\u9762x=0 y=0 x=1 y=1
\u5148\u5206\u6790\u5404\u56fe\u5f62\u7684\u5173\u7cfb\u53ca\u5176\u5728xoy\u5e73\u9762\u4e0a\u7684\u6295\u5f71\u3002(1)\u629b\u7269\u9762\u662f\u5f00\u53e3\u671d\u4e0b\u7684\uff0c\u5176\u5728xoy\u7684\u622a\u9762\u662f\u534a\u5f84\u4e3a2\u7684\u5706\uff0c\u5706\u5fc3\u4e3aO\u3002(2)x\uff1d0~1,y\uff1d0~1\u662f\u4e2a\u6b63\u65b9\u5f62\u7684\u67f1\u9762(3)y\uff1d3z\uff0c\u662f\u4e2a\u8fc7x\u8f74\u7684\u659c\u9762\uff0c\u5176\u548c\u629b\u7269\u9762\u7684\u4ea4\u7ebf\u5728xoy\u4e0a\u7684\u6295\u5f71\u662f\uff1ax²\uff0by²\uff1d4\uff0dy/3\u8be5\u6295\u5f71\u662f\u4e2a\u5706\uff0c\u5e76\u4e14\u5305\u542b\u4e86\u524d\u9762\u7b2c(2)\u6761\u6240\u8bf4\u7684\u6b63\u65b9\u5f62\u3002(4)\u6839\u636e\u4e0a\u8ff0\u5206\u6790\uff0c\u6240\u6c42\u51e0\u4f55\u4f53\u5b9e\u9645\u5c31\u662f\u629b\u7269\u9762\u4e4b\u4e0b\uff0c\u659c\u9762\u4e4b\u4e0a\u7684\u7a7a\u95f4\u88ab\u6b63\u65b9\u5f62\u67f1\u9762\u6240\u622a\u7684\u4f53\u79ef\u3002V\uff1d\u222b[0,1]dx\u222b[0,1](4-x²-y²-y/3)dy\uff1d\u222b[0,1]dx(4y-x²y-y³/3-y²/6)|[0,1]\uff1d\u222b[0,1](4-x²-1/2)dx\uff1d4-1/3-1/2\uff1d19/6
xy平面内的直线：x=0，y=0，y=1- x 所围成一个三角形区域；曲顶柱体下底面：xy平面； 上底面是：z=6- x² - y²
所以，体积：
V=∫∫D[6- x² - y²]dxdy=∫<0,1>dx∫<0,1-x>[6- x² - y²]dy
=∫<0,1>dx[6y- x²y - 1/3y³]|<0,1-x>
=∫<0,1>[6(1-x) - x²(1-x) - 1/3(1-x)³]dx
=∫<0,1>[17/3 - 5x - 2x²+ 4/3 x³ ]dx
=[17/3x - 5/2 x² - 2/3 x³ + 1/3 x^4]|<0,1>
=17/3 - 5/2 - 2/3 + 1/3=16/3 - 5/2=17/6
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绛旓細濡傚浘鎵绀猴細
