求积分数学题

显示全部楼层 · 2011-9-21 20:08:40

[0，R] ∫y√[y(2R-y)]dy=[0，R] {(-1/2)∫√[(2Ry-y2)]d(2Ry-y2)+[0，R]R∫√(2Ry-y2)dy} =(-1/2)[(2/3)(2Ry-y2)^(3/2)]︱[0，R]+[0，R]R∫√[R2-(R-y)2]dy=-(1/3)R3+[0，R]R2∫√{1- [1-(y/R)]2}d[1-(y/R)]
[将[1-(y/R)]看作下面公式中的u]=-(1/3)R3+R2{[1-(y/R)]/2}√{1- [1-(y/R)]2}︱[0，R]+(1/2)R2arc...

千问 · 2011-9-21 20:08:40

此类积分数学题，关键之处在于要懂得二重积分跟二次积分的转化……一般用图解法，也就是微量元素法，看把哪个量作为积分量最好，也就是看其是x型还是y型！...