如题所述
第1个回答 2019-07-01
令x=rsinacosb,y=rsinasinb,z=rcosa
x^2+y^2+z^2<=z
r^2<=rcosa
r<=cosa
原式=∫(0,2π)db*∫(0,π/2)da*∫(0,cosa)r*r^2*sina*dr
=(π/2)*∫(0,π/2)sinada*r^4|(0,cosa)
=(-π/2)*∫(0,π/2)cos^4ad(cosa)
=(-π/10)*cos^5a|(0,π/2)
=(-π/10)*(0-1)
=π/10追问
x^2+y^2+z^2<=z
r^2<=rcosa
r<=cosa
原式=∫(0,2π)db*∫(0,π/2)da*∫(0,cosa)r*r^2*sina*dr
=(π/2)*∫(0,π/2)sinada*r^4|(0,cosa)
=(-π/2)*∫(0,π/2)cos^4ad(cosa)
=(-π/10)*cos^5a|(0,π/2)
=(-π/10)*(0-1)
=π/10追问
怎么知道积分区域是多少
追答看图像
因为球心过z轴,所以0<=b<2π
因为球整体在xoy平面上方,则0<=a<=π/2