求由圆锥面z=4-√(x눀+y눀)与旋转抛物面2z=x눀+y눀所围立体的体积

2025-04-06 11:35:57
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回答1:

z = √(4-x²-y²) 与 3z = x²+y² 消去 z, 得交线在 xOy 坐标平面的投影是
x²+y² = 3,
V = ∫<0, 2π>dt∫<0, √3>[√(4-r^2)-(1/3)r^2]rdr
= 2π∫<0, √3>[√(4-r^2)-(1/3)r^2]rdr
= -π∫<0, √3>√(4-r^2)d(4-r^2) - (2π/3)∫<0, √3>r^3dr
= - (2π/3)[(4-r^2)^(3/2)]<0, √3> - (π/6)[r^4]<0, √3>
= 14π/3 - 9π/6 = 19π/6