Question

用三重积分计算z=√5-x²+y²及x²+y²=4z所围成的立体的体积，求详细过程

Answer 1

由z=√(4-x^2-y^2),z=x^2+y^2
得：z=(√17-1)/2,x^2+y^2=[(√17-1)/2]^2
则立体体积
V=∫∫[√(4-x^2-y^2)-(x^2+y^2)]dxdy
(积分区域D为x^2+y^2<=a^2，a=(√17-1)/2)
=∫∫[√(4-r^2)-r^2]rdrdθ
(设x=rcosθ,y=rsinθ，此时0<=θ<=2π，0<=r<=a)
=∫dθ∫[√(4-r^2)-r^2]rdr
=2π*[-1/3*(4-r^2)^(3/2)-1/4*r^4]|[0,a]
=2π*{[-1/3*(4-a^2)^(3/2)-1/4*a^4]-8/3}
再将a=(√17-1)/2代入即得

Answer 2

就是上面这样了