Question

I=∫∫∫z√(x^2+y^2)dv,其中Ω由y=-√(2x-x^2)和平面z=0,z=1,y=0围成。

Answer 1

y=-√(2x-x²)两边平方后得：x²+y²-2x=0，在xoy面内是以(1,0)为圆心，1为半径的圆，且y<0，因此是下半圆
用柱坐标
∫∫∫z√(x^2+y^2)dv
=∫∫∫zr²dzdrdθ
=∫[-π/2--->0]dθ∫[0--->2cosθ] r²dr∫[0--->1] zdz
=(1/2)∫[-π/2--->0]dθ∫[0--->2cosθ] r²z² |[0--->1]dr
=(1/2)∫[-π/2--->0]dθ∫[0--->2cosθ] r²dr
=(1/6)∫[-π/2--->0] r³[0--->2cosθ] dθ
=(1/6)∫[-π/2--->0] 8cos³θ dθ
=(4/3)∫[-π/2--->0] cos²θ d(sinθ)
=(4/3)∫[-π/2--->0] (1-sin²θ) d(sinθ)
=(4/3)(sinθ-(1/3)sin³θ) [-π/2--->0]
=(4/3)(1-1/3)
=8/9

Answer 2

三重积分，适合柱坐标