D:x²+y²≤2x,y≥0
=> x²-2x+1+y²≤1,y≥0
=> (x-1)²+y²≤1,y≥0
即以(1,0)為圓心,半徑為1的x軸上方的半圓
以(0,0)為極點(diǎn),x軸正方向?yàn)闃O軸建立極坐標(biāo)系,則
x=rcosθ
y=rsinθ
0≤r≤2cosθ,0≤θ≤π/2
∴∫∫ (D) √(4-x²-y²) dxdy
=∫∫ (D) √(4-r²) rdrdθ
=∫(0,π/2)dθ∫(0,2cosθ)√(4-r²)rdr
=∫(0,π/2) (-1/3)[4-(2cosθ)²]^(3/2) dθ
=(-8/3) ∫(0,π/2) sin³θ dθ
=(8/3) ∫(0,π/2) (1-cos²θ)d(cosθ)
=(8/3)(cosθ-cos³θ/3)|(0,π/2)
=-16/9