设空间闭区域Ω由曲面z=√(1-x2-y2)及z=√x2+y2可围成,则三重积分∫∫∫Ω(x2+y2+z2)dxdydz=()
A、∫02πdθ∫0π/2sinφdφ∫01rdr
B、∫02πdθ∫0π/4sinφdφ∫01r4dr
C、∫02πdθ∫0π/2sinφdφ∫01r2dr
D、∫02πdθ∫0π/4sinφdφ∫01r3dr
【正确答案】:B
【题目解析】:采用球坐标系Ω:{0≤θ≤2π, 0≤φ≤π/4, 0≤r≤1故∫∫∫Ω(x2+y2+z2)dxdydz=∫02πdθ∫0π/4dφ∫01r2•r2sinφdr=∫02πdθ∫0π/4sinφdφ∫01r4dr