设∑是球面x2+y2+z2=R2的下半球面，则∫∫∑(x2+y2-z2+1)ds=____.
2R2π。解析：由于z=-√R2-x2-y2,∂z/∂x=x/√R2-x2-y2z/y=y/√R2-x2-y2因此ds=√[1+(∂z/∂x)2+(∂z/∂y)2]dσ=[R/√(R2-x2-y2)Dσ，又∑在Oxy平面上的投影区域为D={(x，y)∣x2+y2≤R2)={(r，θ)∣0≤θ≤2π，0≤r≤R}所以∫∫∑(x2+y2-z2+1)ds=∫∫D[R/√(R2-x2-y2)]dσ=∫02πdθ∫0R[R/(R2-r2)]rdr=2π[-(1/2)]∫0RR(R2-r2)-(1/2)d(R2-r2)=-2Rπ•√(R2-r2)∣0R2π
