Question

计算三重积分∫∫∫zdxdydz,其中V由x2+y2≤z≤1+ √(1-x2-y2)所确定

Answer 1

原式=∫dθ∫rdr∫zdz (作柱面坐标变换)

=2π∫r[(1+√(1-r²))²-r^4]/2dr

=(π/2)∫[2+2√(1-r²)-r²-r^4]d(r²)

=(π/2)[2r²-(4/3)(1-r²)^(3/2)-r^4/2-r^6/3]│

=(π/2)(2+4/3-1/2-1/3)

=5π/4.