∫∫∫(x+y+z^2)dxdydz=∫[0,2pi]dt∫[-1,1]dz∫[0,√(1+z^2)](rcost+rsint+z^2)rdr=∫[0,2pi]dt∫[-1,1] [r^3/3(cost+sint)+r^2/2*z^2)[0,√(1+z^2)]dz=∫[0,2pi]dt∫[-1,1] [(1+z^2)^(3/2)/3(cost+sint)+z^2(1+z^2...截面法，先考虑：由于对称性，∫∫∫(x+y)dxdydz=0，只用考虑被积函数z^2的三重积分就可以了。这样就可以对z截面了。∫∫∫(x+y+z^2)dxdydz=∫∫∫z^2dxdydz=2π∫[0,1]z^2(1+z^2)dz=2π(z^3/3+z^5/5)|[0,1]=16/15* π