If ∑ is the outer side of the sphere x ^ 2 + y ^ 2 + Z ^ 2 = 4, then for the surface integral of coordinates ∫ ∫ x ^ 2dxdy, the answer to this problem is 4 π,

If ∑ is the outer side of the sphere x ^ 2 + y ^ 2 + Z ^ 2 = 4, then for the surface integral of coordinates ∫ ∫ x ^ 2dxdy, the answer to this problem is 4 π,

In Gauss formula, ∫∫∫∫ 2xdxdydz is obtained, then ∫ DX ∫∫ 2xdydz, 2x is advanced, and the cross section area is (16-x ^ 2) π, that is, ∫ 2x (16-x ^ 2) π DX is obtained, and the odd function result is zero