计算曲面积分 ∫∫(x^2+y^2）ds，其中 ∑是上半球面z=根号(4-x^2-y^2)

Answer 1

dz/dx=-x/√(4-x²-y²)，dz/dy=-y/√(4-x²-y²)
dS=√[1+(dz/dx)²+(dz/dy)²] dxdy=2/√(4-x²-y²) dxdy
∫∫ (x²+y²) dS
=2∫∫ (x²+y²)/√(4-x²-y²) dxdy
极坐标
=2∫∫ r²/√(4-r²) *rdrdθ
=2∫[0→2π]dθ∫[0→2] r³/√(4-r²) dr
=4π∫[0→2] [r²/√(4-r²)] *rdr
换元，令√(4-r²)=u，则r²=4-u²，两边微分，rdr=-udu，u：2→0
=-4π∫[2→0] [(4-u²)/u] *udu
=4π∫[0→2] (4-u²)du
=4π(4u-(1/3)u³) |[0→2]
=32π-32π/3
=64π/3