Question

计算曲面积分

∫∫（xy+yz+xz)dS，其中∑为圆锥面z=√x^2+y^2被曲面x^2+y^2=2ax所割下的部分

Answer 1

解：∵dS=√[1+(αz/αx)²+(αz/αy)²]dxdy
=√[1+(x/z)²+(y/z)²]dxdy
=√2dxdy
∴原式=∫<-π/2,π/2>dθ∫<0,2acosθ>(r²sinθcosθ+r²sinθ+r²cosθ)rdr (做极坐标变换)
=4a^4∫<-π/2,π/2>(sinθcosθ+sinθ+cosθ)(cosθ)^4dθ
=4a^4∫<-π/2,π/2>[sinθ(cosθ)^5+sinθ(cosθ)^4+cosθ(1-2sin²θ+(sinθ)^4)]dθ
=4a^4[(-1/6)(cosθ)^6+(-1/5)(cosθ)^5+sinθ-(2/3)sin³θ+(1/5)(sinθ)^5]│<-π/2,π/2>
=4a^4(1-2/3+1/5+1-2/3+1/5)
=(64/15)a^4。