Question

求I=∮Γ(x^2+y^2+2x-z-4)ds,其中,Γ:z=√x^2+5y^2,z=1+2y

Answer 1

Γ:z=1+2y,x^2+(y-2)^2=5.
设x=√5cosu,y=2+√5sinu,则z=5+2√5sinu,
dx=-√5sinudu,dy=√5cosudu,dz=2√5cosudu,
ds=√[(dx)^2+(dy)^2+(dz)^2]
=√[5+20(cosu)^2]du
=√(15+10cos2u)du
所以I=∫<0,2π>[5(cosu)^2+4+4√5sinu+5(sinu)^2+2√5cosu-5-2√5sinu-4)√(15+10cos2u)du
=∫<0,2π>(2√5sinu+2√5cosu)√(15+10cos2u)du
=∫<0,2π>2√10sin(u+π/4)√(15+10cos2u)du,
设v=cos(u+π/4),则cos2u=sin(2u+π/2)=2v√(1-v^2),dv=-sin(u+π/4)du,
此时积分的上下限都是1/√2，故I=0.