x=y
x^2+y^2+4z^2=1
--->2x^2+4z^2=1

设x=y=(根号{2}/2)cost, z=(1/2)sint
0<=t<=2*pi
ds=根号{(1/2)cos^2t+(1/2)cos^2t+(1/4)sin^2t}dt
=根号{1-(3/4)sin^2t}dt

原线积分=积分(t=0-->2*pi) 根号{2}/2|cost| 根号{1-(3/4)sin^2t}dt
=4*积分(t=0-->pi/2) 根号{2}/2|cost| 根号{1-(3/4)sin^2t}dt
=2根号{2} 积分(t=0-->pi/2) cost 根号{1-(3/4)sin^2t}dt
设u=sint, du=costdt
则上面的积分=2根号{2} 积分(u=0-->1) 根号{1-(3/4)u^2}du
=2根号{2}*(根号{3}/2) 积分(u=0-->1) 根号{4/3-u^2}du
=根号{6}*(1/2)(4/3arcsin(根号{3}/2)+根号{4/3-1})
=2根号{6}*pi/9+根号{2}/2