设C是两球面x2+y2+z2=1与x2+y2+z2=2z的交线,方
向与x轴正向符合右手法则,求关于坐标的曲线积∮C∣x-y∣dy+zdz.
由方程组{x2+y2+z2=1, x2+y2+z2=2z, 得 z=1/2,x2+y2=3/4.所以C的参数方程为 {x=(√3/2)cost, y=(√3/2)sint, z=1/2 (0≤t≤2π),取其起点参数为t=0,终点参数为t=2π.于是, ∮C∣x-y∣dx+zdz=∫02π√3/2cost-sint∣[(-√3/2)sint]dt =-3/4[∫0π/4(cost-sint)sintdt-∫π/45π/4(cost-sint)sintdt +∫2π5π/4(cost-sint)sintdt]. 由于∫(cost-sint)sitdt=∫[costsint-1/2(1-cos2t)]dt =(1/2)sin2t-(1/2)2+(1/4)sin2t+C 所以∮C∣x-y∣dx+zdz=-3/4[(1/2)sin2t-(1/2)t+(1/4)sin2t]∣0π/4 -[(1/2)sin2t-(1/2)t+(1/4)sin2t]∣π/45π/4+[(1/2)sin2t-(1/2)t+(1/4)sin2t] ∣5π/42π=-3/4[(1/2-π/8)-(1/2-5π/8)+(1/2-π/8)]+(-π)-(1/2-5π/8)]=0