求曲线积分(x+2y+z^2)ds,曲线L为球面x^2+y^2+z^2=a^2与平面x+y+z=0的交线.

Question

求曲线积分(x+2y+z^2)ds,曲线L为球面x^2 + y^2 + z^2 = a^2与平面x + y + z = 0的交线.

Answer 1

用对称性
曲线L关于坐标轴和坐标面都对称,所以∫xds＝∫yds＝∫zds,∫x^2ds＝∫y^2ds＝∫z^2ds
又平面x+y+z＝0过球面x^2 + y^2 + z^2 = a^2的球心,所以曲线L是一个圆,半径为a
所以,∫(x+2y+z^2)ds＝∫(x+y+z)ds＋1/3×∫(x^2＋y^2＋z^2)ds＝∫ 0ds＋1/3×∫(a^2)ds＝1/3×a^2×2πa＝2πa^3/3