Question

有关第一类曲线积分 的一道题 见图

Answer 1

以y为自变量，x为函数
y^2=2x，则 x=y^2/2， dx/dy=y，所以
∫L yds = ∫L y sqrt(1+(x')^2) dy = ∫_(0<=y<=-√2) y sqrt(1+y^2) dy
令 y=tan a，y=0，对应 a=0，y= -√2，对应 a= - arctan√2，dy=(sec a)^2 da，则
∫L yds = ∫_(0<=y<=-√2) y sqrt(1+y^2) dy
= ∫_(0<=a<= -arctan√2) (tan a) (sec a) (sec a)^2 da
= ∫_(0<=a<= -arctan√2) (sec a)^2 d(sec a)
= (sec a)^2/2 |_(0<=a<= -arctan√2)
= [1+(tan a)^2] / 2 |_(0<=a<= -arctan√2)
= [1+(-√2)^2] / 2 - [1+0^2] / 2
= 1