Question

求曲线积分∫(sinx^2+y)dx,其中L为由y^2=x,x=1所围城区域的边界

求曲线积分∫(sinx^2+y)dx,其中L为由y^2=x,x=1所围城区域的边界

Answer 1

P = sin²x + y、Q = 0
P'y = 1，Q'x = 0
∫(L) (sin²x + y) dx
= ∫∫(D) (0 - 1) dxdy
= - ∫(- 1→1) dy ∫(y²→1) dx
= - 2∫(0→1) (1 - y²) dy
= - 2[ y - (1/3)y³ ]：(0→1)
= - 2[ 1 - (1/3) ]
= - 4/3

L1：x = y²、dx = 2y dy
L2：x = 1，dx = 0
∫(L) (sin²x + y) dx
= ∫(L1) + ∫(L2)
= ∫(1→- 1) (sin²y² + y)(2y) dy + 0
= 2∫(1→- 1) (ysin²y² + y²) dy
= 0 - 4∫(0→1) y² dy
= - 4 • (1/3)[ y³ ]：(0→1)
= - 4 • (1/3)(1 - 0)
= - 4/3