思考题解答: ()1=(x2-3y)dx+(2-x)dy D L+AB JAB B 2l dxdy+ C-元 D (2)2=「,(x2-y+y2)dx+(y2-x)d J (x2-y)dx+(2-xdy+yd L:x= a cos t,y= asin t,t:0→>丌 sin'tdt C 1=-2a 3 HIGH EDUCATION PRESS 机动目录上页下页返回结束思考题解答:  = − + − L I (x y)d x ( y x)d y 2 2 (1) 1 3   = − L+AB AB  = − D 2 d xd y ) 3 2 ( 2 = a a −  = − + − L I (x y )d x ( y x)d y 2 2 2 2 (2) + y  = − + − L (x y)d x (y x)dy 2 2  + L y dx 2 a sin t d t 3 0 3  −  L : x = acost, y = asint, 3 3 2 = − a 3 = −2a t : 0 → 3 3 2 + a = I C o y B A x L D 机动 目录 上页 下页 返回 结束