Question

高等数学第十一章 计算题 1 2

Answer 1

补充直径OA,其中 A(1.0), 则

I =∫

(x^2-y)dx+[-x-(cosy)^2]dy -∫

(x^2-y)dx-[x+(cosy)^2]dy

= -∫∫

{(∂/∂x)[-x-(cosy)^2]-(∂/∂y)(x^2-y)}dxdy +∫

x^2dx

= 0 +∫<0,1>x^2dx = 1/3.

题目有误,应是（2ycosx-x^2siny)dy 吧.

(∂/∂x) (2ycosx-x^2siny)= -2ysinx-2xsiny,

(∂/∂y)(2xcosy-y^2sinx) = -2xsiny-2ysinx = (∂/∂x) (2ycosx-x^2siny),

则该曲线积分与路径无关.

取 A(2,0),B(2,3),取路径 OA+AB,则

∫(2xcosy-y^2sinx)dx+ (2ycosx-x^2siny)dy

=∫<0,2> 2xdx +∫<0,3> (2ycos2-4siny)dy

= 4 +9cos2+4cos3-4 = 9cos2+4cos3.