搜狗做题网高等数学第十一章 计算题 1 2
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高等数学第十一章 计算题 1 2
/>补充直径OA,其中 A(1.0), 则
I =∫<L+AO>(x^2-y)dx+[-x-(cosy)^2]dy -∫<AO>(x^2-y)dx-[x+(cosy)^2]dy
= -∫∫<D>{(∂/∂x)[-x-(cosy)^2]-(∂/∂y)(x^2-y)}dxdy +∫<OA>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
∫<(0,0),(2,3)>(2xcosy-y^2sinx)dx+ (2ycosx-x^2siny)dy
=∫<0,2> 2xdx +∫<0,3> (2ycos2-4siny)dy
= 4 +9cos2+4cos3-4 = 9cos2+4cos3.
I =∫<L+AO>(x^2-y)dx+[-x-(cosy)^2]dy -∫<AO>(x^2-y)dx-[x+(cosy)^2]dy
= -∫∫<D>{(∂/∂x)[-x-(cosy)^2]-(∂/∂y)(x^2-y)}dxdy +∫<OA>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
∫<(0,0),(2,3)>(2xcosy-y^2sinx)dx+ (2ycosx-x^2siny)dy
=∫<0,2> 2xdx +∫<0,3> (2ycos2-4siny)dy
= 4 +9cos2+4cos3-4 = 9cos2+4cos3.