作闭曲线L2=L+L1,其中L1为y轴上从点(0,-a)到点(0,a)的一段直线段,于是有
I1=∮L2{(e^y)siny-y^3}dx+{(e^x)cosy+x^3}dy
=∫∫D(cosy(e^x)+3x^2-(e^y)siny-(e^y)cosy+3y^2)dxdy
=6a^3-sina(e^a+e^(-a))
I2=∫L1{(e^y)siny-y^3}dx+{(e^x)cosy+x^3}dy
=sina-sin(-a)=2sina
所以,I=∫L{(e^y)siny-y^3}dx+{(e^x)cosy+x^3}dy=I1-I2=6a^3-sina(e^a+e^(-a)+2)
没有答案,不知对不对,你自己在看看吧!