设Q(x,y)在xoy平面上有一阶连续偏导数,曲线积分∫L 2xydx+Q(x,y)与路径无关,对任意t恒有∫L 2xydx+Q(x,y)dy从点(0,0)到(t,1)的积分等于从点(0,0)到(1,t)的积分,求Q(x,y)令P(X,Y)=2XY积分
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![设Q(x,y)在xoy平面上有一阶连续偏导数,曲线积分∫L 2xydx+Q(x,y)与路径无关,对任意t恒有∫L 2xydx+Q(x,y)dy从点(0,0)到(t,1)的积分等于从点(0,0)到(1,t)的积分,求Q(x,y)令P(X,Y)=2XY积分](/uploads/image/z/10362286-46-6.jpg?t=%E8%AE%BEQ%EF%BC%88x%2Cy%EF%BC%89%E5%9C%A8xoy%E5%B9%B3%E9%9D%A2%E4%B8%8A%E6%9C%89%E4%B8%80%E9%98%B6%E8%BF%9E%E7%BB%AD%E5%81%8F%E5%AF%BC%E6%95%B0%2C%E6%9B%B2%E7%BA%BF%E7%A7%AF%E5%88%86%E2%88%ABL+2xydx%2BQ%EF%BC%88x%2Cy%EF%BC%89%E4%B8%8E%E8%B7%AF%E5%BE%84%E6%97%A0%E5%85%B3%2C%E5%AF%B9%E4%BB%BB%E6%84%8Ft%E6%81%92%E6%9C%89%E2%88%ABL+2xydx%2BQ%28x%2Cy%29dy%E4%BB%8E%E7%82%B9%EF%BC%880%2C0%EF%BC%89%E5%88%B0%EF%BC%88t%2C1%EF%BC%89%E7%9A%84%E7%A7%AF%E5%88%86%E7%AD%89%E4%BA%8E%E4%BB%8E%E7%82%B9%EF%BC%880%2C0%EF%BC%89%E5%88%B0%EF%BC%881%2Ct%EF%BC%89%E7%9A%84%E7%A7%AF%E5%88%86%2C%E6%B1%82Q%EF%BC%88x%2Cy%EF%BC%89%E4%BB%A4P%28X%2CY%29%3D2XY%E7%A7%AF%E5%88%86)
设Q(x,y)在xoy平面上有一阶连续偏导数,曲线积分∫L 2xydx+Q(x,y)与路径无关,对任意t恒有∫L 2xydx+Q(x,y)dy从点(0,0)到(t,1)的积分等于从点(0,0)到(1,t)的积分,求Q(x,y)令P(X,Y)=2XY积分
设Q(x,y)在xoy平面上有一阶连续偏导数,曲线积分∫L 2xydx+Q(x,y)与路径无关,对任意t恒有
∫L 2xydx+Q(x,y)dy从点(0,0)到(t,1)的积分等于从点(0,0)到(1,t)的积分,求Q(x,y)
令P(X,Y)=2XY
积分与路径无关,所以偏P偏y=偏Q偏x,即2x=偏Q偏x,所以Q=x^2+f(y)
现在要想方法解出f(y)
可用有向直线段依次连接(0,0),(1,t),(t,1).因积分与路径无关,所以在这样的封闭的三角形边上的积分∫L 2xydx+Q(x,y)dy=0,又“∫L 2xydx+Q(x,y)dy从点(0,0)到(t,1)的积分等于从点(0,0)到(1,t)的积分”,所以从(1,t)到(1,t)的直线上的积分∫ 2xydx+Q(x,y)dy=0,
可将x=-y+t+1带入到∫ 2xydx+Q(x,y)dy=0中,
得到∫ [2(-y+t+1)y]d(-y+t+1)+[(-y+t+1)^2+f(y)]dy=0(y从t到1)
怎么求f(y)?
设Q(x,y)在xoy平面上有一阶连续偏导数,曲线积分∫L 2xydx+Q(x,y)与路径无关,对任意t恒有∫L 2xydx+Q(x,y)dy从点(0,0)到(t,1)的积分等于从点(0,0)到(1,t)的积分,求Q(x,y)令P(X,Y)=2XY积分
本解答从这一步出发:
得到∫ [2(-y+t+1)y]d(-y+t+1)+[(-y+t+1)^2+f(y)]dy=0(y从t到1)
也即∫ [-2(-y+t+1)y]dy+[(-y+t+1)^2+f(y)]dy=0(y从t到1)
由此求f(y)的方法是,等式两边对t求导【在此只需将被积函数中的y换成t,取负号】,得到
2t-1-f(t)=0.