∫L(x^2+2xy)dx+(x^2+y^4)dy,L是y=sin(π/2)从(0,0)到(1,1)
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/06/16 13:42:15
∫L(x^2+2xy)dx+(x^2+y^4)dy,L是y=sin(π/2)从(0,0)到(1,1)
求积。
求积。
![∫L(x^2+2xy)dx+(x^2+y^4)dy,L是y=sin(π/2)从(0,0)到(1,1)](/uploads/image/z/4859924-68-4.jpg?t=%E2%88%ABL%28x%5E2%2B2xy%29dx%2B%28x%5E2%2By%5E4%29dy%2CL%E6%98%AFy%3Dsin%28%CF%80%2F2%29%E4%BB%8E%280%2C0%29%E5%88%B0%281%2C1%29)
是求曲线积分吗?
取O(0,0),B(1,0),A(1,1)三点,
连结BA,
设P=x^2+2xy,Q=x^2+y^4,
∂P/∂y=2x,
∂Q/∂x=2x,
∴∂P/∂y=∂Q/∂x,
∴曲线积分和路径无关,只与其起讫点有关,
可取OB,y=0,0≤x≤1,
BA,x=1,0≤y≤1,
原式=∫[(0,0),(1,1)](x^2+2xy)dx+(x^2+y^4)dy
=∫[0,1](x^2+2x*0)dx+∫[1,2](1^2+y^4)dy
=x^3/3[0,1]+(y+y^5/5)[1,2]
=1/3+(2+32/5-1-1/5)
=113/15.
取O(0,0),B(1,0),A(1,1)三点,
连结BA,
设P=x^2+2xy,Q=x^2+y^4,
∂P/∂y=2x,
∂Q/∂x=2x,
∴∂P/∂y=∂Q/∂x,
∴曲线积分和路径无关,只与其起讫点有关,
可取OB,y=0,0≤x≤1,
BA,x=1,0≤y≤1,
原式=∫[(0,0),(1,1)](x^2+2xy)dx+(x^2+y^4)dy
=∫[0,1](x^2+2x*0)dx+∫[1,2](1^2+y^4)dy
=x^3/3[0,1]+(y+y^5/5)[1,2]
=1/3+(2+32/5-1-1/5)
=113/15.
∫L(x^2+2xy)dx+(x^2+y^4)dy,L是y=sin(π/2)从(0,0)到(1,1)
∫L[y^2+sin^2(x+y)]dx-[x^2+cos^2(x+y)]dy,其中L是从点(1,0)沿y=根号下(1-
计算∫L(x+y)dx+(y-x)dy,其中L是y=x^2上从点(0,0)到点(1,1)的一段弧
求∫L(x^2-y)dx-(x+sin^2y)dy,L是y=根号下1-x^2以A(-1,0)到B(1,0)
∫L(x+y)dx+(x-y)dy,L为从(1,1)到(2,3)的直线.
计算曲线积分∫L (x^2+2xy)dx+(x^2+y^4)dy,其中L为点(0,0)到点(1,1)的曲线弧y=sin(
计算曲线积分∫L(sin2x+xy)dx+2(x^2-y^2)dy,其中L是曲线y=sinx上从(π,0)到(2π,0)
计算∫L((x+y)dx+(x-y)dy),其中L是抛物线y=x^2从点(0,0)到(1,1)的一段弧.
计算曲线积分I=∫(e^y+x)dx+(xe^y-2y)dy,L为从(0,0)到(1,2)的圆弧
计算曲面积分∫(y+2xy)dx+(x^2+2x+y^2)dy,其中L是由A(4,0)沿上半圆周y=√(4x-x^2)到
求曲线积分∫L(x^2+2xy-y^2)dx+(x^2-2xy-y^2)dy,其中L是沿着椭圆x^2/4+y^2/4=1
计算∫L(3xy+sinx)dx+(x^2-ye^y)dy,其中L是从点(0,0)到点(4,8)的抛物线段y=x^2-2