对坐标的曲线积分问题计算∫(L) (x+y)dy+(x-y)dx / x^2+y^2-2x+2y ,其中L为圆周(x-1)^2 + (y+1)^2 =4正向
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 02:23:05
![对坐标的曲线积分问题计算∫(L) (x+y)dy+(x-y)dx / x^2+y^2-2x+2y ,其中L为圆周(x-1)^2 + (y+1)^2 =4正向](/uploads/image/z/14068437-69-7.jpg?t=%E5%AF%B9%E5%9D%90%E6%A0%87%E7%9A%84%E6%9B%B2%E7%BA%BF%E7%A7%AF%E5%88%86%E9%97%AE%E9%A2%98%E8%AE%A1%E7%AE%97%E2%88%AB%28L%29+%28x%2By%29dy%2B%28x-y%29dx+%2F+x%5E2%2By%5E2-2x%2B2y+%2C%E5%85%B6%E4%B8%ADL%E4%B8%BA%E5%9C%86%E5%91%A8%EF%BC%88x-1%29%5E2+%2B+%28y%2B1%29%5E2+%3D4%E6%AD%A3%E5%90%91)
对坐标的曲线积分问题计算∫(L) (x+y)dy+(x-y)dx / x^2+y^2-2x+2y ,其中L为圆周(x-1)^2 + (y+1)^2 =4正向
对坐标的曲线积分问题
计算∫(L) (x+y)dy+(x-y)dx / x^2+y^2-2x+2y ,其中L为圆周(x-1)^2 + (y+1)^2 =4正向
对坐标的曲线积分问题计算∫(L) (x+y)dy+(x-y)dx / x^2+y^2-2x+2y ,其中L为圆周(x-1)^2 + (y+1)^2 =4正向
1.使用参数法.
令(x-1)/2=cost,(y+1)/2=sint,得:
x=1+2cost,y=-1+2sint,dx=-2sintdt,dy=2costdt,代入积分式得:
∫(L) (x+y)dy+(x-y)dx/(x^2+y^2-2x+2y)
=∫(L) (x+y)dy+(x-y)dx/[(x-1)²+(y+1)²-2]
=(下限0,上限2π)∫[4(cost+sint)cost-4(1+cost-sint)sint]dt/(4-2)
=(下限0,上限2π)∫2(1-sint)dt=4π
2.使用格林理论.
∫(L) (x+y)dy+(x-y)dx/(x^2+y^2-2x+2y)
=∫(L) (x+y)dy+(x-y)dx/[(x-1)²+(y+1)²-2] ...由于圆周是(x-1)²+(y+1)²=4.在圆的周边线上积分时,上面分母中的(x-1)²+(y+1)²=4.所以:
∫(L) (x+y)dy+(x-y)dx/[(x-1)²+(y+1)²-2]
=∫(L) (x+y)dy+(x-y)dx/(4-2)
=(1/2)∫(L) (x+y)dy+(x-y)dx
使用格林理论将上面的线积分转化为面积分:
=(1/2)∫∫(S)[∂(x+y)/∂x-∂(x-y)/∂y]dxdy
=(1/2)∫∫(S)(1+1)dxdy=(1/2)∫∫(S)(2)dxdy
=∫∫(S)dxdy
上面的面积分积分就是这个圆的面积.由于这个圆的半径是2,所以,其面积为πr²=π2²=4π.