对坐标的曲线积分的计算方法
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 展开
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 展开
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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.
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.
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