Question

计算曲线积分I=∫L ydx-xdy x^2+y^2,其中L：(x-1)^2+(y-1)^2=1(逆时针)

Answer 1

【答案】：用格林公式：奇点(0,0)不在积分域内.
I = ∮L (ydx - xdy)/(x^2 + y^2)
= ∫∫D [(x^2 - y^2)/(x^2 + y^2)^2 - (x^2 - y^2)/(x^2 + y^2)^2] dxdy
= 0
用参数方程.
{ x = 1 + cost、dx = - sint dt
{ y = 1 + sint、dy = cost dt
0 ≤ t ≤ 2π
∮L (ydx - xdy)/(x^2 + y^2)
= ∫(0→2π) [(1 + sint)(- sint) - (1 + cost)(cost)]/[(1 + cost)^2 + (1 + sint)^2] dt
= - ∫(0→2π) (sint + cost + 1)/(2sint + 2cost + 3) dt
令u = tan(t/2)、dt = 2/(1 + u^2) du,sint = 2u/(1 + u^2)、cost = (1 - u^2)/(1 + u^2)
∫ (sint + cost + 1)/(2sint + 2cost + 3) dt
= ∫ [2u/(1 + u^2) + (1 - u^2)/(1 + u^2) + 1]/[2 * 2u/(1 + u^2) + 2 * (1 - u^2)/(1 + u^2) + 3] * 2/(1 + u^2) du
= 4∫ (u + 1)/[(u^2 + 1)(u^2 + 4u + 5)] du
= ∫ du/(u^2 + 1) + ∫ du/(u^2 + 4u + 5)
= ∫ du/(u^2 + 1) + ∫ du/[(u + 2)^2 + 1]
= arctan(u) + arctan(u + 2) + C
= arctan[tan(t/2)] + arctan[2 + tan(t/2)] + C
于是I = - arctan[tan(t/2)] - arctan[2 + tan(t/2)]：(0→2π)
将区间分为：0→π⁻,π⁺→2π
I = (- π/2 - π/2) - (- π/2 - π/2)
= 0