曲线积分∫L(ye^xsiny-xe^xcosy)ds,其中L为x^2+y^2=1位于第一象限部分

Question

曲线积分∫L( ye^xsiny-xe^xcosy)ds,其中L为x^2+y^2=1位于第一象限部分

Answer 1

L：x² + y² = 1
用参数方程化简：
{ x = cost、dx = - sint dt
{ y = sint、dy = cost dt
0 ≤ t ≤ π/2
ds = √[(dx)² + (dy)²] = √[(- sint)² + (cost)²] dt = dt
∫L (ye^x * siny - xe^x * cosy) ds
= ∫(0,π/2) [sint * e^(cost) * sin(sint) - cost * e^(cost) * cos(sint)] dt
= ∫(0,π/2) e^(cost) * (- 1)[cost * cos(sint) - sint * sin(sint)] dt
= - ∫(0,π/2) e^(cost) * cos(t + sint) dt
= - sin(sint) * e^(cost)：(0,π/2)
= - sin(1)
——————————————————————————————————————————
其中∫ e^(cost) * cos(t + sint) dt = sin(sint) * e^(cost)
第一个式：
∫ e^(cost) * cost * cos(sint) dt
= ∫ e^(cost) * cos(sint) d(sint)
= ∫ e^(cost) d[sin(sint)]
= sin(sint) * e^(cost) - ∫ sin(sint) * e^(cost) * (- sint) dt
= sin(sint) * e^(cost) + ∫ e^(cost) * sint * sin(sint) dt
而另一边就是- ∫ e^(cost) * sint * sin(sint) dt,后面两个积分刚好抵消了.
剩下sin(sint) * e^(cost)