∫（x^2+y^2）ds,其中L为曲线x=a(cost+tsint),y=a(sint-tcost),(0

题目

∫（x^2+y^2）ds,其中L为曲线x=a(cost+tsint),y=a(sint-tcost),(0

答案

x = a(cost + tsint),y = a(sint - tcost)
dx/dt = a(- sint + sint + tcost) = atcost
dy/dt = a(cost - cost + tsint) = atsint
ds = √[(dx/dt)² + (dy/dt)²] dt = √[(atcost)² + (atsint)²] dt = √(a²t²cos²t + a²t²sin²t) dt = at dt
∫_L (x² + y²) ds
= ∫(0-->2π) [a²(cost + tsint)² + a²(sint - tcost)²] · at dt
= ∫(0-->2π) a³(t³ + t) dt
= a³ · (t⁴/4 + t²/2) |(0-->2π)
= 2a³π²(1 + 2π²)