Question

∫（x^2+y^2）ds,其中L为曲线x=a(cost+tsint),y=a(sint-tcost),(0<=t<=2kai) 求对弧长的积分曲线

Answer 1

∫（x^2+y^2）ds=∫((a(cost+tsint)）^2+（a(sint-tcost)）^2）sqrt((dx/dt)^2+(dy/dt)^2)dt

下面是matlab代码实现；0<=t<=2kai， 在这里由于没有看明白t的区间范围，所以就默认把t区间范围改成0<=t<=2*pi计算的结果
syms x y t a;
x=a*(cos(t)+t*sin(t));
y=a*(sin(t)-t*cos(t));
sqrt(diff(x,t)^2+diff(y,t)^2)
fun=simple((a^2*(cos(t)+t*sin(t))^2+a^2*(sin(t)-t*cos(t))^2)*sqrt(diff(x,t)^2+diff(y,t)^2))
% simple 在这里起到化简到最简形式的作用
I=simple(int(fun, 0,2*pi))
% I = 2*pi^2*(2*pi^2 + 1)*(a^2)^(3/2)即为输出结果

Answer 2

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)