Question

高等数学怎么做 第六题

Answer 1

六。 0 ≤ t ≤ 2π, 则
ds =√[(x')^2+(y')^2]dt
= √[(-sint)^2+(-2sin2t)^2]dt
= sint√[1+16(cost)^2]
I = ∫<0, π/2> costsint√[1+16(cost)^2]dt
- ∫<π/2，π> costsint√[1+16(cost)^2]dt
= - ∫<0, π/2> cost√[1+16(cost)^2]dcost
+ ∫<π/2，π> cost√[1+16(cost)^2]dcost
= -(1/32)∫<0, π/2>√[1+16(cost)^2]d[1+16(cost)^2]
+ (1/32)∫<π/2，π>√[1+16(cost)^2]d[1+16(cost)^2]
= (-1/48)[1+16(cost)^2]^(3/2)<0, π/2>
+ (1/48)[1+16(cost)^2]^(3/2)<π/2, π>
= (1/24)(17√17-1)