计算曲线积分l(x^2+y^2)ds,其中L是圆心在（2,0），半径为2的上半圆周。

Question

如题所述

Answer 1

半圆周参数方程为：
x=2+2cost,
y=2sint,(0≤t≤π/2）
x'(t)=-2sint,
y'(t)=2cost,
ds=√[x'(t)^2+y'(t)^2]dt=2dt,
I=∫[0，π/2] [（2+2cost)^2+(2sint)^2]*2dt
=2∫[0，π/2] [4+8cost+4(cost)^2+4(sint)^2]dt
=16∫[0，π/2] (1+cost)dt
=16(t+sint)[0,π/2]
=16（π/2+1-0）
=8π+16。本回答被提问者和网友采纳

Answer 2

(x - 2)² + y² = 2
{ x - 2 = √2cost ==> x = 2 + √2cost
{ y = √2sint

∫L (x² + y²) ds
= ∫(0→π) [(2 + √2cost)² + 2sin²t]√[(- √2sint)² + (√2cost)²] dt
= √2∫(0→π) (4 + 4√2cost + 2cos²t + sin²t) dt
= √2∫(0→π) [5 + 4√2cost + (1 + cos2t)/2] dt
= √2∫(0→π) (11/2 + 4√2cost + (1/2)cos2t) dt
= √2 * [11t/2 + 4√2sint + (1/4)sin2t] |(0，π)
= √2 * (11π/2 + 0 + 0)
= 11π/√2

Answer 3

16π追问

过程！！！