∫x^2ds 其中C为曲线x^2+y^2+z^2=4与z=根号3的交线

2025-05-15 07:17:48
推荐回答(3个)
回答1:

把z = √3代入x² + y² + z² = 4
x² + y² + 3 = 4 得 x² + y² = 1
于是令x = cost,y = sint,0 ≤ t ≤ 2π
ds = √[(x'(t))² + (y'(t))²] dt = √(sin²t + cos²t) dt = dt
∮ x² ds
= ∫(0→2π) cos²t dt
= ∫(0→2π) (1 + cos2t)/2 dt
= (1/2)[t + (1/2)sin2t]:(0→2π)
= π

回答2:

x^2+y^2=1.x=cos(s),∫x^2ds=∫(0to2π)[cos(s)]^2ds=4∫(0toπ/2)[cos(s)]^2ds=4*1/2*π/2=π。

回答3:

利用在曲线C上X和Y的对称性做,不知道你现在会了没有,不会的话请追问。。。