曲线积分计算∮Γ(x^2+y^2+z^2)dL,其中Γ为曲线x^2+y^2+z^2=4,x+y+z=0的交线
1个回答
直接代入方程：
∮Γ (x^2 + y^2 + z^2) ds
= ∮Γ 4 ds
= 4 * 2π(2)
= 16π
或将方程参数化然后计算：
{ x^2 + y^2 + z^2 = 4
{ x + y + z = 0
将z = - x - y代入x^2 + y^2 + z^2 = 4中
==>x^2 + y^2 + xy = 2
(x + y/2)^2 + (√3y/2)^2 = 2
{ x + y/2 = √2cost
{ √3y/2 = 2sint
==>
{ x = √2cost - (√6/3)sint、dx = [- √2sint - (√6/3)cost] dt
{ y = (2√6/3)sint、dy = (2√6/3)cost dt
{ z = - √2cost - (√6/3)sint、dz = [√2sint - (√6/3)cost] dt
0 ≤ t ≤ 2π
ds = √[(dx)^2 + (dy)^2 + (dz)^2] dt = √4 dt = 2 dt
∮Γ (x^2 + y^2 + z^2) ds
= ∫(0→2π) {[√2cost - (√6/3)sint]^2 + [(2√6/3)sint]^2 + [- √2cost - (√6/3)sint]^2} * 2 dt
= ∫(0→2π) 4 * 2 dt
= 8 * 2π
= 16π
相关问题
