高数 计算曲线积分
2个回答
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答:
(1):2√2 π
(2):6√2 π
(3):0
x² + 2y² = 2
z = 1 + y
参数化:
x = √2 cost、dx/dt = - √2 sint
y = sint、dy/dt = cost
z = 1 + sint、dz/dt = cost
ds = √( x'² + y'² + z'²) dt
= √(2sin²t + cos²t + cos²t) dt
= √2 dt
0 ≤ t ≤ 2π
(1):
∮_(Γ) x² ds
= ∫(0,2π) 2cos²t * √2 dt = 2√2 π
(2):
∮_(Γ) (x² + y² + z²) ds
= ∫(0,2π) (2cos²t + sin²t + (1 + sint)²) √2 dt
= ∫(0,2π) (2sint + 3) * √2 dt
= 6√2 π
(3):
∮_(Γ) x² dx + y² dy + z² dz
= ∫(0,2π) [ (2cos²t)(- √2sint) + (sin²t)(cost) + (1 + sint)²(cost) ] dt
= 0
(1):2√2 π
(2):6√2 π
(3):0
x² + 2y² = 2
z = 1 + y
参数化:
x = √2 cost、dx/dt = - √2 sint
y = sint、dy/dt = cost
z = 1 + sint、dz/dt = cost
ds = √( x'² + y'² + z'²) dt
= √(2sin²t + cos²t + cos²t) dt
= √2 dt
0 ≤ t ≤ 2π
(1):
∮_(Γ) x² ds
= ∫(0,2π) 2cos²t * √2 dt = 2√2 π
(2):
∮_(Γ) (x² + y² + z²) ds
= ∫(0,2π) (2cos²t + sin²t + (1 + sint)²) √2 dt
= ∫(0,2π) (2sint + 3) * √2 dt
= 6√2 π
(3):
∮_(Γ) x² dx + y² dy + z² dz
= ∫(0,2π) [ (2cos²t)(- √2sint) + (sin²t)(cost) + (1 + sint)²(cost) ] dt
= 0
追问
第3个是怎么得到0的呀
追答
把那个积分拆开后逐个计算就可以了
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