大师作文网6.3第一型曲面积分有问：
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/13 01:57:10
6.3第一型曲面积分有问：
取Σ：z = √(x² + y²)
z'x = x/√(x² + y²)、z'y = y/√(x² + y²)
取D：x² + y² = 2x ==> (x - 1)² + y² = 1 ==> r = 2cosθ with - π/2 ≤ θ ≤ π/2
D只关于x轴对称 ==> ∫(- b→b) (关于y的奇函数) dy = 0
∫∫Σ (xy + yz + zx) dS
= ∫∫D (xy + yz + zx)√[1 + x²/(x² + y²) + y²/(x² + y²)] dxdy
= 0 + 0 + ∫∫D x√(x² + y²) * √2 dxdy
= ∫(- π/2,π/2) dθ ∫(0,2cosθ) √2 * rcosθ * r * r dr
= 2√2∫(0,π/2) cosθ * (1/4)[r⁴：0,2cosθ] dθ
= (2√2)(1/4)(16)∫(0,π/2) cos⁵θ dθ
= 8√2 * 4!/5!
= 8√2 * (4 * 2)/(5 * 3 * 1)
= (64√2)/15
z'x = x/√(x² + y²)、z'y = y/√(x² + y²)
取D：x² + y² = 2x ==> (x - 1)² + y² = 1 ==> r = 2cosθ with - π/2 ≤ θ ≤ π/2
D只关于x轴对称 ==> ∫(- b→b) (关于y的奇函数) dy = 0
∫∫Σ (xy + yz + zx) dS
= ∫∫D (xy + yz + zx)√[1 + x²/(x² + y²) + y²/(x² + y²)] dxdy
= 0 + 0 + ∫∫D x√(x² + y²) * √2 dxdy
= ∫(- π/2,π/2) dθ ∫(0,2cosθ) √2 * rcosθ * r * r dr
= 2√2∫(0,π/2) cosθ * (1/4)[r⁴：0,2cosθ] dθ
= (2√2)(1/4)(16)∫(0,π/2) cos⁵θ dθ
= 8√2 * 4!/5!
= 8√2 * (4 * 2)/(5 * 3 * 1)
= (64√2)/15