∫∫∫1/根号(x²+y²+²)dxdydz,其中区域为z=根号(x²+y²)和z=1围成的
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 22:03:27
![∫∫∫1/根号(x²+y²+²)dxdydz,其中区域为z=根号(x²+y²)和z=1围成的](/uploads/image/z/6942875-59-5.jpg?t=%E2%88%AB%E2%88%AB%E2%88%AB1%2F%E6%A0%B9%E5%8F%B7%28x%26%23178%3B%2By%26%23178%3B%2B%26%23178%3B%29dxdydz%2C%E5%85%B6%E4%B8%AD%E5%8C%BA%E5%9F%9F%E4%B8%BAz%3D%E6%A0%B9%E5%8F%B7%28x%26%23178%3B%2By%26%23178%3B%29%E5%92%8Cz%3D1%E5%9B%B4%E6%88%90%E7%9A%84)
∫∫∫1/根号(x²+y²+²)dxdydz,其中区域为z=根号(x²+y²)和z=1围成的
∫∫∫1/根号(x²+y²+²)dxdydz,其中区域为z=根号(x²+y²)和z=1围成的
∫∫∫1/根号(x²+y²+²)dxdydz,其中区域为z=根号(x²+y²)和z=1围成的
{ x² + y² = z² --> r = z
{ z = 1 --> r = 1 --> r = secφ
球面坐标法:
∫∫∫ 1/√(x² + y² + z²) dxdydz
= ∫(0→2π) dθ ∫(0→π/4) sinφ dφ ∫(0→secφ) 1/r • r² dr
= 2π • ∫(0→π/4) sinφ dφ • (1/2)[ r² ]:(0→secφ)
= π • ∫(0→π/4) sinφ • (sec²φ) dφ
= π • ∫(0→π/4) secφtanφ dφ
= π • [ secφ ]:(0→π/4)
= (√2 - 1)π
切片法:
∫∫∫ 1/√(x² + y² + z²) dxdydz
= ∫(0→1) dz ∫(0→2π) dθ ∫(0→z) 1/√(r² + z²) • r dr
= 2π • ∫(0→1) dz • (1/2)[ 2√(r² + z²) ]:(0→z)
= 2π • ∫(0→1) (√2 - 1)z dz
= 2(√2 - 1)π • (1/2)[ z² ]:(0→1)
= (√2 - 1)π
投影法:
∫∫∫ 1/√(x² + y² + z²) dxdydz
= ∫(0→2π) dθ ∫(0→1) r dr ∫(r→1) 1/√(r² + z²) dz
= 2π • ∫(0→1) r dr • ln[ z + √(r² + z²) ]:(r→1)
= 2π • ∫(0→1) r • {ln[1 + √(r² + 1)] - ln[r + √2r]} dr
= 2π • { (1/2)√(r² + 1) - (1/2)r²ln[(1 + √2)r] + (1/2)r²ln[1 + √(1 + r²)] - 1/4 }:(0→1)
= 2π • { [(1/2)√2 - (1/2)ln(1 + √2) + (1/2)ln(1 + √2) - 1/4] - (1/2 - 1/4)}
= 2π • [(1/√2 - 1/4) - 1/4]
= (√2 - 1)π