三重积分问题~
发布网友
发布时间：2024-10-22 20:31
共1个回答
热心网友
时间：2024-10-24 20:16
= ∫(0→1) z dz ∫∫ Dxy dxdy
= ∫(0→1) z • π(2z) dz
= 2π • (1/3)[ z³ ] |(0→1)
= 2π/3
或
∫∫∫Ω z dV
= ∫∫Dxy dxdy ∫(r²/2→1) z dz
= ∫(0→2π) dθ ∫(0→√2) r dr • (1/2)[ z² ] |(r²/2→1)
= π • ∫(0→√2) r • [ 1 - r⁴/4 ] dr
= (π/4)∫(0→√2) (4r - r⁵) dr
= (π/4) • [ 2r² - (1/6)r⁶ ] |(0→√2)
= (π/4) • [ 4 - (1/6)(8)]
= 2π/3
