计算三重积分∫∫∫Ω(x+z)dv 其中Ω由曲面z=√(x2+y2)与z=√(1-x2-y2)所围成的区域,请用球坐标求积分
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利用对称性:∫∫∫Ω(x+z)dv = ∫∫∫Ω(z)dv
换成球坐标:z = r cosφ, x = r sinφ cos θ, y = r sinφ sin θ
r from 0 to 1; φ from 0 to pi/4; θ from 0 to 2pi
∫∫∫Ω(z)dv = ∫[0,2pi] d θ ∫[0, pi/4] dφ ∫[0,1] rcosφ r^2sinφ dr
= (pi/4) ∫[0, pi/4] sin φ cos φ dφ
= (pi/4)(1/2)sin^2φ from 0 to pi/4
= pi/16
换成球坐标:z = r cosφ, x = r sinφ cos θ, y = r sinφ sin θ
r from 0 to 1; φ from 0 to pi/4; θ from 0 to 2pi
∫∫∫Ω(z)dv = ∫[0,2pi] d θ ∫[0, pi/4] dφ ∫[0,1] rcosφ r^2sinφ dr
= (pi/4) ∫[0, pi/4] sin φ cos φ dφ
= (pi/4)(1/2)sin^2φ from 0 to pi/4
= pi/16
追问
φ为什么是0到4/π
追答
φ是与z轴的夹角。从零度开始,到pi/4,因为在锥面上,取x = 0, 则 z = y。tanφ = y/z = 1, 所以 φ = pi/4.
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