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步骤:
1) 化为polar functions: r1 = √2; r2 = 2cosθ
交点处:r1 = r2 ==> cosθ = √2/2 ==> θ = π/4
2) 利用积分区域的对称性, y 的积分为零, x 的积分只算一半,结果乘 2可得:
原积分 = 2[∫[π/4,π/2] dθ ∫[2cosθ, √2] r cosθ rdr + ∫[π/2,π] dθ ∫[0, √2] r cosθ rdr
= 2[∫[π/4,π/2] (1/3)[2√2-8cos^3θ]cosθ dθ + ∫[π/2,π] (1/3)(2√2) cosθ dθ]
= 2[-2/3 - ∫[π/4,π/2] (1/3)8cos^4θ dθ ]
= 2[-2/3 - ∫[π/4,π/2] (1/3)8[(1/2)(1+cos2θ)]^2 dθ ]
= 2[-2/3 - ∫[π/4,π/2] (2/3)(1+2cos2θ+cos^22θ)]^2 dθ ]
= 2(-2/3 + 2/3 - π/4)
= -π/2
1) 化为polar functions: r1 = √2; r2 = 2cosθ
交点处:r1 = r2 ==> cosθ = √2/2 ==> θ = π/4
2) 利用积分区域的对称性, y 的积分为零, x 的积分只算一半,结果乘 2可得:
原积分 = 2[∫[π/4,π/2] dθ ∫[2cosθ, √2] r cosθ rdr + ∫[π/2,π] dθ ∫[0, √2] r cosθ rdr
= 2[∫[π/4,π/2] (1/3)[2√2-8cos^3θ]cosθ dθ + ∫[π/2,π] (1/3)(2√2) cosθ dθ]
= 2[-2/3 - ∫[π/4,π/2] (1/3)8cos^4θ dθ ]
= 2[-2/3 - ∫[π/4,π/2] (1/3)8[(1/2)(1+cos2θ)]^2 dθ ]
= 2[-2/3 - ∫[π/4,π/2] (2/3)(1+2cos2θ+cos^22θ)]^2 dθ ]
= 2(-2/3 + 2/3 - π/4)
= -π/2
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