Question

求二重积分∫∫|x^2+y^2-2x|dQ,区域D:X^2+Y^2<=4求详细过程

Answer 1

这个题太麻烦了，不管有没做对，都要采纳啊。

将积分区域分为两部分
D1：x²+y²≤2x，即(x-1)²+y²≤1
D2：2x≤x²+y²≤4
在D1内
∫∫|x²+y²-2x|dxdy
=∫∫(2x-x²-y²)dxdy
=∫∫ (2rcosθ-r²)r drdθ
=∫[-π/2→π/2] dθ∫[0→2cosθ] (2rcosθ-r²)r dr
=∫[-π/2→π/2] dθ∫[0→2cosθ] (2r²cosθ-r³) dr
=∫[-π/2→π/2] ((2/3)r³cosθ-(1/4)r⁴) |[0→2cosθ] dθ
=∫[-π/2→π/2] ((16/3)cos⁴θ-4cos⁴θ) dθ
=(4/3)∫[-π/2→π/2] cos⁴θ dθ
用奇偶对称性
=(8/3)∫[0→π/2] cos⁴θ dθ
该积分可以降幂计算，也可以直接套公式，我下面直接套公式
=(8/3)*(3/4)(1/2)*(π/2)
=π/2

在D2内：
∫∫|x²+y²-2x|dxdy
=∫∫ (x²+y²-2x) dxdy
=∫∫ (r²-2rcosθ)r drdθ
=∫[0→2π] dθ ∫[2cosθ→2] (r³-2r²cosθ) dr
=∫[0→2π] ((1/4)r⁴-(2/3)r³cosθ) |[2cosθ→2]dθ
=∫[0→2π] (4-(16/3)cosθ-4cos⁴θ+(16/3)cos⁴θ) dθ
=∫[0→2π] (4-(16/3)cosθ+(4/3)cos⁴θ) dθ
=4θ-(16/3)sinθ+(4/3)∫[0→2π] cos⁴θ dθ
=4θ-(16/3)sinθ+(2/3)∫[0→2π] (1+cos2θ)² dθ
=4θ-(16/3)sinθ+(2/3)∫[0→2π] (1+2cos2θ+cos²2θ) dθ
=4θ-(16/3)sinθ+(2/3)∫[0→2π] (1+2cos2θ+(1/2)(1+cos4θ)) dθ
=4θ-(16/3)sinθ+(2/3)∫[0→2π] (3/2+2cos2θ+(1/2)cos4θ) dθ
=4θ-(16/3)sinθ+(2/3)((3/2)θ+sin2θ+(1/8)sin4θ) |[0→2π]
=8π+2π
=10π

最后结果=π/2+10π=21π/2

Answer 2

楼上结果应该是正确的。π/2+10π=21π/2