x²+y²=1的极坐标方程为：r=1
x²+y²=2x的极坐标方程为：r²=2rcosθ，即r=2cosθ
2cosθ=1，则：cosθ=1/2，θ=π/3
请自己画图
因此两曲线所围区域可分为两部分，
第一部分θ：0-->π/3，r：0-->1
第二部分：θ：π/3-->π/2，r：0-->2cosθ
∫∫xydxdy
=∫∫rcosθ*rsinθ*rdrdθ
=∫[0-->π/3]∫[0-->1] r³cosθsinθ drdθ+∫[π/3-->π/2]∫[0-->2cosθ] r³cosθsinθ drdθ
=∫[0-->π/3]cosθsinθdθ*∫[0-->1] r³dr+∫[π/3-->π/2] cosθsinθdθ∫[0-->2cosθ] r³dr
=1/4∫[0-->π/3]sinθd(sinθ)*r⁴ | [0-->1] + 1/4∫[π/3-->π/2] cosθsinθ*r⁴｜[0-->2cosθ]dθ
=1/8sin²θ |[0-->π/3] + 4∫[π/3-->π/2] cosθsinθ*cos⁴θdθ
=(1/8)(3/4) - 4∫[π/3-->π/2] cos⁵θd(cosθ)
=3/32 - 2/3cos⁶θ |[π/3-->π/2]
=3/32 + (2/3)(1/64)
=9/96 + 1/96
=10/96
=5/48
感觉这个题应该是你的答案错了。
