x^2+y^2 = x+y, 化为极坐标， r = cost+sint
I = ∫<-π/4, 3π/4>dt ∫<0, cost+sint>r(cost+sint) rdr
= ∫<-π/4, 3π/4>(cost+sint)dt ∫<0, cost+sint>r^2 dr
= (1/3)∫<-π/4, 3π/4>(cost+sint)^4dt = (1/3)∫<-π/4, 3π/4>(1+sin2t)^2dt
= (1/3)∫<-π/4, 3π/4>[1+2sin2t+(sin2t)^2]dt
= (1/3)∫<-π/4, 3π/4>[3/2+2sin2t+(1/2)cos4t]dt
= (1/3)[3t/2 - cos2t + (1/8)sin4t]<-π/4, 3π/4> = π |
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