利用极坐标求x²+y²-3的二重积分，其中D为闭区域x²+y²=2x？
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x²+y²=2x 极坐标 是 r = 2cost
∫∫<D>(x²+y²-3)dxdy = ∫<-π/2, π/2>dt∫<0, 2cost> (r^2-3)rdr
= ∫<-π/2, π/2>dt∫<0, 2cost> (r^3-3r)dr
= ∫<-π/2, π/2>dt[r^4/4-3r^2/2]<0, 2cost>
= ∫<-π/2, π/2>[4(cost)^4-6(cost)^2]dt
= 2∫<0, π/2>[(1+cos2t)^2-3(1+cos2t)]dt
= 2∫<0, π/2>[1+2cos2t+(cos2t)^2-3(1+cos2t)]dt
= 2∫<0, π/2>[1+2cos2t+1/2+(1/2)cos4t-3(1+cos2t)]dt
= 2∫<0, π/2>[-3/2-cos2t+(1/2)cos4t]dt
= 2[-3t/2-(1/2)sin2t+(1/8)sin4t]<0, π/2> = -3π/2
∫∫<D>(x²+y²-3)dxdy = ∫<-π/2, π/2>dt∫<0, 2cost> (r^2-3)rdr
= ∫<-π/2, π/2>dt∫<0, 2cost> (r^3-3r)dr
= ∫<-π/2, π/2>dt[r^4/4-3r^2/2]<0, 2cost>
= ∫<-π/2, π/2>[4(cost)^4-6(cost)^2]dt
= 2∫<0, π/2>[(1+cos2t)^2-3(1+cos2t)]dt
= 2∫<0, π/2>[1+2cos2t+(cos2t)^2-3(1+cos2t)]dt
= 2∫<0, π/2>[1+2cos2t+1/2+(1/2)cos4t-3(1+cos2t)]dt
= 2∫<0, π/2>[-3/2-cos2t+(1/2)cos4t]dt
= 2[-3t/2-(1/2)sin2t+(1/8)sin4t]<0, π/2> = -3π/2
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