曲线积分问题，我怎么算都和答案不一样，答案是20√2π 10
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（5）C : (x-1)^2 + (y-1)^2 = 2, 令 x = 1+√2cost，y = 1+√2sint，则
I = ∮<C> (2x^2+3y^2)ds
= ∫<-π/4, 3π/4> [2+4√2cost+4(cost)^2+3+6√2sint+6(sint)^2]√2dt
= √2∫<-π/4, 3π/4> [9+4√2cost+6√2sint+2(sint)^2]dt
= √2∫<-π/4, 3π/4> (10+4√2cost+6√2sint-cos2t)dt
= √2[10t+4√2sint-6√2cost-(1/2)sin2t)]<-π/4, 3π/4>
= √2[10π+8+12-0] = 10√2(π+2)
I = ∮<C> (2x^2+3y^2)ds
= ∫<-π/4, 3π/4> [2+4√2cost+4(cost)^2+3+6√2sint+6(sint)^2]√2dt
= √2∫<-π/4, 3π/4> [9+4√2cost+6√2sint+2(sint)^2]dt
= √2∫<-π/4, 3π/4> (10+4√2cost+6√2sint-cos2t)dt
= √2[10t+4√2sint-6√2cost-(1/2)sin2t)]<-π/4, 3π/4>
= √2[10π+8+12-0] = 10√2(π+2)
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