Re: [微積] 積分

[air11 (拿出一張白紙...)]

※ 引述《jacky7987 (憶)》之銘言：
: ∫∫  exp(x^2+2xy+5y^2)dA, where E={(x,y)|x^2+2xy+5y^2≦1}
:     E
: 我配方之後let x+y=rcosθ  2y=rsinθ
: r的範圍應該是0-1 那請問一下θ的範圍要怎麼看呢?
: 感謝:D


let x^2+2xy+5y^2 = (x^2+2xy+y^2)+4y^2

    = (x+y)^2 + (2y)^2

    let u =x+y, v=2y, then |J|=1/2

    so ∫∫ exp(x^2+2xy+5y^2)dA = 1/2 * ∫∫   exp(u^2+v^2)dA
         E                             u^2+v^2=1

          2pi 1                    pi
    = 1/2∫ ∫  r exp(r^2)drdθ = ----(e-1)
          0   0                    2       #

   (Since u^2+v^2=1 is unit disk, so 0<=r<=1 and 0<=θ<=2pi )

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