Re: [微積] 積分
※ 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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09/21 19:55, , 1F
09/21 19:55, 1F
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