Re: [考古] 重積分

[Frobenius (i^(-i)= e^(π/2))]

※ 引述《GayerDior (最近色團越來越忙o(＞＜:)》之銘言：
: E:[83交大]
: Let R ={ (x，y)|(x-1)^2 + (y-1)^2 ≦ 2 }. Fill the following blanks:

                            1+√2    1+√(1+2y-y^2)
                           ╴╴╴    ╴╴╴╴╴╴╴ √(x^2+y^2)
 ∫∫ √(x^2+y^2) dxdy = ∫        ∫               ╴╴╴╴╴╴ dxdy
     R                      1-√2    1-√(1+2y-y^2)
                           ╴╴╴    ╴╴╴╴╴╴╴

                             2π       √2
                           ╴╴╴    ╴╴╴         r
                       = ∫        ∫           ╴╴╴╴ r drdΘ
                              0         0
                           ╴╴╴    ╴╴╴

                           2π  r^3  |r = √2
                       = ∫   (-----)|        dΘ
                            0    3   |r =  0

                           2π 2√2
                       = ∫   (----) dΘ
                            0   3

                              4√2
                              ----π
                               3
                       =  ╴╴╴╴╴╴
: 如果有人會寫，
: 幫我寫出詳細過程，不要只填空格
: 拜託了，謝謝

(x-1)^2 + (y-1)^2 = 2 => x = 1 ± √(1+2y-y^2)

(1-1)^2 + (y-1)^2 = 2 => y = 1 ± √2

                ╴╴╴╴╴╴╴╴╴╴╴╴
√(x^2+y^2) = √(rCosΘ)^2 + (rSinΘ)^2  = r

    dxdy    = rdrdΘ

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