高靛女 發問時間: 科學數學 · 1 0 年前

高等微積分......急急急

let S be the region in the first quadrant bounded by the curves xy=1,xy=4,and lines y=x,y=4x. find the area and the centriod of S using the transmation u=xy,v=y/x.

九摳逼 算出來之 後

輩積分函數 是什麼

麻煩各位大大

請把過程寫詳細一點

感激 不進...

3 個解答

評分
  • 1 0 年前
    最佳解答

    1. the area of S= double integral [over S] {1} dA; call it A(S)

    2. the the centriod of S=(x0, y0), where x0 = 1/(A(S)) * double integral [over S] {x} dA; y0 = 1/(A(S)) * double integral [over S] {y} dA.

    以上是你要的[被積分函數 是什麼]

    Using the transmation u=xy,v=y/x => = x=(u/v)^(1/2); y=(uv)^(1/2) => x_u= (1/2) * u^(-1/2) * v^(-3/2); x_v = (-1/2) * u^(1/2) * v^(-5/2);

    y_u= (1/2) * u^(-1/2) * v^(1/2); x_v = (1/2) * u^(1/2) * v^(-1/2); => The transformation Jacobian |J|=(1/2)*v^(-2) ----大概是大大說的九摳逼吧?

    Thus dA=dxdy= |J| dudv and S--->R after the change of variables.

    Calculating A(S): (x,y)-->(u,v) A(S) = double integral [over R] {1* (1/2)*v^(-2) } dudv, where R is the rectangle {(u,v)| 1<=u<=4, 1<=v<=4} on uv-plane. ------ => A(S)=9/8;

    Using formulas in 2, you should be able to finish finding the centroid.

  • 1 0 年前

    利用Jacobian變換之後的圖形較易算,自變數改為u,v,積分上下限也都是介於1和4之間

  • linch
    Lv 7
    1 0 年前

    被積分函數就是 1 啊!

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