線性代數一題

Let T:R^2→R^3 be given by T(x,y)=(2x+3y, x-y, 2y)

Find the area of the image in R^3 under T of the unit disk in R^2.

小弟直接做法就是矩陣化,但接下來就不知道要怎麼做了!

請教一下高手!謝謝

4 Answers

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  • 1 decade ago
    Favorite Answer

    Solution method uses concepts from surface geometry:

    Position vector of any point on mapped surface is R = (2x+3y, x-y, 2y).

    The infinitesimal area at any point on the mapped surface is given by:

    dA = |Rx Ry| dx dy,

    where denotes cross product and Rx, Ry are partial derivatives of R

    w.r.t. x and y (NB these are vectors).

    [The infinitesimal area comprises a parallelogram with sides |Rx|dx, |Ry| dy,

    and the area of a p'gram is product of side lengths times the sin of the angle

    between them.]

    So, A = ∫ ∫ |Rx Ry| dx dy}

    = ∫ ∫ { |(2,1,0)(3,-1,2)| dx dy }

    = ∫ ∫ { |(2,-4,-5)| dx dy }

    = |(2,-4,-5)| ∫ ∫ {dx dy }

    = √(45).π, since ∫ ∫{dx dy } = area of the unit disc = π(standard integral).

    Therefore, area of image under T is 3√5.π.

    I think the image is an ellipse.

    My answer and 煩惱即是菩提 's are the same.

    Source(s): myself
  • Anonymous
    1 decade ago

    底下在昨天已經貼在 無名 但忘記貼在 知識家, 請參考看看:

    http://www.wretch.cc/album/show.php?i=gspgraph&b=1...

    http://www.wretch.cc/album/show.php?i=gspgraph&b=1...

  • L
    Lv 7
    1 decade ago

    令 A =

    (2 3)

    (1 -1)

    (0 2)

    B =

    (1/5 3/5 0)

    (1/5 -2/5 0)

    ( 2 4 -5)

    首先注意到 T 的矩陣表現為 A, 且 T 把 xy 平面送到 |R^3 中的某一斜平面 (其法向量為 (2,-4,-5)), 故 xy 平面上的單位圓會被 T 映至此斜平面上的某一封閉曲線.

    取 B 如此是因 BA =

    (1 0)

    (0 1)

    (0 0)

    這表示 |R^2 上的單位圓盤通過 BA 作用後仍是 |R^3 中 xy 平面上的單位圓盤 (令為 C), 其面積為 π. 而 R^3 中 xy 平面上的單位圓通過 B^-1 後即是前言的封閉曲線 (令為 Γ), 我們要求此封閉曲線所圍成的面積, 令所求面積為 A, 透過雙重積分的變數變換我們有

    π = ∫∫_C 1 dxdy = ∫∫_Γ 1*(1/5) dudv = A/5

    故 A = 5π

  • 1 decade ago

    3√5 *π

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