發問時間: 科學數學 · 6 年前

微積分高手20點~~

【1】

(1) Let R be the region bounded by the curves x=y^2 and x =2y - y^2

. Express the area of the region R as an iterated double integral and evaluate the integral.

(2) Let D be the ice cream cone cut from the solid sphere r(有些書寫 讀法:lo)≦1 by the cone Φ=π/3 in the spherical coordinates. Express the volume of the solid D as an iterated triple integral and evaluate the integral.

【2】

(1) Find the volume of the solid bounded above the paraboloid z =x^2 +y^2 and below by the square R ={(x, y) -1≦x≦1,-1≦y≦1 } in the xy plane.

【3】

Let f (x, y) = x ^2+ x cos y ,u =i + j = (1,1) , and a point P (1, π/2)

Is gradient f(1,π/2 ) normal to the level curve f (x, y) =1? Why?

3 個解答

評分
  • 麻辣
    Lv 7
    6 年前
    最佳解答

    (1) Let R be the region bounded by the curves x=y^2 and x=2y-y^2.

    Express the area of the region R as an iterated double integral and

    evaluate the integral.

    Ans:

    x=y^2=-y^2+2y => 0=2y(y-1) => y=0,1

    x=y^2=0,1 => Limit points: A=(0,0), B=(1,1)

    dA=dx*dy =>

    A=∫∫(y^2~2y-y^2)dx*dy

    =∫x*dy

    =∫(0~1)(2y-y^2-y^2)dy

    =∫(2y-2y^2)dy

    =y^2-2y^3/3

    =1-2/3

    =1/3

    (3) Find the volume of the solid bounded above the paraboloid z=x^2

    +y^2 and below by the square R={(x, y)|-1≦x≦1,-1≦y≦1} in the xy

    plane.

    Ans:

    Projectional method: z=x^2 on x.z plane

    dV=πx^2*dz

    V=π∫x^2*dz

    =π∫(0~1)zdz

    =πz^2/2

    =π/2

    (4) Let f(x,y)=x^2+x*cos(y), u=i+j=(1,1), and a point P=(1,π/2).

    Gradient f(1,π/2) normal to the level curve f(x,y)=1? Why?

    Ans:

    Grad f(x,y)=[2x+cos(y),-x*sin(y)]

    Grad f(1,π/2)=(2,-1)

    f(1,π/2)=x^2+x*cos(y)=1+1*0=1

    => Grad[F(P)] is normal f(P)=1

  • 匿名使用者
    6 年前

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  • 6 年前

    第【2】有問題

    被限制在在拋物圓椎面上面

    xy平面下面

    沒這種東西

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