# 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

• 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

到下面的網址看看吧

• 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