SKKU-Calculus-Sec-15-9 Divergence Theorem 최주영

15.9    Divergence Theorem    by SGLee, 최주영

Point. Gauss Divergence Theorem는 vector version of Green's Theorem 의 벡터영역을 3차원으로 확장한 것입니다.

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발산을 이용한 쉬운 삼중적분 (쉬움)

1-4. Using the Divergence Theorem, evaluate the surface integral:

1.  where .

Define "Div" function

 var('x,y,z');    def Div(F):     assert(len(F) == 3)     return (diff(F[0],x)+diff(F[1],y)+diff(F[2],z))

 var('x,y,z'); Div([x^3,x^2*y,x*y])

0

Since divergence is 0, the given integral value is .

2.  where  is a closed surface consisting of the circular cylinder

and the circular disks  and .

 var('x,y,z');    def Div(F):     assert(len(F) == 3)     return (diff(F[0],x)+diff(F[1],y)+diff(F[2],z)) Div([x^3,x^2*y,x*y])

4*x^2

Parametrized by

 var('t') x=cos(t) y=sin(t) z=z w=integral(4*cos(t)^2,t,0,2*pi) var('a,b'); w*a*b

4*pi*a*b

3.  where  is a parallelepiped of ,　,.

 var('x,y,z'); Div([sin(x),(2-cos(x))*y,0])

2

 integral(integral(integral(2,x,0,3),y,0,2),z,0,1)

12

4.  where : cube of side  and three of whose edges are along the axes.

 var('x,y,z,r,t'); def Div(F):     assert(len(F) == 3)     return (diff(F[0],x)+diff(F[1],y)+diff(F[1],z)) Div([x^2-y*z,-2*x^2*y,z])

-2*x^2 + 2*x

Parametrized by

 x=cos(t) y=sin(t) z=z integral(integral(integral(-2*cos(t)^2+2*cos(t),t,0,2*pi),r,0,2),z,0,3)

-12*pi

5-13. Verify the Gauss Divergence Theorem for :

5.  taken over the region bounded by  and .

 var('x,y,z,t') p1 = implicit_plot3d(x^2+y^2==4, (x,-2,2), (y, -2,2), (z, -5,5), opacity=0.2, color="red", mesh=True); p2 = implicit_plot3d(z==0, (x,-2,2), (y, -2,2),(z, -5,5), opacity=0.3, color="blue", mesh=True); p3 = implicit_plot3d(z==3, (x,-2,2), (y, -2,2), (z, -5,5),   opacity=0.5, color="orange", mesh=True); show(p1+p2+p3, aspect_ratio=1)

 var('x,y,z') def Div(F):     assert(len(F) == 3)     return (diff(F[0],x)+diff(F[1],y)+diff(F[1],z)) Div([4*x,-2*y^2,z^2])

-4*y + 4

 var('x,y,z,t') integral(integral(2*(-8*sin(t)+2*z+4),t,0,2*pi),z,0,3)

84*pi

,  ,

,

.

Therefore .

6.  taken over the entire surface of the cube ,,.

,

,

,

,

(Since  on ),

,

Therefore

※ div. Hence div.

7.  taken over the entire surface of the sphere of radius  and centered at the origin.

http://matrix.skku.ac.kr/cal-lab/cal-14-8-7.html

 var('a,b,c,x,y,z') def Div(F):     assert(len(F) == 3)     return (diff(F[0],x)+diff(F[1],y)+diff(F[1],z)) Div([a*x,b*y,c*z])

a+b+c

So, integral of a+b+c over the entire surface of the sphere of radius  is .

Compute the second part, and show they are equal.

8.  and  is the total surface of the rectangular parallelepiped bounded by the coordinate planes and .

 var('x,y,z,t') p1 = implicit_plot3d(x==1, (x,0,5), (y, 0,5), (z, 0,5), opacity=0.2, color="red", mesh=True); p2 = implicit_plot3d(y==2, (x,0,5), (y, 0,5), (z, 0,5), opacity=0.3, color="blue", mesh=True); p3 = implicit_plot3d(z==3, (x,0,5), (y, 0,5),(z, 0,5),   opacity=0.5, color="orange", mesh=True); show(p1+p2+p3, aspect_ratio=1)

 def Div(F):     assert(len(F) == 3)     return (diff(F[0],x)+diff(F[1],y)+diff(F[1],z)) d10=Div([2*x*y,y*z^2,x*z])

 integral(integral(integral(d10,x,0,1),y,0,2),z,0,3)

48

Compute the second part, and show they are equal.

9.  over the upper half of the sphere .

A parameterization of sphere is

,

,

,

.

Hence  (See the figure).

,

.

Therefore .

Compute the second part, and show they are equal.

10.  taken over the rectangular parallelepiped bounded by the coordinate planes and  and .

.

11.  taken over the surface of the ellipsoid .

Let  be a parameterization of .

,

.

Hence

.

Compute the second part, and show they are equal.

12.  taken over the upper half of the unit sphere .

,  ,

,

.

Therefore .

Compute the second part, and show they are equal.

13.  taken over the closed region of the cylinder , bounded by the planes   and .

,  ,

.

,

Therefore .

Compute the second part, and show they are equal.

14. (a)  Prove Green's first identity :

(b) Let  and . Find .

15. Let  be a solid surrounded by  and  and vector field

. Find the flux of , that is,

div,

div

16. Evaluate .

Here , and  is the boundary of the solid surrounded by .

div.

Let .

.

17. Let  be a surface  between  and  and . Find a flux .

div.

div

18. Evaluate the surface integral , where  and  is the part of surface of the paraboloid  above -plane.

Let  and  be the region bounded by  and .

Then, with the aid of the Divergence Theorem,

. Since  and  on ,

we can see that . On the other hand, , and

therefore, we have  = volume of ,

which can be computed as follows:

.

19. Let . Evaluate , where  is the part of the sphere

.

Note that

. Using the Divergence    Theorem and polar coordinates , where

,

we compute  .

div

.

.

:

.

20. If , evaluate   over the volume of a cube of side .

21. Evaluate  over the solid region of the sphere  when   where  are constants.