15.6 Gauss Divergence Theorem: Transformation between surface integral and volume integral

1-4. Using divergence theorem, evaluate the surface integral:


1. where : .

http://matrix.skku.ac.kr/cal-lab/cal-14-8-1.html

 Sol) 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 : Closed surface consisting of the circular cylinder

          and the circular disks and .

http://matrix.skku.ac.kr/cal-lab/cal-15-6-2.html

 Sol)

 

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 : parallelopiped ,,.

http://matrix.skku.ac.kr/cal-lab/cal-15-6-3.html

 Sol)

 

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.

http://matrix.skku.ac.kr/cal-lab/cal-15-6-4.html

 Sol)

 

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 Gauss divergence theorem for :


5. taken over the region bounded by , and .

http://matrix.skku.ac.kr/cal-lab/cal-14-8-5.html

 Sol)

 

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 ,,.

 Sol)

   


  ,

  ,

  ,

  ,

  ,

  ,

  Therefore .


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

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

 Sol)

 

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 a+b+c over the entire surface of the sphere of radius d is .


8. and is the total surface of the rectangular pallelepiped bounded by the coordinate planes and , , .
http://matrix.skku.ac.kr/cal-lab/cal-15-6-8.html

 Sol)

 

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


9. over the upper half of the sphere .

 Sol)

    

 ,

 ,

 ,

  

               ,

  .

  Therefore .


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

 Sol) .


11. taken over the surface of the ellipsoid .

 Sol) ,

     ,

     ,

     

                  .


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

 Sol)

   


 ,

 ,

  

  .

  Therefore .


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

 Sol)

    


 ,

 .


  

 ,

 

  Therefore .


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

(b) Let and . Find .


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

    . Find flux of , that is,

 Sol) div,


     div

     


16. Evaluate .

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

 Sol) div.

     Let .

      

                 

                 .


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

 Sol) div.


     div

     


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

 Sol) Let and be the region bounded by and .

     Then, with the aid of 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

    , .

 Sol) Note that

     . Using the divergence    Theorem and polar coordinates , , where 
,
we compute .

       div 

                      

                      .


       

       

                   

                  .


       :

       .


  (The End)