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)