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.
곡면에서의 면적분 ( 어려움)
발산을 이용한 쉬운 삼중적분 (쉬움)
![]()
1-4. Using the Divergence Theorem, evaluate the surface integral:
1.
where
:
.
http://matrix.skku.ac.kr/cal-lab/cal-14-8-1.html
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
.
http://matrix.skku.ac.kr/cal-lab/cal-15-6-2.html
|
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
,
,
.
http://matrix.skku.ac.kr/cal-lab/cal-15-6-3.html
|
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
|
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
.
http://matrix.skku.ac.kr/cal-lab/cal-14-8-5.html
|
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.