[2008][07-1] Engineering Mathematics 2ocw.snu.ac.kr/sites/default/files/NOTE/4694.pdf · 2018. 1....
Transcript of [2008][07-1] Engineering Mathematics 2ocw.snu.ac.kr/sites/default/files/NOTE/4694.pdf · 2018. 1....
-
2008_Vector Calculus(4)
Naval A
rch
itectu
re &
Ocean
En
gin
eerin
g
Engineering Mathematics 2
Prof. Kyu-Yeul Lee
Department of Naval Architecture and Ocean Engineering,
Seoul National University of College of Engineering
[2008][07-1]
October, 2008
-
2008_Vector Calculus(4)
Naval A
rch
itectu
re &
Ocean
En
gin
eerin
g
Vector Calculus (4) : Green’s, Stokes’and
Divergence Theorem
Green’s, Theorem
Stokes’ Theorem
Divergence Theorem
-
2008_Vector Calculus(4)
Green’s Theorem
Green’s Theorem in the Plane
Suppose that C is piecewise smooth simple closed curve bounding a region R.
If are continuous on R, then
Theorem 9.13
xQyPQP /and,/,,
RC
dAy
P
x
QQdyPdx
Positive direction
C
Negative direction
C
R R
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2008_Vector Calculus(4)
Green’s Theorem
Green’s Theorem in the Plane
Suppose that C is piecewise smooth simple closed curve bounding a region R.
If are continuous on R, then
Theorem 9.13
xQyPQP /and,/,,
RC
dAy
P
x
QQdyPdx
Proof)
R
)(2 xgy
)(1 xgy
ba x
y
bxaxgyxgR ),()(: 21
C
b
a
b
a
b
a
b
a
xg
xgR
dxyxP
dxxgxPdxxgxP
dxxgxPxgxP
dydxy
PdA
y
P
),(
))(,())(,(
))](,())(,([
21
12
)(
)(
2
1
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2008_Vector Calculus(4)
Green’s Theorem
Green’s Theorem in the Plane
Suppose that C is piecewise smooth simple closed curve bounding a region R.
If are continuous on R, then
Theorem 9.13
xQyPQP /and,/,,
RC
dAy
P
x
QQdyPdx
R
)(2 xgy
)(1 xhx
b
c
x
y
)(1 yhx d
dycyhxyhR ),()(: 21
C
d
c
d
c
d
c
d
c
xh
xhR
dyyxQ
dyyyhQdyyyhQ
dyyyhQyyhQ
dxdyy
QdA
y
Q
),(
)),(()),((
)]),(()),(([
21
12
)(
)(
2
1
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2008_Vector Calculus(4)
Green’s Theorem
Example 1Using Green’s Theorem
Evaluatewhere C consists of the boundary of the region in the first quadrant that is bounded by the graphs of y=x2 and y=x3.
,)2()( 22 C dyxydxyx
420
11)(
)(
)21(
)21(
)()2(
)2()(
1
0
2346
1
0
2
1
0
22
22
2
3
2
3
dxxxxx
dxyy
dxdyy
dAy
dAy
yx
x
xy
dyxydxyx
x
x
x
x
R
R
C
Solution)
x
y
2xy 3xy
)1,1(
R
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2008_Vector Calculus(4)
Green’s Theorem
Example 2Using Green’s Theorem
Evaluatewhere C is the circle (x-1)2+(y-5)2=4.
,)2()3(35
Cy dyexdxyx
Solution)
x
y
4)5()1( 22 yx
4isarea
R
4
)32(
)3()2(
)2()3(
5
5
3
3
R
R
R
y
C
y
dA
dA
dAy
yx
x
ex
dyexdxyx
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2008_Vector Calculus(4)
Green’s Theorem
Example 3Work Done by a Force
Find the work done by the force
acting along the simple closed curve Cshown in Figure below.
jiF )34()sin16( 22 xexy y
Solution)
R
R
y
C
y
C
dAx
dAy
xy
x
xe
dyxedxxy
dW
)166(
)sin16()34(
)34()sin16(
22
22
rF
4
3
4,10:
rR
4)8cos2(
)8cos2(
)16cos6(
4/3
4/
4/3
4/
1
0
23
4/3
4/
1
0
d
drr
drdrrW
x
y
xyC :1
1: 222 yxC
xyC :3
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2008_Vector Calculus(4)
Green’s Theorem
Example 4Green’s Theorem Not Applicable
Let C be the closed curve consisting of the four straight line segments C1, C2, C3, C4 shown in Figure. Green’s theorem is not applicable to the line integral
Cdy
yx
xdx
yx
y2222
x
y
2:1 yC
R
2:2 xC
2:3 yC
2:4 xC
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2008_Vector Calculus(4)
Green’s Theorem
Example 4Green’s Theorem Not Applicable
Let C be the closed curve consisting of the four straight line segments C1, C2, C3, C4 shown in Figure. Green’s theorem is not applicable to the line integral
Cdy
yx
xdx
yx
y2222
x
y
2:1 yC
R
2:2 xC
2:3 yC
2:4 xC
since P, Q, ∂P/∂y, and ∂Q/∂y are not continuous at the origin.
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Green’s Theorem
Region with Holes
C
CC
RRR
QdyPdx
QdyPdxQdyPdx
dAy
P
x
QdA
y
P
x
QdA
y
P
x
Q
21
21
1C 2C
1C
2C
2R
1R
(C=C1∪C2 )
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2008_Vector Calculus(4)
Green’s Theorem
Example 5Region with a Hole in It
Solution)
Evaluatewhere C=C1∪C2 is the boundary of the shaded region R shown in Figure
,2222
Cdy
yx
xdx
yx
y
x
y
R
1C
1: 222 yxC
2222),(,),(
yx
xyxQ
yx
yyxP
222
22
222
22
)(,
)( yx
xy
x
Q
yx
xy
y
P
1
2
2 2 2 2
2 2 2 2
2 2 2 2
2 2 2 2
2 2 2 2 2 20.
( ) ( )
C
C
C
R
y xdx dy
x y x y
y xdx dy
x y x y
y xdx dy
x y x y
y x y xdA
x y x y
since P, Q, ∂P/∂y, and ∂Q/∂y are continuous on the region R bounded by C, it follows from the above discussion that
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Green’s Theorem
Example 6Example 4 Revisited
Evaluate the line integral in Example 4.
Example 4.
Cdy
yx
xdx
yx
y2222
Solution)
C CCCC 4321
x
y
R
C
1: 22 yxC
2222),(,),(
yx
xyxQ
yx
yyxP
222
22
222
22
)(,
)( yx
xy
x
Q
yx
xy
y
P
2 2 2 2
2
0
22 2
0
2
0
sin ( sin ) cos (cos )
(sin cos )
2
C
y xdx dy
x y x y
t t t t dt
t t dt
dt
2 2 2 2C
y xdx dy
x y x y
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2008_Vector Calculus(4)
Stokes’ Theorem
Vector Form of Green’s Theorem
kFF
y
P
x
Q
QP
zyx
kji
0
curl
)1() curl( R
CCdAdsd kFTFrF
i
j
x
y
z
k
C
R
T
n
Green’s Theorem in 3-Space
( Stokes’ theorem)
Green’s theorem in 3-space relates a line integral around a
closed curve C forming the boundary of a surface S with a
surface integral over S.
Equation (1) relate a line integral around a closed curve C forming the
boundary of a plane region R to a double integral over R.
R
)(2 xgy
)(1 xhx
b
c
x
y
)(1 yhx d
Green Theorem in 2-D space
Green Theorem in 3-D space
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2008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ Theorem
Let S be a piecewise smooth orientable surface bounded by a piecewise smooth
simple closed curve C. Let be a
vector filed for which P,Q, and R are continuous and have continuous first partial
derivatives in a region of 3-space containing S. if C is traversed in the positive
direction, then
Where n is a unit normal to S in the direction of the orientation of S.
Theorem 9.14
kjiF ),,(),,(),,(),,( zyxRzyxQzyxPzyx
S
CCdSdsd nFTFrF ) curl()(
i
j
x
y
z
k
C
R
T
n
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2008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ Theorem
Let S be a piecewise smooth orientable surface bounded by a piecewise smooth
simple closed curve C. Let be a
vector filed for which P,Q, and R are continuous and have continuous first partial
derivatives in a region of 3-space containing S. if C is traversed in the positive
direction, then
Where n is a unit normal to S in the direction of the orientation of S.
Theorem 9.14
kjiF ),,(),,(),,(),,( zyxRzyxQzyxPzyx
S
CCdSdsd nFTFrF ) curl()(
i
j
x
y
z
k
C
R
T
n
See also
Streeter V.L., Fluid Mechanics,
McGraw-Hill, 1948,
p47, ‘24. Stokes’ Theorem’
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Stokes’ Theorem
Stokes’ TheoremTheorem 9.14
S
CCdSdsd nFTFrF ) curl()(
Partial Proof)
kjiF
y
P
x
Q
x
R
z
P
z
Q
y
R curl
22
1
y
f
x
f
y
f
x
fkji
n
),( yxfz S is oriented upward and is defined by a function
kjiF ),,(),,(),,(),,( zyxRzyxQzyxPzyx
i
j
x
y
z
k
C
R
T
n
0),(),,( yxfzzyxgIf we write
If S is defined by g(x,y,z)=0,
gg
||||
1n
(R.H.S)
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2008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ TheoremTheorem 9.14
S
CCdSdsd nFTFrF ) curl()(
kjiF
y
P
x
Q
x
R
z
P
z
Q
y
R curl
22
1
y
f
x
f
y
f
x
fkji
n
kjiF ),,(),,(),,(),,( zyxRzyxQzyxPzyx
i
j
x
y
z
k
C
R
T
n
RS
dAy
P
x
Q
y
f
x
R
z
P
x
f
z
Q
y
RdSnF) curl(
Hence,
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2008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ TheoremTheorem 9.14
S
CCdSdsd nFTFrF ) curl()(
kjiF ),,(),,(),,(),,( zyxRzyxQzyxPzyx
i
j
x
y
z
k
C
R
T
n
R
C
b
a
CC
dAx
fRP
yy
fRQ
x
dyy
fRQdx
x
fRP
dtdt
y
y
f
dt
dx
x
fR
dt
dyQ
dt
dxP
RdzQdyPdxd
xy
rF
Chain rule
Green’s theorem
btatytxfztyytxx ))(),((),(),(
btatyytxx )(),(
:C
:xyC (Projection of C onto the xy-plane)(L.H.S)
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Stokes’ Theorem
Stokes’ TheoremTheorem 9.14
S
CCdSdsd nFTFrF ) curl()(
kjiF ),,(),,(),,(),,( zyxRzyxQzyxPzyx
i
j
x
y
z
k
C
R
T
n
x
f
y
f
z
R
y
f
x
R
yx
fR
x
f
z
Q
x
Q
x
f
z
R
x
R
y
f
yx
fR
x
f
z
Q
x
Q
y
fyxfyxRyxfyxQ
xy
fRQ
x
2
2
)),(,,()),(,,(
Chain and
Product Rules
Similarly,
y
f
x
f
z
R
x
f
y
R
xy
fR
y
f
z
P
y
P
x
fRP
y
2
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Stokes’ Theorem
Stokes’ TheoremTheorem 9.14
S
CCdSdsd nFTFrF ) curl()(
kjiF ),,(),,(),,(),,( zyxRzyxQzyxPzyx
i
j
x
y
z
k
C
R
T
n
x
f
y
f
z
R
y
f
x
R
yx
fR
x
f
z
Q
x
Q
y
fRQ
x
2
∴ (L.H.S)=(R.H.S)
y
f
x
f
z
R
x
f
y
R
xy
fR
y
f
z
P
y
P
x
fRP
y
2
R
RC
dAy
P
x
Q
y
f
x
R
z
P
x
f
z
Q
y
R
dAx
fRP
yy
fRQ
xdrF
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Stokes’ Theorem
Example 1Verifying Stokes’ Theorem
Let S be the part of the cylinder z=1-x2 for 0≤x≤1, -2≤y≤2.Verify Stokes’ theorem if F=xyi+yzj+xzk.
Solution) 1) Surface IntegralkjiF xzyzxy
kji
kji
F xzy
xzyzxy
zyx
curl
01),,( 2 xzzyxg
14
2
2
x
x
g
g kin
2)4(
)2(
)2(
14
2)curl(
1
0
1
0
2
2
2
1
0
2
2
2
dxx
dxxyxy
dxdyxxy
dAxxy
dSx
xxydS
R
SS
nF
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Stokes’ Theorem
Example 1Verifying Stokes’ Theorem
Let S be the part of the cylinderz=1-x2 for 0≤x≤1, -2≤y≤2.Verify Stokes’ theorem if F=xyi+yzj+xzk.
Solution) 2) Line Integral
C CCCC 43210,0,0,1:1 dzdxzxC
00)0()0(1
C dyyy
xdxdzdyxzyC 2,0,1,2: 22
15
11)222(
)2)(1(0)1(22
0
1
42
22
2
dxxxx
xdxxxxxdxC
0,0,1,0:3 dzdxzxC
0002
23
ydyydyC
xdxdzdyxzyC 2,0,1,2: 24
15
19)222(
)2)(1(0)1(22
1
0
42
22
4
dxxxx
dxxxxxxdxC
215
190
15
110
C
xzdzyzxydx
C Cd Pdx Qdy Rdz F r
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Stokes’ Theorem
Example 2Using Stokes’ Theorem
Evaluatewhere C is the trace of the cylinder x2+y2=1 in the plane y+z=2.Orient C counterclockwise as viewed from above. See the Figure below
C ydzxdyzdx ,
Solution)
kjiF yxz
kji
kji
F
yxz
zyxcurl
02),,( zyzyxg
kjn2
1
2
1
g
g
2222
2
1
2
1)(
RS
SC
dAdS
dSd kjkjirF
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2008_Vector Calculus(4)
Stokes’ Theorem
Physical Interpretation of Curl
(curl )r
r
CS
d dS F r F n
For a small but fixed value of r,
0 0
0 0
0 0
(curl ( )) ( )
(curl ( )) ( )
(curl ( )) ( )
r
r
r
CS
S
r
d P P dS
P P dS
P P A
F r F n
F n
F n
0 00
1(curl ( )) ( ) lim
rCrr
P P dA
F n F r
0 0
1(curl ( )) ( )
rCr
P P dA
F n F r
0P
rC
rS
0( )Pn
By Stokes’
theorem,
Cr is Small circle of
radius r centered at P0
Because radius of circle Cr is small, then we
approximate curl F(P) ≈ curl F(P0),
Roughly then, the curl of F is the circulation of F
per unit area.
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Stokes’ Theorem
Physical Interpretation of Curl
(curl )r
r
CS
d dS F r F n
For a small but fixed value of r,
0 0
0 0
0 0
(curl ( )) ( )
(curl ( )) ( )
(curl ( )) ( )
r
r
r
CS
S
r
d P P dS
P P dS
P P A
F r F n
F n
F n
0 00
1(curl ( )) ( ) lim
rCrr
P P dA
F n F r
0 0
1(curl ( )) ( )
rCr
P P dA
F n F r
0P
rC
rS
0( )Pn
By Stokes’
theorem,
Cr is Small circle of
radius r centered at P0
Because radius of circle Cr is small, then we
approximate curl F(P) ≈ curl F(P0),
Roughly then, the curl of F is the circulation of F
per unit area.
See also
Streeter V.L., Fluid Mechanics,
McGraw-Hill, 1948,
p49, ‘25. Circulation’
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2008_Vector Calculus(4)
Divergence Theorem of Gauss
(Transformation Between Triple and Surface Integrals)
Let T be a closed bounded region in space whose boundary
is a piecewise smooth orientable surface S.
x
z
yR
n
n
n2S
1S
3S
Fig. Example of a special region
),,( zyxF
T S
dAdV nFFdiv
: a vector function that is continuous
and has continuous first partial
derivatives in T
(2)
],,,[ 321 FFFF ]cos,cos,[cos nUsing component,
S
ST
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
)(
)coscoscos()(
321
321321 (2’)
z
F
y
F
x
F
321divF
Divergence Theorem*
*Erwin Kreyszig, Advanced Engineering Mathematics 9th ,2006, Wiley,Ch10.7(p458~463)
Ref. Divergence Theorem
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2008_Vector Calculus(4)
Divergence Theorem
)(
)coscoscos()(
321
321321
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
S
ST
(proof) we can start with (2*)
This equation is true if and only if the integrals of each component
on both sides are equal
SST
SST
SST
dxdyFdAFdxdydzz
F
dxdzFdAFdxdydzy
F
dydzFdAFdxdydzx
F
333
222
111
cos
cos
cos
(3)
(4)
(5)
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Divergence Theorem(proof continue)
dAFdxdydzz
F
ST
cos33
(5)
We first prove (5) for a special region T that is bounded by a piecewise smooth
orientable surface S and has the property that any straight line parallel to
any one of the coordinate axes and intersecting T has at most one segment
(or a single point)
It implies that T can be represented in the form
),(),( yxhzyxg (6)
x
z
yR
n
n
n2S
1S
3S
),(:
),(:
2
1
yxgS
yxhS
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Divergence Theorem
(proof continue)
x
z
yR
n
n
n2S
1S
3S
),(:
),(:
2
1
yxgS
yxhS
dAFdxdydzz
F
ST
cos33
(5)
R R
R
yxh
yxgT
dxdyyxgyxFdxdyyxhyxF
dxdydzz
Fdxdydz
z
F
)],(,,[)],(,,[ 33
),(
),(
33
RRS S
dxdyyxgyxFdxdyyxhyxFdxdyFdAF )],(,,[)],(,,[cos 3333
We can decide the sign of the integral because on S2, and
on S10cos
0cos
Therefore, we prove (5). In the same manner, (3),(4) can be proven.
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Divergence Theorem
T S
dAdV nFFdiv(2)
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2008_Vector Calculus(4)
Divergence Theorem
(Example 1) Evaluate S
zdxdyxydzdxxdydzxI 223
x y
z
a a
b
2 2 2
2 2 2
: (0 )
0 ( )
S x y a z b
z and z b x y a
T S
dAdV nFFdiv(2)
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2008_Vector Calculus(4)
Divergence Theorem
(Example 1) Evaluate S
zdxdyxydzdxxdydzxI 223
x y
z
a a
b
2 2 2
2 2 2
: (0 )
0 ( )
S x y a z b
z and z b x y a
zxFyxFxF 232
2
3
1 ,, (sol)
2222 53 div xxxx F
T S
dAdV nFFdiv(2)
33/50
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2008_Vector Calculus(4)
Divergence Theorem
(Example 1) Evaluate S
zdxdyxydzdxxdydzxI 223
x y
z
a a
b
2 2 2
2 2 2
: (0 )
0 ( )
S x y a z b
z and z b x y a
zxFyxFxF 232
2
3
1 ,, (sol)
2222 53 div xxxx F
T S
dAdV nFFdiv(2)
Changing to the polar coordinate,
zzryrx ,sin,cos
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2008_Vector Calculus(4)
Divergence Theorem
(Example 1) Evaluate S
zdxdyxydzdxxdydzxI 223
x y
z
a a
b
2 2 2
2 2 2
: (0 )
0 ( )
S x y a z b
z and z b x y a
zxFyxFxF 232
2
3
1 ,, (sol)
2222 53 div xxxx F
T S
dAdV nFFdiv(2)
2
2 2 2
0 0 0
2 42
2 4
0 0 0
5 5 cos
55 cos 5
4 4 4
b a
z rT
b b
z z
I x dxdydz r r dr d dz
a ad dz dz a b
Changing to the polar coordinate,
zzryrx ,sin,cos
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Divergence Theorem
2 2 2: 4S x y z
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2008_Vector Calculus(4)
(Example 2) Evaluate over the spheredAzxS
nki )7(
Divergence Theorem
2 2 2: 4S x y z
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2008_Vector Calculus(4)
(Example 2) Evaluate over the spheredAzxS
nki )7(
(a) by Divergence Theorem
6423
466div 3
TTS
dVdVdA FnF
617,0,7divdiv zxF
Divergence Theorem
2 2 2: 4S x y z
38/50
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2008_Vector Calculus(4)
(Example 2) Evaluate over the spheredAzxS
nki )7(
(a) by Divergence Theorem
6423
466div 3
TTS
dVdVdA FnF
617,0,7divdiv zxF
Divergence Theorem
2 2 2: 4S x y z
39/50
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2008_Vector Calculus(4)
(Example 2) Evaluate over the spheredAzxS
nki )7(
vvvuv sin2,sincos2,coscos2r
(a) by Divergence Theorem
6423
466div 3
TTS
dVdVdA FnF
617,0,7divdiv zxF
(b) directly
dudvdA Nn
on spherical coordinate,
,0,coscos2,sincos2 uvuvu r vuvuvv cos2,sinsin2,cossin2 r
vvuvuvvu sincos4,sincos4,coscos4 22 rrN vuvzx sin2,0,coscos14,0,7 F
vvuvvvvuvuv 2232 sincos8coscos56sincos4)sin2(coscos4coscos14 NF
We shall use
643/216)3/22(56sincos82cos56
sincos8coscos56)7(
2
0
23
2
0
2
0
223
dvvvv
dvduvvuvdAzxS
nki
Divergence Theorem
2 2 2: 4S x y z
40/50
-
2008_Vector Calculus(4)
Divergence Theorem
Example 1Verifying Divergence Theorem
Solution)
Let D be the region bounded by the hemisphere x2+y2+(z-1)2=9, 1≤z≤4, and the plane z=1. Verify the divergence theorem if F=xi+yj+(z-1)k
1) Triple Integral
2) Surface Integral
kjiF yxz
3div F
542
34
2
1333div
3
DDD
dvdvdvF
S1 : hemisphere
S2 : z=1
1:,:, 2121
zShemisphereSSSS
222 )1(),,( zyxzyxg
kjikji
n3
1
33)1(
)1(
222
zyx
zyx
zyx
g
g
33
)1(
33
222
zyx
nF
54)9(9
9
3)3()(
2
0
3
0
2/12
22
1
drdrr
dAyx
dSRS
nF
kn
1 znF
0)0()1()(
222
SSS
dSdSzdSnF
54)( S
dSnF
41/50
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2008_Vector Calculus(4)
Divergence Theorem
Example 2Using Divergence Theorem
If F=xyi+y2zj+z3k, evaluate ∬S(F∙n)dS, where the unit cube defined by 0≤x≤1, 0≤y≤1, 0≤z≤1.
232div zyzy F
Solution)
22
1
2
1
32
1
32
)32(
)32(
)32(
1
0
32
1
0
2
1
0
1
0
222
1
0
1
0
2
1
0
1
0
1
0
2
2
zzz
dzzz
dzyzzyy
dzdyzyzy
dzdydxzyzy
dVzyzydSDS
nF
42/50
-
2008_Vector Calculus(4)
Naval A
rch
itectu
re &
Ocean
En
gin
eerin
g
Reference slides
Divergence Theorem
-
2008_Vector Calculus(4)
Divergence Theorem
S
ST
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
)(
)coscoscos()(
321
321321
3 3
3
cos cosF dA F dx dy
F dxdy
44/50
-
2008_Vector Calculus(4)
Divergence Theorem
S
ST
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
)(
)coscoscos()(
321
321321
①
3 3
3
cos cosF dA F dx dy
F dxdy
45/50
-
2008_Vector Calculus(4)
Divergence Theorem
S
ST
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
)(
)coscoscos()(
321
321321
①
dx
3F
n
x
z①
3 3
3
cos cosF dA F dx dy
F dxdy
46/50
-
2008_Vector Calculus(4)
Divergence Theorem
S
ST
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
)(
)coscoscos()(
321
321321
①
dx
3F
n
x
z①
3 3
3
cos cosF dA F dx dy
F dxdy
3 cosF
47/50
-
2008_Vector Calculus(4)
Divergence Theorem
S
ST
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
)(
)coscoscos()(
321
321321
①
②
dx
3F
n
x
z①
3 3
3
cos cosF dA F dx dy
F dxdy
3 cosF
48/50
-
2008_Vector Calculus(4)
Divergence Theorem
S
ST
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
)(
)coscoscos()(
321
321321
①
②
dx
3F
n
x
z① dx
3F
n
x
z②
3 3
3
cos cosF dA F dx dy
F dxdy
3 cosF
49/50
-
2008_Vector Calculus(4)
Divergence Theorem
S
ST
dxdyFdzdxFdydzF
dAFFFdxdydzz
F
y
F
x
F
)(
)coscoscos()(
321
321321
cosdx dx
①
②
dx
3F
n
x
z① dx
3F
n
x
z②
3 3
3
cos cosF dA F dx dy
F dxdy
3 cosF
50/50