[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


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


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

3/50


Green’s Theorem




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

4/50


Green’s Theorem




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

5/50


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

6/50


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

7/50


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

8/50


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

9/50


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.

10/50


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 )

11/50


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

12/50


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

13/50


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

14/50


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

15/50


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


S


i

j

x

y

z

k

C

R

T

n

See also

Streeter V.L., Fluid Mechanics,

McGraw-Hill, 1948,

p47, ‘24. Stokes’ Theorem’

16/50


Stokes’ Theorem

Stokes’ TheoremTheorem 9.14

S


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


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)

17/50


Stokes’ Theorem


S


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


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,

18/50


Stokes’ Theorem


S



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)

19/50


Stokes’ Theorem


S



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

20/50


Stokes’ Theorem


S



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

21/50


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

22/50


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

23/50


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

24/50


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.

25/50


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’

26/50


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

27/50


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)

28/50


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

29/50


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.

30/50


Divergence Theorem

T S

dAdV nFFdiv(2)

31/50


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)

32/50


Divergence Theorem



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


Divergence Theorem



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

34/50


Divergence Theorem



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

35/50


Divergence Theorem

2 2 2: 4S x y z

36/50


(Example 2) Evaluate over the spheredAzxS

nki )7(

Divergence Theorem

2 2 2: 4S x y z

37/50



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



nki )7(


6423

466div 3

TTS

dVdVdA FnF

617,0,7divdiv zxF

Divergence Theorem

2 2 2: 4S x y z

39/50



nki )7(

vvvuv sin2,sincos2,coscos2r


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


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


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


Naval A

rch

itectu

re &

Ocean

En

gin

eerin

g

Reference slides

Divergence Theorem


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


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


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


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


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


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


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

[2008][07-1] Engineering Mathematics 2ocw.snu.ac.kr/sites/default/files/NOTE/4694.pdf · 2018. 1....

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Transcript of [2008][07-1] Engineering Mathematics 2ocw.snu.ac.kr/sites/default/files/NOTE/4694.pdf · 2018. 1....