2008_Vector Calculus(4)
Naval Architecture & Ocean Enginee ring
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 Architecture & Ocean Enginee ring
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
x Q y
P Q
P , , / , and /
R
C
dA
y P x
Qdy Q Pdx
Positive direction
C
Negative direction
C
R R
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
x Q y
P Q
P , , / , and /
R
C
dA
y P x
Qdy Q Pdx
Proof)
R
)
2(x g y
)
1(x g y
a b x
y
b x a x
g y x
g
R :
1( )
2( ),
C
b a b
a b a
b a
x g
x g R
dx y x P
dx x
g x P dx
x g x P
dx x
g x P x
g x P
y dydx dA P
y P
) , (
)) ( , ( ))
( , (
))]
( , ( ))
( , ( [
2 1
1 2
) (
) (
2
1
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
x Q y
P Q
P , , / , and /
R
C
dA
y P x
Qdy Q Pdx
R
)
2(x g y
)
1(x h x
b c
x y
)
1(y h x d
d y c y
h x y
h
R :
1( )
2( ),
C
d c d
c d c
d c
x h
x h R
dy y x Q
dy y y h Q dy
y y h Q
dy y y h Q y
y h Q
y dxdy dA Q
y Q
) , (
) ), ( ( )
), ( (
)]
), ( ( )
), ( ( [
2 1
1 2
) (
) (
2
1
2008_Vector Calculus(4)
Green’s Theorem
Example 1
Using Green’s Theorem
Evaluate
where C consists of the boundary of the region in the first quadrant that is bounded by the graphs of y=x
2and y=x
3.
, ) 2
( )
(
2 2
Cx y dx y x dy
420 ) 11
(
) (
) 2 1 (
) 2 1 (
) (
) 2
(
) 2
( )
(
1 0
2 3
4 6
1 0
2 1
0
2 2
2 2
2 3 2
3
dx x
x x
x
dx y
y
dx dy y dA y
y dA y x
x x y
dy x y dx
y x
x x x
x R R C
Solution)
x y
x
2y
x
3y
) 1 , 1 (
R
2008_Vector Calculus(4)
Green’s Theorem
Example 2
Using Green’s Theorem
Evaluate
where C is the circle (x-1)
2+(y-5)
2=4.
, ) 2
( )
3
(
5 3
C
y
dy e
x dx
y x
Solution)
x y
4 ) 5 ( ) 1
(x 2 y 2
4 is area R
4
) 3 2 (
) 3 (
) 2
(
) 2
( )
3 (
5 5
3
3
R R R
y C
y
dA
dA
y dA y x
x e x
dy e
x dx
y
x
2008_Vector Calculus(4)
Green’s Theorem
Example 3
Work Done by a Force
Find the work done by the force
acting along the simple closed curve C shown in Figure below.
j i
F ( 16 y sin x
2) ( 4 e
y 3 x
2)
Solution)
R R
y C
y C
dA x
y dA x y
x x e
dy x
e dx
x y
d W
) 16 6
(
) sin 16
( )
3 4
(
) 3 4
( )
sin 16
(
2 2
2 2
r F
4 3 , 4
1 0
: r
R
4 )
8 cos
2 (
) 8 cos
2 (
) 16 cos
6 (
4 / 3
4 /
4 / 3
4 /
1 0 2 3
4 / 3
4 /
1 0
d
d r
r
d rdr r
W x
y
x y C
1:
1 :
2 22
x y
C
x y
C
3:
2008_Vector Calculus(4)
Green’s Theorem
Example 4
Green’s Theorem Not Applicable
Let C be the closed curve consisting of the four straight line segments C
1, C
2, C
3, C
4shown in Figure. Green’s theorem is not applicable to the line integral
C dy
y x
dx x y x
y
2 2
2 2
x y
2
1
: y C
R
2
2
: x C
2
3
: y C
2
4
: x
C
2008_Vector Calculus(4)
Green’s Theorem
Example 4
Green’s Theorem Not Applicable
Let C be the closed curve consisting of the four straight line segments C
1, C
2, C
3, C
4shown in Figure. Green’s theorem is not applicable to the line integral
C dy
y x
dx x y x
y
2 2
2 2
x y
2
1
: y C
R
2
2
: x C
2
3
: y C
2
4
: x C
since P, Q, ∂P/∂y, and ∂Q/∂y are not
continuous at the origin.
2008_Vector Calculus(4)
Green’s Theorem
Region with Holes
C
C C
R R
R
Qdy Pdx
Qdy Pdx
Qdy Pdx
y dA P x
dA Q y
P x
dA Q y
P x
Q
2 1
2 1
C
1C
2C
1C
2R
2R
1(C=C
1∪C
2)
2008_Vector Calculus(4)
Green’s Theorem
Example 5
Region with a Hole in It
Solution) Evaluate
where C=C
1∪C
2is the boundary of the shaded region R shown in
Figure
2 ,
2 2
C 2 dy y x dx x y xy
x y
R
C1
1 : 2 2
2 x y
C
2 2
2
2
, ( , )
) ,
( x y
y x x y Q
x y y
x
P
2 2 2
2 2 2
2 2
2 2
) , (
)
( x y
x y x
Q y
x
x y 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 2
0.
( ) ( )
C
C
C
R
y x
dx dy
x y x y
y x
dx dy
x y x y
y x
dx dy
x y x y
y x y x
x y x y dA
since P, Q, ∂P/∂y, and ∂Q/∂y are
continuous on the region R bounded by C, it follows from the above
discussion that
2008_Vector Calculus(4)
Green’s Theorem
Example 6
Example 4 Revisited
Evaluate the line integral in Example 4.
Example 4.
C dyy x dx x y x
y
2 2 2
2
Solution)
C
C
1
C
2
C
3
C4x
yR
C
1 : 2 2
x y C
2 2
2
2
, ( , )
) ,
( x y
y x x y Q
x y y
x
P
2 2 2
2 2 2
2 2
2 2
) , (
)
( x y
x y x
Q y
x
x y y
P
2 2 2 2
2 0
2 2 2
0 2 0
sin ( sin ) cos (cos ) (sin cos )
2
C
y x
dx dy
x y x y
t t t t dt
t t dt
dt
2 2 2 2
C
y x
dx dy
x y x y
2008_Vector Calculus(4)
Stokes’ Theorem
Vector Form of Green’s Theorem
k F
F
y
P x
Q Q
P
z y
x
k j
i
0 curl
) 1 ( )
curl
(
R C
C
F d r F T ds F k dA
i
j
x
y
z
k
C
R
T
Green’s Theorem in 3-Space
n( 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(x g y
)
1(x h x
b c
x y
)
1(y h x d
Green Theorem in 2-D space
Green Theorem in 3-D space
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
k j
i
F ( x , y , z ) P ( x , y , z ) Q ( x , y , z ) R ( x , y , z )
S C
C
F d r ( F T ) ds ( curl F ) n dS
i
j
x
y
z
k
C
R
T n
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
k j
i
F ( x , y , z ) P ( x , y , z ) Q ( x , y , z ) R ( x , y , z )
S C
C
F d r ( F T ) ds ( curl F ) n dS
i
j
x
y
z
k
C
R
T n
See also
Streeter V.L., Fluid Mechanics, McGraw-Hill, 1948,
p47, ‘24. Stokes’ Theorem’
2008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ Theorem Theorem 9.14
S C
C
F d r ( F T ) ds ( curl F ) n dS
Partial Proof)
k j
i
F
y P x
Q x
R z
P z
Q y
curl R
2 2
1
y f x
f
y f x
f i j k n
) , ( x y f
z
S is oriented upward and is defined by a function
k j
i
F ( x , y , z ) P ( x , y , z ) Q ( x , y , z ) R ( x , y , z )
i
j
x
y
z
k
C
R
T n
0 ) , ( )
, ,
( x y z z f x y g
If we write
If S is defined by g(x,y,z)=0,
g g
||
||
n 1 (R.H.S)
2008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ Theorem Theorem 9.14
S C
C
F d r ( F T ) ds ( curl F ) n dS
k j
i
F
y P x
Q x
R z
P z
Q y
curl R
22
1
y f x
f
y f x
f i j k n
k j
i
F ( x , y , z ) P ( x , y , z ) Q ( x , y , z ) R ( x , y , z )
i
j
x
y
z
k
C
R
T n
R S
y dA P x
Q y
f x R z
P x
f z Q y
dS R n F) curl (
Hence,
2008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ Theorem Theorem 9.14
S C
C
F d r ( F T ) ds ( curl F ) n dS
k j
i
F ( x , y , z ) P ( x , y , z ) Q ( x , y , z ) R ( x , y , z )
i
j
x
y
z
k
C
R
T n
C b a C C
x dA R f
y P y
R f x Q
y dy R f Q x dx
R f P
dt dt y y f dt
dx x R f dt
Q dy dt
P dx
Rdz Qdy
Pdx d
xy
r F
Chain rule
Green’s theorem
b t a t
y t x f z t y y t x
x ( ), ( ), ( ( ), ( )) b
t a t
y y t x
x ( ), ( ) :
C
xy
:
C
(Projection of C onto the xy-plane)(L.H.S)
2008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ Theorem Theorem 9.14
S C
C
F d r ( F T ) ds ( curl F ) n dS
k j
i
F ( x , y , z ) P ( x , y , z ) Q ( x , y , z ) R ( x , y , z )
i
j
x
y
z
k
C
R
T n
x f y f z R y
f x R y
x R f x f z Q x
Q
x f z R x
R y
f y
x R f x f z Q x
Q
y y f
x f y x R y
x f y x x Q
y R f x Q
2 2
)) , ( , , ( ))
, ( , , (
Chain and Product Rules
Similarly,
y f x f z R x
f y R x
y R f y f z P y
P x
R f
y P
22008_Vector Calculus(4)
Stokes’ Theorem
Stokes’ Theorem Theorem 9.14
S C
C
F d r ( F T ) ds ( curl F ) n dS
k j
i
F ( x , y , z ) P ( x , y , z ) Q ( x , y , z ) R ( x , y , z )
i
j
x
y
z
k
C
R
T n
x f y f z R y
f x R y
x R f x f z Q x
Q y
R f
x Q
2∴ (L.H.S)=(R.H.S)
y f x f z R x
f y R x
y R f y f z P y
P x
R f
y P
2
R R C
y dA P x
Q y
f x R z
P x
f z Q y
R
x dA R f
y P y
R f x Q
dr
F
2008_Vector Calculus(4)
Stokes’ Theorem
Example 1
Verifying Stokes’ Theorem
Let S be the part of the cylinder z=1-x
2for 0≤x≤1, -2≤y≤2.
Verify Stokes’ theorem if F=xyi+yzj+xzk.
Solution) 1) Surface Integral
kj i
F xy yz xz
k j i k
j i
F y z x
xz yz
xy
z y
x
curl
0 1 )
, ,
( x y z z x
2 g
1 4
2
2
x x g
g i k
n
2 )
4 (
) 2
(
) 2
(
1 4
) 2 curl
(
1 0
1 0
2 2 2
1 0
2 2
2
dx x
dx xy xy
dx dy x xy
dA x xy
dS x
x dS xy
R S S
n
F
2008_Vector Calculus(4)
Stokes’ Theorem
Example 1
Verifying Stokes’ Theorem
Let S be the part of the cylinder z=1-x
2for 0≤x≤1, -2≤y≤2.
Verify Stokes’ theorem if F=xyi+yzj+xzk.
Solution) 2) Line Integral
C
C
1
C
2
C
3
C40 ,
0 ,
0 ,
1
1
: x z dx dz
C
0 0 )
0 ( ) 0 (
1
C y y dyxdx dz
dy x
z y
C
2: 2 , 1
2, 0 , 2
15 ) 11
2 2
2 (
) 2 )(
1 ( 0 ) 1 ( 2 2
0 1
4 2
2 2
2
dx x x
x
xdx x
x x
C xdx
0 ,
0 ,
1 ,
0
3
: x z dx dz
C
0 0
0 2
3 2
C ydy ydyxdx dz
dy x
z y
C
4: 2 , 1
2, 0 , 2
15 ) 19
2 2
2 (
) 2 )(
1 ( 0 ) 1 ( 2 2
1 0
4 2
2 2
4
dx x x
x
dx x x
x x
C xdx
15 2 0 19 15
011
xydx yz xzdzC
d
CPdx Qdy Rdz
F r
2008_Vector Calculus(4)
Stokes’ Theorem
Example 2
Using Stokes’ Theorem
Evaluate
where C is the trace of the cylinder x
2+y
2=1 in the plane y+z=2.
Orient C counterclockwise as viewed from above. See the Figure below
Czdx xdy ydz ,
Solution)
k j
i
F z x y
k j i k j
i
F
y x
z
z y
curl x
0 2 )
, ,
( x y z y z g
k j
n 2
1 2
1
g g
2 2
2 2
2 1 2
) 1 (
R S
S C
dA dS
dS
d r i j k j k
F
2008_Vector Calculus(4)
Stokes’ Theorem
Physical Interpretation of Curl
(curl )
r
r
C
S
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
C
S
S
r
d P P dS
P P dS
P P A
F r F n
F n
F n
0 0
0
(curl ( )) ( ) lim 1
Cr
r r
P P d
A
F n F r
0 0
(curl ( )) ( ) 1
Cr
r
P P d
A
F n F r
P0
Cr
Sr
(P0) n
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.
2008_Vector Calculus(4)
Stokes’ Theorem
Physical Interpretation of Curl
(curl )
r
r
C
S
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
C
S
S
r
d P P dS
P P dS
P P A
F r F n
F n
F n
0 0
0
(curl ( )) ( ) lim 1
Cr
r r
P P d
A
F n F r
0 0
(curl ( )) ( ) 1
Cr
r
P P d
A
F n F r
P0
Cr
Sr
(P0) n
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’
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
R y n n
n S2
S1
S3
Fig.Example of a special region
) , , ( x y z F
T S
dA dV F n F
div
: a vector function that is continuous and has continuous first partial
derivatives in T (2)
], ,
,
[ F
1F
2F
3
F n [cos , cos , cos ] Using component,
S S T
dxdy F
dzdx F
dydz F
dA F
F F
dxdydz z
F y
F x
F
) (
) cos cos
cos (
) (
3 2
1
3 2
1 3
2
1
(2’)
z F y
F x
F
1 2 3Divergence Theorem* divF
Ref. Divergence Theorem
2008_Vector Calculus(4)
Divergence Theorem
) (
) cos cos
cos (
) (
3 2
1
3 2
1 2 3
1
dxdy F
dzdx F
dydz F
dA F
F F
dxdydz z
F y
F x
F
S S T
(proof) we can start with (2*)
This equation is true if and only if the integrals of each component on both sides are equal
S S
T
S S
T
S S
T
dxdy F
dA F
dxdydz z
F
dxdz F
dA F
dxdydz y
F
dydz F
dA F
dxdydz x
F
3 3
3
2 2
2
1 1
1
cos cos cos
(3)
(4)
(5)
2008_Vector Calculus(4)
Divergence Theorem
(proof continue)
dA F
dxdydz z
F
S
T
3
3cos
(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
) , ( )
,
( x y z h x y
g
(6)
x
z
R
y n nn S2
S1
S3
)
, ( :
) , ( :
2 1
