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Transcript of ANU MATH2405 Vector Notes
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VECTOR CALCULUS
2
Part 1 (Syllabus) Curves and Surfaces -Euclidean Spaces-Coordinate transformations in 2D Euclidean Spaces and the Jacobian-Area integrals in the plane -Vector valued functions of a scalar variable -Parametric curves in 3D Euclidean Spae. Curvature and torsion.
-Parametric surfaces in 3D Euclidean Space and curvilinear coordinates-Elements of length and area in curvilinear coordinates-Scalar and vector Fields -Line, surface and volume integrals involving scalar and vector fields Part 2 (Syllabus) Vector Differential Operators-Vector differential operators: grad, div and curl-Identities involving vector differential operators -Integral Theorems (Greens, Gausss, Stokess) -Irrotational and Solenoidal Fields-Scalar and Vector potentials -Grad, div and curl (cylindrical and spherical polar coordinates)
3
Part 1Curves and Surfaces
in 3 D Euclidean Space
444
3-D Euclidean SpacesSpaces whose geometry satisfies the postulates of Euclid arose from
the first attempts to describe the geometry of the space we live in.
Such spaces are called Euclidean spaces and are examples
of "intrinsically flat" spaces. These spaces satisfy in prticular
Euclid's 5th postulate: "Given a straight line, and a point not on
the line, there exists exactly one parallel line".
In 3-D Euclidean spaces (E3) a rectangular cartesian coordinate system OXYZ
can be found such that the "metric element of length" dsbetween neighbouring
points (x, y, z), (x+dx,y+ dy,z+dz) is given by
ds2= dx
2+ dy
2+ dz
2 (in essence, Pythogoras's Theorem)
The vector algebra developed in first year was based on
such spaces and coordinate systems.
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NOTATION: RH: Right Handed RC: Rectangular CartesianOXYZ or OX1X2X3 : RH, RC coordinate systems in 3 D Euclidean Space E
3
Position Vector r = x i + y j + z k x = x1e1 + x2e2 + x3e3= (x, y, z) = (x1, x2, x3)
Z X3X2
X1
r =(x, y, z)
x=(x1, x2, x3)
XY
We use (x,y,z) o r (x1 , x2, x3) in different contexts depending on convenience of notation.
O O
6
If we use x = (x1,x2.x3) to be the position vector of a point in a
RH, RC coordinate system OX1X2X3, the vector element of length
between neighbouring points (the displacement vector)
is
d x = dl = dx1e1 + dx2 e2 + dx3e3
The scalar element of length is
dl = d x
If we use r = (x,y,z) to be the position vector of a point in a
RH, RC coordinate system OXYZ, the vector element of length
between neighbouring points is dr = ds = dxi + dy j+ dzk
Scalar element of length is ds = d r
dx
drO
Or
x
77
Vectors should be viewed as mathematical objects that have an existence that is independent of the coordinate system used for its representation.Thus, the displacement vectorjoining two adjacent points remains the same even though its components change under coordinate transformations.So does the position vector of a point.
To illustrate this point we show below the displacement vector
and its components relative to two RH RC coordinate systems: OX1X2X3
and OY1Y2Y3obtained by rotating the original coordinate system.
Y1X1
X2X3Y3
Y2 dl = dr = (dx1,dx2,dx3) = (dy1,dy2,dy3 )O
88
Differentiation
Consider a vector vwhich is function of the scalar u.
We define its derivative in the standard way with the
difference in vbetween adjacent values of utaken as a
vectorial difference (see figure).
dv
du= lim!u"0
!v
!u
= lim!u"0
v(u+!u)# v(u)
!u
= lim!u"0
v1(u+!u)#v1(u)
!ui+ lim
!u"0
v2(u+!u)# v2(u)
!uj+ lim
!u"0
v3(u+!u)# v3(u)
!uk
=dv1
dui+
dv2
duj+
dv3
duk
That is, simply differentiate the components, and sum vectorially.
v(u+!u)
v(u)
!vVector valued functions
of a scalar variable
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Differential of a vector function v(u)
dv =dv
dudu
and is in the limiting direction of !vas
!u" 0.
Differential
v(u+!u)
v(u)
!v
10
X
Y
Z
P
Q
r=r(t) r=r(t+dt)
drA particle moves so that its position vector
is given by
r(t) = ti + t2 j+ t3 k!!!!! a vector valued function of t.
The first derivative gives the velocity
v(t) =dr
dt
= i +2t j+3t2 k --another vector valued function of t.
v = v = 1+ 4t2 +9t4 ------ speed
Note:
dr =dr
dtdt = vdt
so the velocity is in the limiting direction of
!r.
That is the velocity is tangent to the space curve
r = r(t).
Speed, Velocity and Acceleration (revision)
11X
Y
Z
PQ
r=r t r=r t+ t
v t
v (t+dt)
Y
Z
v(t)
v (t+dt)
dv
It is often useful to visualise derivatives of vectors in a coordinate
free way as we have done here.
The derivative of the velocity vwith respect to the scalar tis
the acceleration
a(t) =dv
dt= 2j+ 6tk ---- another vector valued function
12
Differentiation of sums and productsa(u), b(u), c(u) are differentiable vector valued functions (i.e. the components
are differentiable), and f(u) is a differentiable scalar valued function of u.
The following can be proved .
d
du
(a(u) + b(u))= d
du
a(u) + d
du
b(u)
d
du(f(u)a(u))=
df
dua(u)+ f(u)
d
dua(u)
d
du(a(u) . b(u))=
da(u)
du.b(u) +a(u).
db(u)
du
d
du(a(u) ! b(u))=
da(u)
du!b(u) +a(u)!
db(u)
du
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Example:
To proved
du(f(u)a(u))=
df(u)
dua(u) +f(u)
da (u)
du, we take components
and prove the result for each component.
d
du(f(u)ak(u)) (k= 1,2,3)
lim!u!0
f(u+!u)ak(u+!u)-f(u)ak(u)
!u
= lim!u!0
(f(u)+ "f (u)!u+O(!u2 ))((ak(u)+ "ak(u)!u+O(!u2 ))-f(u)ak(u)
!u(assuming differentiability)
= "f (u)ak(u)+f(u) "ak(u)
Example:
p(u) =p1(u)e1 +p2(u)e2 + p3(u)e3
d
dup(u)=
dp1(u)
due1 +p1(u)
de1
du+....... ( only because basis vectors are constant vectors)
=dp1(u)
due1 +
dp2 (u)
due2 +
dp3(u)
due3..(i.e. can simply differentiate components)
14
d
du(a.( b"c)) =
d a
du. (b"c) +a. (
db
du"c) +a.( b"
dc
du)
d
du
(a"( b"c))=d a
du
" (b"c) +a" (db
du
"c) +a"( b"dc
du
)
Rule : diffe rentiate term by term preserving order of vectors
in each expression.
Differentiating triple products
(see appendix for revision of scalar product, vector product,scalar triple product and vector triple product)
15
Example
Suppose p =p(t) where t is the scalar variable (time),
!p =d
dtp(t) etc.
(1)d
dt(p.!p! !!p) = !p.!p! !!p+p.!!p! !!p+p.!p!!!!p
=p.!p!!!!p
(2)d
dt(p!!p) =p! !!p+ !p!!p
= p! !!p
16
If a vector is a function of more than one scalar variable,
we can define partial derivatives.
Suppose p(u, v)
pu=
!p
!u= lim
"u!0
p(u+"u, v)"p(u, v)
"u
provided the limit exists.
The rules we had previously carry over also to partial derivatives:
Thus
!
!u(a(u,v) . b(u, v))=
!a(u,v)
!u. b(u,v) +a(u, v).
!b(u,v)
!u
etc.
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Chain Rule for Vector Valued Functions If
a(u1,u
2,....un )
and
uj = uj(v1,v2.....vn ) (j= 1,n)
then
!a
!vi=
!a
!ukk=1
n
" !uk
!vi
[Follows from the chain rule applied to the pth
component
ap (u1,u2,....un ) of a, which is a scalar function.
!ap
!vi=
!ap
!ukk=1
n
" !uk
!vi (from first year) ]
18
Sum implicitly over repeated indices. The above can be written
more compactly as
!a
!vi
=!a
!uk
!uk
!vi
kis called the dummy index andk=1
n
" is implied.
Einstein Summation Convention
1919
Integration of a vector function of a scalar variable Suppose b(u) is the vector function of a scalar variable u
possessing a derivative
a(u) =db
du
Then the indefinite integral of a(u) isa(u)! du = b(u)+ c
where cis an arbitrary constant vector.
If a(u)= a1(u)e1 + a2(u)e2 + a1(u)e3
b(u) = e1 a1(u)! du+ e2 a2 (u)! du+ e3 a3(u)! du20
Example: Given a(u), evaluate
c! "da
dudu
where cis a constant vector.
Nowd
du(c"a) =
dc
du"a+ c"
da
du= c"
da
du (since cis a constant)
Hence
c"da
du! du = c"a+ d
where dis a constant vector.
Note that we could also have written
c! "da
dudu = c"
da
du! du = c"a+ d
because cis independent of uand can therefore be taken "out of
the integral". (Strictly, this has to be justified by taking
components - try to show this!)
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Notation:
A function fdefined in an open domain Dis said to be
-- of class C0if it exists and is continuous in D
-- of class C1if its first partial derivatives exist and are continuous
in D
---of class C2if its second partial derivatives exist and are continuous
in D
2222
Consider a RH, RC coordinate system (x1, x2 ) which specifies points in
R2. Introduce a new coordinate system (x1!, x2!) by rotating the coordinate axes
through an angle !. By construction, this is also a RH, RC coordinate system.
The components of the position vector OP! "!!
are different in the two
coordinate systems and are related by
x1 = !x1 cos!" !x2 sin!= f1(x1!, x2!)
x2 = !x1 sin!+ !x2 cos!= f2 (x1!, x2!)
This is an example of an "orthogonal" coordinate
transformation (the matrix L =cos! "sin!
sin! cos!
#
$%
&
'(
connecting x and x! is orthogonal" in the linear algebra sense
i.e. L"1
=LT)
Orthogonal Transformations in 2D Euclidean Space
O
P
P1P1
/X1
X1/
x2x1
1x2
1
X2X2/
x1!
!
23
Consider a RH, RC coordinate system (x1, x2 ) which specifies points in
R2. Introduce a new coordinate system (u,v) through the transformation
x1 = f1(u,v), x2 = f2(u,v)
or x = f(u,v)
We assume that f !C1in some domain Ruv. We can look at this as a mapping from
the region Ruv in (u,v) coordinate space to a regionRin (x1,x2 ) space.
The Jacobian of the transformation is
J =
!f1
!u
!f1
!v
!f2
!u
!f2
!v
(definition)
General Coordinate Transformations in 2D Euclidean Space
x1
R
u
v
Ruv
x2
f
24
From the transformation, and the differentiabili ty
of f1(u,v) and f2 (u,v) [because any C1 function is differentiable] , we have
dx1 =!f1
!udu+
!f1
!vdv
dx2 =!f2
!udu+
!f2
!vdv
For these equations to be invertible, (solvable for (du, dv))
we require
J=
!f1
!u
!f1
!v
!f2
!u
!f2
!v
! 0
We choose transformations that are (1-1) on Ruv- that is J! 0.
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Note that the Jacobian of the inverse transformation is
!(u,v)
!(x,y)=
!u
!x
!u
!y
!v
!x
!v
!y
It can be shown that
!(x,y)!(u,v)
=1 !(u, v)!(x,y)
, !(x,y)!(x,y)
=1 ----- (show using chain rule)
We also have transitivity under successive coordinate transformations
(x,y)! (s, t)! (u, v)
!(x,y)
!(s, t)=
!(x,y)
!(u, v)!(u, v)
!(s, t)
[See Adams for examples on Jacobians]
26
C2: x=f(uP,v)
C1: x=f(u,vp)
v coordinatecurves
u coordinatecurves
X2
X1
P
By keeping one parameter fixed and varying the other,
we get coordinate curves
C1 :x = f(u,vP )= f1(u,vP )e1 + f2(u,vP )e2
C2 :x = f(uP,v)= f1(uP,v)e1 + f2(up,v)e2
Along C1, u varies. This is the u-coordinate curve through P.
(u,v) are sometimes called curvilinear coordinates.
Curvilinear coordinate and coordinate curves
27
Example: Plane polar coordinates
x = f1(r,")=rcos"
y = f2(r,") =r sin"
r =c = const ("coordinate curves)
# x =ccos", y =csin"
# x2
+y2
=c2
$$$$circle
"=%= const (rcoordinate curves)
# x =rcos%, y =r sin%
# y =x tan%$$$$$ straight lines through origin
x0
r-coordinatecurves!"coordinate
curves
y
28
Example: Orthogonal transformation
x1 = !x1 cos!" !x2 sin!= f1(x1!, x2!)
x2 = !x1 sin!+ !x2 cos!= f2 (x1!, x2!)
!x2 = c = const --> !x
1coordinate curve
x1= !x
1cos!" csin!
x2 =
!x1sin!+ ccos!
# x1 + csin! = (x2 " ccos!)cot!
# x2 = (tan!)x
1+ c /sin!
O X1
X2
!x1coordinate curves
!x2coordinate curves
!
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Elements of length and area
x = f(u,v)! dx = fudu+ f
vdv (because fis differentiable)
where fu=
!f
!u, f
v=
!f
!v
Distance between two points P(u,v), Q(u+du, v+dv)
is given by
dl2 = dx. dx = dx12+ dx2
2= f
udu+ f
vdv( ). fudu+ f vdv( )
= fu. f
udu2 +2f
u. f
vdudv+ f
v. f
vdv2
Elements of length along u and vcoordinate curves are
dlu
2= f
u. f
udu2 = h1
2du2 (with dv = 0)
dlv
2= f
v. f
vdv2 = h2
2dv2 (with du = 0)
h1 = fu =!f
!u, h2 = fv =
!f
!v
X2
X1
h2dv
h1du
P(u,v)Q(u+du,v+dv
R(u+du, v)
Arc Lengths
T(u,v+dv)
Element of length
P
x=f(u,v)
Q
x=f(u+du,v+dv)
=x+dx
dx
30
We can form unit tangents !1 to coordinate curve C1 and !2 to coordinate
curve C2 at P:
!1 =1
h1
"f
"u
!
"#
$
%&P
,!2 =1
h2
"f
"v
!
"#
$
%&P
h1 ="f
"uP
, h2 ="f
"vP
(The subscript Pis usually omitted)
!1'!2 =1
h1h2
"f
"u'"f
"v
()*
+,-
=
1
h1h2("f1
"u
"f2
"v."f1
"v
"f2
"u)e3 =
1
h1h2
"f1
"u
"f1
"v
"f2
"u
"f2
"v
e3
Note that for a non-degenerate transformation (i.e distinct families of
of coordinate curves required for a 1-1 transformation),we require
!1'!2/ 0 . Since the Jacobian is non-zero, this is guaranteed.
C2: x=f(uP,v)
C1: x=f(u,vp)
X2
X1
!2
!1
P
31
If !1 and !2 satisfy
!1!!2 = e3
or equivalently if
"f1
"u
"f1
"v
"f2
"u
"f2
"v
=
"(f1,f2 )
"(u, v)= h1h2
we have an orthogonal curvilinear coordinate system.
Through any point, the two coordinate curves are
orthogonal to each other.
Orthogonal curvilinear coordinate systems
x0
r-coordinatecurves!"coordinate
curves
y
Example
32
dA = (h1du!1)!(h2dv!2 ) = h1h2dudv!1!!2
= h1h21
h1h2f
u!f
vdudv = f
u!f
vdudv
= Abs value of
"f1
"u
"f1
"v
"f2
"u
"f2
"v
dudv
dA ="(f1,f2 )
"(u,v)dudv
or equivalently
dx1dx2 ="(f1,f2 )
"(u,v)dudv
(Jacobian measures local magnification of area dudv)
For an orthogonal system,
dA = h1h2dudv
X2
X1
h2dv
h1du
P(u,v)Q(u+du,v+dv)
R(u+du, v)
Arc Lengths
T(u,v+dv)
Element of area
u
v
(u,v) (u+du,v)
(u+du,v+dv)(u,v+dv)
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Example: x = rcos!, y = rsin! (plane polar coordinates)
"(x,y)
"(r,!)=
"x
"r
"x
"!
"y
"r
"y
"!
=cos! !rsin!
sin! rcos!= r
"(r,!)
"(x,y) =1
r (check directly)
dA ="(x,y)
"(r,!)drd!= rdrd!
Note:
h1 ="f
"r= (
"f1
"r)
2+ (
"f2
"r)
2= cos
2!+ sin
2! =1
h2 ="f
"!= (
"f1
"!)
2+ (
"f2
"!)
2= r
2sin
2!+ r
2cos
2! = r
dA = (h1dr)(h2d!)!!!!!because coordinate curves orthogonal
x0
r-coordinatecurves!"coordinate
curves
y
r
!
drd!
dr rd!
34
Single integrals: x =x(u)
f(x)dx = f(x(u))dx
duu1
u2
"x1
x2
" du
Double Integrals; x =x(u,v)
f(x1,x2 )dAR
"" = f((x1(u,v),x2(u,v))Ruv
"" #(x1,x2 )
#(u,v)dudv
or using (x,y) as variables
f(x,y)dxdyR
"" = f((x(u,v),y(u,v))Ruv
"" #(x,y)
#(u,v)dudv
Area integrals by coordinate transformation
35
X2
X1 u
v
Ruv
X1
X2
R
uu
u
Ruv
No Obviousadvantage
Clear
Advantage!
Choose a transformation which simplifies the boundariesor the integrand or both
36
EX: Calculate area bounded by the curves
y=0, y = x2, y =1! x
2, y = 4! x
2.
The form of the boundaries suggests that we
cook up a transformation that makes the two
families of curves coordinate curves: So we
take
y = ux2(1" u "1), y = v! x
2(1" v " 4)
The four boundaries are u=
0, u=
1, v=
1, v=
4u = y x
2, v = x
2+ y
#
!(u, v)
!(x,y)=
!
2y
x3
1
x2
2x 1
= !
2
x(
y
x2+1)= !2
(u+1)3
2
v
1
2
!(x,y)
!(u, v)= !
v
1
2
2(u +1)3
2
y=x2
y=4-x2
y=1-x2y=0v=1 v=4u=1u=0
Y
X
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dA = dxdy ="(x,y)
"(u,v)###### dudv
$ dA =1
2(
v=1
v=4
### v
1
2
(u +1)3
2
du )dvu=0
u=1
#
= 12
( v1
2
v=1
v=4
# dv)( 1(u+1)
3
2
duu=0
u=1
# )
=1
2
2
3v
3
2%
&'
(
)*1
4
+2(u+1)+1
2%
&'
(
)*0
1
=7
3( 2 +1)
u
v
0 1
4
0
38
Scalar and Vector Fields Consider a doman D" R
3
ScalarField#(x)
For every point P $ Dassign a scalar #$ R
#(x)= #(x1,x2,x3)
Ex:#
(x) = x1
2+ x
2
2+ x
3
2=r (distance from origin)
%(x) = density at position x in a fluid
Vector Field v(x)
For every point P $ Dassign a vector v $ R
v(x) =v1(x1,x2,x3)e1 +v2(x1,x2,x3)e2 +v3(x1,x2,x3)e3
Ex: v(x) = x (Big bang model of universe)
v = v = x = r
39
Example : v = (x,0,0)
Velocity has only a "x" component.
The magnitude is
v = x2+0+0 = x
Example : v = (y,0,0)
Velocity has only a "x" component.
The magnitude of the velocity is
v(x,y,z)= y2+0+0 = y
Example : v = ( x
x2+ y
2,
y
x2+ y
2,0)
The magnitude of the velocity is
v =1
x2+ y
2 x
2+y
2=
1
x2+ y
2=
1
r
X
Y
X
Y
Y
X 40
Curves and Surfaces in R3
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A curve Cis a set (" R3) of points having
at least one parametric representation
x = f(t)= f1(t)e1 +f2(t)e2 + f3(t)e3
where fis continuous in [a,b] (that is f#C0 )
We say that fmaps Ct :[a,b] onto C.
Cis an arc if f(a)$ f(b)
Cis a closed curve if f(a)= f(b)
a b
t
X1
X2
X3 t=a
t=b
Ct
C
Curves
42
Cis simple if there exists at least one C0mapping f(t) of [a,b] on to C
which is one to one on [a,b]. That is, a simple curve cannot intersect
itself
C is smooth if there exists at least one C1mapping
(
dfj
dt= "f j continuous) with
"f (t) = f3"2
+ f2"2
+ f3"2
# 0 on [a,b]
(the latter is the requirement for the curve to have a tangent)
A simple curve Cis piecewise smooth if the images C1,C2,C3....CN
(Nfinite) of [a,t1],[t1,t2],........[tN$1,b] are smooth.
X1
X2
t=a
t=b
C
A simple
piecewise-
smooth curve
43
A plane curve (e.g. x3 = f3 = 0)
x(t)= f1(t)e1 + f2(t)e2
is said to be positively directed if tincreasing
corresponds to counter clockwise
passage in the x1 !x2 plane.
Each simple curve has two orientations. A simple curve with a sense ofdirection is said to be an oriented simple curve or a directed simple
curve. The orientation is induced by the parametrisation.Ex: P to Q , or Q to P in the simple curve shown below.
PQ Q
Pu increasing v increasingx = f(u)
x = g(v)
44
Suppose C is a simple smooth curve given by x=f(t) a ! t! b.
Let P and Q be two adjacent points on C
P: x = x P = f(t)
Q:x = xQ = f(t+"t)
Then PQ = "x = f(t+"t)#f(t)
Thus "x "tis a vector in direction PQ. In limit"t$ 0,
PQ becomes tangent to curve at P. Thus
T =d x
dt= lim
"t$0
"x
"t=
d f
dt= %f (t)
is a vector which is tangent to the curve at P.
Unit Tangent
!=dx dt
dx dt=
%f (t)
%f (t)=
1
h%f (t)
h = %f (t) is related to arc length along curve (see later)
O
P(t)
Q(t +!t)xP
xQ"
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45
dl2= d x
2= dx1
2+dx2
2+dx3
2
=dx1
dt
!
"#
$
%&
2
+
dx2
dt
!
"#
$
%&
2
+
dx3
dt
!
"#
$
%&
2'
())
*
+,,
dt2= -f1
2+ -f2
2+ -f3
2'( *+dt2
dl = -f12
+ -f22
+ -f32
= -f dt= hdthis called the length scale factor that converts parametric coordinate difference
to metric (physical) length. Sometimes we use dsinstead of dl. The arc length is
s(t) = l(t) = dl =. h(-t)d-tt0
t
.
Note that d x ( = dr ) can be looked upon as a vectorial element of length
along the curve. We sometimes write
dl = ds = d x = (hdt)!
O
P(t)
x
x +dx
Q(t+dt)
dl=IdxI = hdtElement of Length
4646
CurvatureSuppose that C is a simple smooth curve parametrised as
x = f(s) a " s " b
where sis the arc length along the curve (for an arbitrary parametrisation t,
useds
dtto change variables from t# s)
With this parametrisation the tangent
T=d x
ds
is now also the unit tangent because the
scale factor
h = d x
ds=1 (! dx =ds)
$ % =d x
ds
X1
X2
X3
P(x(s)
s
P0
Unit tangent s)
4747
Since "."=1
".d"
ds+d"
ds."= 0 # 2".
d"
ds= 0
which implies that "is perpendicular tod"
ds.
We define the curvature of x(s) to be
$ = d"
ds
and the principal normal nby
d"
ds=$n
The plane containing " and n at any point P of the curve
is called the osculating plane.
Note that
n =1
$
d"
ds=
1
$
d2x(s)
ds2
4848
x1
X3
!"
Unit tangent#(s)
X2
Unit tangent#(s+!s)
P[x(s)]Q[x(s+!s)]
P0
s
d"
ds= lim#s$0
#"
#s= lim#s$0
#%
#s= &
So curvature measures the
rate of rotation of the
unit tangent as we move along the curve.
!(s)
!(s+"s)
"#
"!
"! = "#Radius of Curvature: "=
1
#
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4949
Ex : Consider the circular helix defined by
x(t) = coste 1 + sinte 2+te 3
T=d x
dt= "sinte 1 +coste 2+e3 """" tangent
ds = d x=dx
dtdt= 2dt# scale factor h = 2
andds
dt
= 2
$=d x
dt/d x
dt=
1
h
d x
dt=
1
2("sinte 1 + coste 2+e3) - - - unit tangent
d$
dt=
1
2("coste 1 "sin te 2 )
%d$
ds=dt
ds
d$
dt=
1
2
1
2("coste1 " sinte 2 )
%&=d$
ds=
1
2 and
Principal normal is n = "(coste 1 +sinte 2 )
X3
X1
X2
1
5050
TorsionDefine the unit bi-normal vector by
b(s) = !(s)!n(s) (i.e. complete the orthonormal triad).
b(s).b(s)=1" 2b(s).db(s)
ds= 0
"db
dsis perpendicular to b
Also
!.b = 0"d!
ds. b+!.
db
ds= 0
"n. b+!.db
ds= 0
!.db
ds= 0
"db
dsis perpendicular to !
#db
dsis in the direction of n
b(s
)
!(s)n(s)
5151
We define torsion by
db
ds= !!n
That is,
! =db
ds
Since bis a unit vector, we see as before that !
is the rate of rotation ofb, or equivalently of
the osculating plane with distance along the curve.
!is positive if brotates positively (in a right handed
sense) around "as sincreases.
The torsion of a curve can also be seen to be the extent
to which a curve fails to be planar.
b(s)
b(s+!")
!"
!b
!b =!"
b(s)
!(s)n(s)
5252
Example : Calculate the torsion of the circular helix x(t) = coste1 +sin te 2 + te3
We had
ds
dt= 2
!=1
2(!sin te1 +coste 2 + e3) ---unit tangent
n =!(coste1 +sin te 2 ) ---------unit normal
"b = !# n =1
2
e1 e2 e3
!sin t cost 1
!cost !sin t 0
=1
2
(sin te1 ! coste 2 + e3)
----unit binormal
db
dt=
1
2(coste1 +sin te 2 )
$db
ds=
db
dt
dt
ds=
1
2(coste1 +sin te 2 )
Usingdb
ds= !"n, torsion "=
1
2
X3
X1
X2
1
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5353
Example: Show that
dn
ds=!!"+#b
(Hint: differentiate n = b"")
Collecting all the results, we have a set of three
equations which are known as the Frenet-Serret Relations.
d"
ds=!n
dn
ds=!!"+#b
db
ds=!#n
54
Line IntegralsLine integrals along a path C given by
x = x1(t)e1 +x2(t)e2+x3(t)e3 = f1(t)e1 +f2(t)e2 +f3(t)e3
may be of one of several forms;
A(x). dxC
! , A(x)"dxC
! , !(x)dlC
! , A1(x)dx2C
! etc.
where A(x) is a vector field and !(x) is a scalar field.
Now
A(x).dxC
! = (e1A1 + e2A2 + e3A3). (e1 dx1 + e2dx2 + e3dx3)C
!
= A1(x)dx1 +C
! A2(x)dx2 +C
! A3(x)dx3C
!
X1
X2
X3
C
a
bCx1
O
55
I = A!dxC
" = (e1A1 + e2A2 + e3A3)!(e1dx1 + e2dx2 + e3dx3)C
"
= e1 (A2dx3 # A3C
" dx2 )+ e2 (A3dx1 # A1C
" dx3)+ e3 (A1dx2# A2C
" dx1)
and
[ A(x)!dx
C
" ]1 = (A2dx3C
" # A3dx2 ) etc.
56
Noting that dxk = !fk(t)dt, dl =d x =dx
dtdt = !f (t)dt
all such line integrals can be built up from the following
two basic forms:
!C
" (x)dl = !(f(t)) !f (t)dta
b
"
and
!C
" (x)dxk = !a
b
" (f(t)) !fk(t)dt (k=1,2,3)
where !(x) is some scalar field.
So we now focus on how to evaluate such integrals.
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57
Note:
It is important to visualise any integral as the limit of a
sum (Riemann Sum) formed as follows:
Partition the variable of integration into N discrete intervals.
Weight each interval by the value of the integrand evaluated
at a point in the interval
.!
C
! (x)dx1 = LimN"#"x1
p"0
!(x1p)"x1
p
p=1
N
$
!C
! (x)dl =L imN"#"l"0
!(xkp)"lp
p=1
N
$
X1
X2
X3
C
a
bCx1
O
58
The following theorem can easily be proved (left as an excercise!)
Theorem:
If Cis an oriented simple curve or an oriented simple closed curve
we can unambigously evaluate
!(x)C! dx
kor !(x)dlC!
using any orientation preserving parametrisation x = f(t) of C. The sign
of the integrals will be reversed if we change the orientation.
59
If we use one of the coordinates in E3as an parameter - say x1 (an "explicit representation")
C: x2 = g(x1), x3 = h(x1) a!x1!b
where h and g are continuous functions in [a,b],
the parametric equation can be written as
x = e1x1 + e2x2 + e3x3
= e1x1 + e2g(x1)+ e3h(x1) (here parameter t= x1)
dx = e1dx1 + e2 "g (x1)dx1 + e3 "h (x1)dx1
dl = ds = d x =dx
dx1dx1 = 1+ "g
2+ "h 2dx1
Then these basic elements become
!C
# (x)dl = !(a
b
# x1,g(x1), h(x1)) 1+ "g 2 + "h 2dx1
!C
# (x)dx1 = !(a
b
# x1,g(x1), h(x1))dx1
!C
# (x)dx2 = !(a
b
# x1,g(x1), h(x1)) "g (x1)dx1 ...etc.
Explicit representation
X1
X2
X3
C
a
bCx1
O
60
Example: Evaluate xy2 dsC
! where Cis the quarter circle
x=cos t, y=sin t (0"t"!
2)
r = (x,y) = (cost,sin t)
ds =dr
dtdt= sin
2 t+ cos2 tdt= dt
xy2 dsC
! = costsin20
!/2
! tdt= sin20
!/2
! td(sin t) =sin
3 t
3
#
$%
&
'(
0
!/2
=1
3
Example: Evaluate xy2 dxC
! , xy2 dyC
! along the parabolay =x 2 from (0,0) to (2,4).
r = (x,y) = (x,x2 ) (using explicit representation withx as parameter)
dr = (dx,dy) = (dx,2xdx)
xy2 dxC
! = x. x40
2
! dx =x6
6
#
$%
&
'(
0
2
=
32
3
xy2 dyC
! = x.x4.2x0
2
! dx = 2 x7
7
#
$%
&
'(
0
2
=256
7
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61
Given a force field F(r) acting on a particle and a curve C,
the work done in moving the particle along Cis
W= F. drC
!
Example: Calculate work done by the force of gravity F=(0,0,-mg) in
moving a particle of mass mfrom a = (x0,y0, z0 ) to b = (x,y, z).
Work done: W= F. drC1
!
= (C1
! Fx,Fy,Fz ). (dx,dy,dz)
= (Fxdx+C1
! Fydy+Fzdz)
=- mgdzz0
z
! ="mg(z"z0 )
(Note: The path C1 never came into the calculation!)
B(x,y,z)
A(x0,y0,z0)
CF
X
Z
Y
C1
Work Integral
62
To calculate the work integral along a general path, first
write the equation of the path as a parametrised curve.
In the above example, the equation of path C(a straight line) is
r = a+ t(b!a) = (x0,y0,z0 )+ (x -x0,y!y0,z -z0 )t
(0 " t"1)
W = F. drC
# = F(r(t)). drdt0
1
# dt
= (0
1
# (0,0,-mg)).(x -x0,y!y0,z -z0 )dt= !mg(z!z0 ) dt0
1
# = !mg(z!z0 )
as before
In fact, W = F.drC1
# = F.drC
# independently of path because
the force field is "conservative" (see later)
63
If Cis a closed curve,
I = A. d x!! or A. dl or A. dr !! or A. ds !! !!
is called the CIRCULATION of Aabout Cand written
Example: Calculate the circulation of v = (x1, 0,0) over unit circle.Parametrise circle:x = (cost,sint,0) (0 " t" 2!)
v. dxunit circle!! = v.
d x
dtdt
unit circle!! = (cos t,0,0).(#sint,cost, 0)dt
0
2!
!
=- sin tcost dt0
2!
! = 0
CirculationIntegral
x1
x2
64
Another type of integral is
I= A!dxC
"
This is interpreted as
I= (e1A1 +e2A2 + e3A3)!(e1dx1 + e2dx2 + e3dx3)C
"
= e1 (A2dx3 # A3C
" dx2 )+e2 (A3dx1 # A1C
" dx3)+ e3 (A1 dx2# A2C
" dx1)
Components areI1 = A2dx3 # A3dx2
C
" etc. of the standard form.C
"
Another type of circulation integral would be
I= A!dxC
!"
where Cis a closed curve.
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65
Example : In the previous example with v = (x1,0,0) calculate
I= v
unit circle
" # dx
Now x = (cos t,sin t,0) and dx = ($sin t,cos t,0)
I1 = (v2dx3$ v3C
" dx2) = 0
I2 = (v3dx1$ v1C
" dx3) = 0
I3 = v1unit circle" dx2$ v2
unit circle" dx1
v1unit circle
" dx2 = x1unit circle
" dx2 = costd(sin t)0
2%
" = cos2 tdt0
2%
"
=
1
2[1+ cos2t] dt
0
2%
" = %
and v2unit circle
" dx1 = 0
&I = %e3
x1
x2
66
Not all surfaces can be represented as a graph of a function
of 2 variables using rectangular cartesian coordinatesz = f(x,y)
Thus, it is not possible to represent the surface of a doughnutby such an equation. We are therefore led to consider more general parametric
representations of surfaces. Just as curves can be seen as mappings of a portion of a straight linesurfaces can be seen as being mappings of a region of the plane.
Surfaces
67
Cuv is positively directed (simple piecewise smooth) curve in the (u,v) plane.
Suvis the union of Cuvand its interior.
A surface S is a set (" R3) of points having at least one parametric representation
x =g(u,v) = e1g1(u,v)+ e2g2(u,v)+ e3g3(u,v) u,v# Suv
g(u,v) is continuous on Suv . As u,vvary, the tip of xdescribes a
surface inx1x2x3space. gis a mapping of Suvonto S.
C
S
x =g(u,v)
x1
x2
x3
P
Suv
u
v
Cuv
Mapping
Parametric Surfaces
g (u,v)
68
Sis simple if there exists at least one mapping x = g(u,v)
(of Suv
onto S) that is one to one.
If Sis simple, Cuv
maps onto the edge of C.
If g(u,v)!C1, we also say that the surface is differentiable,
or a C1 surface.
Sis smooth if the functions!g
!u and
!g
!v are continuous
on Suv
(that is, g(u,v)!C1) with
!g
!u"
!g
!v# 0 ( u, v ! S)
(This is the condition for there to be a tangent plane at
every point -int uitively,no corners)
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69
Example: Consider the parametrised surface
x = g(r,!) = rcos!e1 + rsin!e2 + re3 (r ! 0, 0 "!" 2")
Here
g1(r,!) = rcos!, g2 (r,!) = rsin!, g3(r,!) = r
This is the surface of a cone of semi vertex angle 450because
x3 = x12+x2
2
with (0,0)#
(0,0,0) (....the vertex)Clearly, all partial derivatives of gare continuous everywhere
in (r,!) plane. But
T1 =#g
#r= cos!e1 + sin!e2 + e3
T2 =#g
#!=$rsin!e1 + rcos!e2
and
T1%T2 = r($cos!e1 $ sin!e2 + e3) = 0 at (0,0)
This is a differentiable surface, but it is not smooth (by definition).
r!
x3
x1
x245
70
Let Sbe a simple smooth surface. Consider a point P! (uP,vp ) on S.
The curves
C1 :x = g(u,vP )
C2 :x = g(uP,v)
which clearly lie on Sand pass through P, are
called the coordinate curves through P.
We can form tangents T1 to C1 and T2 to C2 at P.
T1 =!g
!u
"
#$
%
&'uP,vP
,T2 =!g
!v
"
#$
%
&'uP,vP
!g
!u(
!g
!vP
) 0 (smooth* coordinate curves through P are distinct).
Note: The directions that we get for the tangent vectors (i.e. which way
they point) depends on the parametrisation.
Coordinate Curvesn+
C1: x=g(u,vP)
C2: x=g(uP,v)
n-
7171
We can form everywhere on S two unit normals
n+
=
gu!g
v
gu!g
v
, n"
=-n+
Note that what we define as the + and - directions is arbitrary
depending on our parametrisation.
For a given parametrisation, the positive sign is used for the
normal direction that forms a RH triad with respect to !1 and !2.
This is turn depends on the sense of travel long the bounding curve
of the surface (assuming an open surface) as uand vincrease.
!1
!2
P
n+
n-
A surface has two normals at every point
72
Unit tangents to coodinate curves:
!1 ="g
"u
"g
"u=
1
h1
"g
"u, !2 =
"g
"v
"g
"v=
1
h2
"g
"v
h1 = g u , h2 = g v can be identified as the length scales
along the coordinate curves (see below)
The plane containing!
1,!
2 is called the tangent plane at P.Every tangent to the surface at P lies in the tangent plane
through P.
Now for any two adjacent points P,Q on surface,
dl = dx ="g
"udu+
"g
"vdv = g
udu+ g
vdv
!1
!2
P
n+
n-
x1
x2
x3
P
Q
S
g(u,v)
g(u+!u,v+!v)
dl
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73
The metric distance between two points P(u,v), Q(u+du, v+dv) is given by
dl2 = dx1
2+ dx2
2+ dx3
2= dx. dx = g
udu+ g
vdv( ) . gudu+ gvdv( )
= gu.g
udu2 +2g
u. g
vdudv+ g
v. g
vdv2
dl is called the "metric element" on the surface.
Length element between two points P(u,v), S(u+du, v)
along a ucoordinate curve is
dlu = gu. gudu =!g
!udu = h1du (dv = 0)
Similarly, along a vcoordinate curve
dlv = gv. gvdv =!g
!vdv = h2dv (du = 0)
R (u+du,v)
h1du
h2dv
P(u,v)
T(u,v+dv)
74
If every ucurve is perpendicular to every v - curve, we have an orthogonal
curvilinear coordinate system. A necessary condition is that
gu.g
v= 0 ("1."2 = 0 - tangent curves orthogonal)
In this case
dl 2 =h1
2du2 +h
2
2dv2
75
Example: Parametric representation on the surface of a sphere
using co- latitude "and longitude #.
x =g(",#) =R sin"cos#e1 +Rsin"sin#e2 +Rcos"e3
T1 =$g
$"=R(cos"cos#e1 + cos"sin#e2 % sin"e3)
T2 =$g
$#=R(%sin"sin#e1 +sin"cos#e2 )
T1.T2 = 0& orthogonal curvilinear coordinates
h1 =$g
$"=R
h2=
$g
$#=R sin"
& dl 2 =R 2d"2 +R2 sin2"d#2
X1
X2
X3
R sin
R sin cos!
76
Example : Parametric representation on the surface of a cylinder :
(",z) : "is plane polar angle, and z in vertical coordinate.
x =g(",z)=R cos"e1 +Rsin"e2 +ze3
T1 =#g
#"=R($sin"e1 + cos"e2 )
T2 =#g
#z=Re 3
T1.T2 = 0% orthogonal curvilinear coordinates
h1 =#g
#"=R
h2 =#g
#z=1
dl 2 =R 2d"2 +dz2
P
z
! R
X
Y
Z
!coordinatecurve
R coordinatecurve
z coordinatecurve
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Example: On surface of sphere
x = g(!,") =R sin!cos"e1 +Rsin!sin"e2 + Rcos!e3
g!=R(cos!cos"e1 + cos!sin"e2 ! sin!e3)
g"=R(!sin!sin"e1 +sin!cos"e2 )
dS= (g!"g
")d!d"= R
2sin!d!d"(sin!cos"e1 +sin!sin"e2 + cos!e3)
er = sin!cos"e1 +sin!sin"e2 + cos!e3
is unit vector in outward radial direction (by projection)
# dS =R2sin!d!d"er
h1 =#g
#!=R
h2 =#g
#"=R sin!
Note: Since this is an orthogonal curvilinear coordinate system, we also
could have written directly
dS= h1h2d!d"= R2sin!d!d"
X1
X2
X3
R sin
R sin cos!
78
Element of area dS Now, between adjacent points on surface,
dl = dx =!g
!udu+
!g
!vdv = g
udu+g
vdv
and along coordinate curves
dl u = gudu, dl v = gvdv
dSis the area of the parallelogram formed by dlu and dlv
This can be seen to be
dS= d lu"dl v = (dlu!1)"(dlv!2 ) = g u "gv dudv
dS= dl u !dl v
= (gu!g
v)dudv = g
u!g
vdudvn
+
X1
X3
P(u,v) Q(u+du,v+dv)
X2
dlg(u,v)
g(u+dv,v+dv)
Vector element of area dS S
n+
uu+du vv+dv
X1
X2
X3
d
79C
n+
S
Orientable
An orientable surface is a two sided surface with one side specified as the outside
or positive side and the other side the inside or the negative side.
Consider a surface with a bounding curve C.
For a two sided surface we can assign a unit normal nat
every point of the surface such that the resulting vector field n(x) of
normals is continuous everywhere. For such a surface, when the foot of ndescribes
any closed curve not crossing C, the initial and final directions of nare the same.
Such a surface has two orientations (two sides) determined
by the normals n(x), -n(x).
80
If the direction of the bounding curve C ("the positive
orientation") induced by the parametrisation
and n(x) form a RH system, C is said to be positively directed with
respect to n(x). This choice of direction n(x) (written as n+
(x)) determines
the positive side (that is the positive orientation) of the surface. It is
arbitrary to the extent that it depends on the prametrisation.
C is positively directed with respect ton
+
(x
) in accompanying figure.
C
n+
S
One orientation of a
two sided surface
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81
The other orientation (negative side) of the same surface.The field of normals is now n-
n-
Note: We have assigned a direction to C2to make it positively
directed with respect to the normal n- . The direction of C
which is the natural direction that corresponds to the parametisationis now negatively directed with respect to n- .
8282
If a surface is one sided, any two points on the surface can be connected
by a curve without crossing the boundary and when one returns to the starting point the normals are in the opposite direction!
Twist
n1
n2
Not all surfaces are two sided, even though we can assign two normals to every
point on the surface.
Non orientable-Mobius strip-
83
A piecewise smooth surface need not be orientable even if it can be
divided into components, each of which is separately orientable.
The Mobius strip is such an example (see next slide).A sumof surfaces is orientable(i) If each component is orientable (ii) An orientation can be induced on each piece such that positive
direction on common boundaries are opposite.n+
C1
The theorems that we derive will be applicable to orientable smooth surfacesand more generally to orientable sums of smooth surfaces. 84
Two Orientations of Sn+
P
n-
C1 C2
n+
n+
Orientable Sumsn-
n+
Twist!
Mobius Strip- -non orientablesum
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85
Surface integrals can be of one of many types:
a(x). dSS
! , a(x)" dSS
! , !(x)dSS
!
where a(x), is a vector field and!(x) is a scalar field.
Here
a(x). dSS! = a(x).(ndS) =
S!! a1(x)n1 dS
S! + a2(x)n2dS
S! + a3(x)n3 dS
S!
and
a(x)" dSS
!#
$%
&
'(
1
= (a2S
! dS3) a3dS2 ) = (a2S
! n3)a3n2 )dS
where dS= ndS
All integrals can be built up from the following two basic types:
"(x)dSS
! and "(x)n+3 dSS
!
Surface Integrals
86
Now, for a parametric surface
x = g(u,v)
the unit normal is
n =gu!g
v
gu!g
v
=
1
gu!g
v
e1 e2 e3
g1,u
g2,u
g3,u
g1,v g2,v g3,v
"
#
$$$
$
%
&
'''
'
This is also taken to be n+
(the direction induced by the parametrisation).
The vector element of area is
dS= ndS= (gu!g
v)dudv
dS= d S = gu!g
vdudv and
n3
+dS=(g
u!g
v)3
gu!g
v
gu!g
vdudv=(g1,ug2,v( g2,ug1,v )dudv
87
Using the above
!S
!! (x)dS= ![g(u,v)]"g
"u""g
"vSuv!! dudv
and
!S
!! (x)n3+dS= ![g(u,v)]"g1
"u
"g2
"v#"g2
"u
"g1
"v
$
%&'
()Suv
!! dudv
+two similar integrals (with n1+
,n2+
)
If we have an explicit representation of a surface,
x3 = f(x1,x2 ) x1,x2 * Sx1x2
the parametric representation is
x = g(x1,x2 )= e1x1 + e2x2 + e3f(x1,x2 ) (x1 = u, x2 = v)
where Sx1x2 is simply the projection of S onto the OX1X2 plane.
S: x =g(u,v)
x1
x2
x3
P
n+
C
Sx1x2
88
"g
"x1
#"g
"x2
=e1$"f
"x1
%
&'
(
)*+e2
"f
"x2
%
&'
(
)*+e3
"g
"x1
#"g
"x2
=
"f
"x1
%
&'
(
)*
2
+
"f
"x2
%
&'
(
)*
2
+1
dS="g
"x1
#"g
"x2
%
&'
(
)*dx1dx2
dS
x1
x2
x3 n+
dScos !3
=dSn3+
=dx1dx2
!3
dS
!(xS
!! )dS= !Sx1x2
!! (x1,x2,f(x1,x2 )) "f
"x1
"
#$
%
&'
2
+
"f
"x2
"
#$
%
&'
2
+1 dx1dx
2
!(xS
!! )n3+dS= !Sx1x2
!! (x1,x2,f(x1,x2 ))dx1dx2
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8989
As for parametric curves, the following theorem can easily
be proved for surfaces (left as an excercise!)
Theorem:
If Sis an simple smooth orientable surface, we can unambigously evaluate
!(x
S
!! )dS, !(xS
!! )n3+dS
using any orientation preserving parametrisation of S. The sign
of the integrals will be reversed if we change the orientation.
90
EX : Find the surface area of the paraboloid of revolutionz = 4 "x 2 "y2
about the z axis for z > 0.
We use the explicit parametrisation (x,y) since the equation of the
surface is given in the form z = f(x,y). Any point on this surface
can be written in vector notation as
r(x,y)=xi +y j+ (4 "x 2 "y2 )k
#r
#x= i " 2xk
#r
#y= j" 2yk
Element of area is
dS= (#r
#x$
#r
#y)dxdy = (2xi +2y j+ k)dxdy""" (outward)
dS= 4x 2 + 4y2 +1 dxdy """ Note: chosen to be positive
S= 4x 2 +4y2 +1 dxdySxy
%%
Z
X
Y
r(x,y)
Sxy
(0,0,4)
91
We do this integral by changing to
plane polar coordinates;
Sxy" Sr#
S= 1+ 4r2
0
2
$0
2%
$ rdrd#
= 2%1
12 (1+ 4r2
)3/ 2&
'()
*+0
2
=
%
6 (173/ 2
,1)
X
Y
r
!
r
!
0 2
2 "
Sr!
Sxy
2
92
n
a
P
SurfaceS
a.n
volumeV
S: Closed surface about P with outward
unt outward normal n
V : Interior of S
a(x) : Vector field
(a. n)dS="flux" of aacross dS
a. n = flux of aper unit area of surface
Flux of aout of S is
a. dSS
"" = aS
"" . ndS (definition)
If ais velocity (m s-1), flux is rate of outflow of volume of fluid.
Dimension of a.dSS
"" :m
sm2 =
m3
s =
volume
time
Outward flux of a vector field across a closed surface
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93
If !(r) is the density of a fluid and v(r) is the velocity of a fluid,
the "mass flux" across element of area dSis !v. ndS.
!v. ndSS
!!
represents the total rate of outflow of mass across S and has
dimension [Mass
time
].
One can also define upward and downward fluxes across open surfaces.
Ex: calculate the upward flux of the vector field
F=(i+ j+ (x2+y
2)k) across the unit discx
2+y
2=1.
Of the two normal directions of the disc we select n=k(upward).
Funit disc
!! .ndS= (i+ j+ (x2 +y2 )k)unit disc
!! .kdS= (x2 +y2 )unit disc
!! dS= r20
2"
!0
1
! (rdrd#) =1
42"
94
Coordinate Transformations in 3D Euclidean Space A coordinate transformation is a mapping
x = f(u) = e1f1(u1,u2,u3)+...+ e3f3(u1,u2,u3) (fcontinuous)
with u!D u " x = f(u)!D
where f and D u are so chosen such that for every point u!D u
(1) The functions!fj
!uk
are continuous
(2) The mapping is one to one ( f(u)= f(u0 )# u = u0 )
with the Jacobian J
J f1, f2, f3
u1,u2,u3
$
%&
'
()=
!f1
!u1
!f1
!u2
!f1
!u3
!f3
!u1
!f3
!u3
* 0
X1
X2
X3
u1
u2
u3f(u)
D Du
95
Curvilinear coordinates, coordinate surfaces
and coordinate curves Set u
1= const =k
1, (and let u
2,u
3vary).
x = f(k1,u
2,u
3) describes a two parameter (u
2,u
3)" coordinate surface.
Set u1=k
1,u
2=k
2(and let u
3vary)
x = f(k1,k
2,u
3) describes a one parameter u
3" coordinate curve.
Suppose P # Dand has parametric coordinates (u1P
,u2P
,u3P
)
S12P : (u1 " u2surface) x = f(u1,u2 ,u3P )
S31P
: (u1"u
3surface) x = f(u
1,u
2P,u
3)
S23P
: (u2" u
3surface) x = f(u
1p,u
2,u
3)
P
u1
u2
u3
S12P
1P
C2P
C3P
96
C1P :(u1curve) x = f(u1,u2P ,u3P ) "1 = #f
#u1
#f#u
1
C2P : (u2curve) x = f(u1P ,u2 ,u3P ) "2 =
#f
#u2
#f
#u2
C3P :(u3curve) x = f(u1P ,u2P ,u3) "3 =
#f
#u3
#f
#u3
P
u1
u2
u3
S12P
C1P
C2P
C3P
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97
Element of length is given by
dl2= dx. dx =
!f
!u1du1 +
!f
!u2du2 +
!f
!u3du3
!
"#
$
%&.
!f
!u1du1 +
!f
!u2du2 +
!f
!u3du3
!
"#
$
%&
=
!f
!u1
2
du12+
!f
!u2
2
du22+
!f
!u3
2
du32+ (....)du1du2 + (.........)du2du3 + (.........)du3du1
Note that in the (u1,u2,u3) coordinate system, the form of the metric element
of length may be such that it is not immediately obvious that the underlying
space is Euclidean ( flat).
That is, it may not be obvious that there is a transformation
(u1,u2,u3)' (w, s, t) such that
dl2= ds
2+ dt
2+ dw
2
The Element of Length
98
Element of length along uk
curve is
dkl =!f
!ukduk = hkduk (k= 1,2,3)
hk =!f
!uk= (
!f1
!uk)2 +..+ (
!f3
!uk)2 ! 0
"k =1
hk
!f
!uk
Here hk =!f
!ukis the length scale factor.
"1.("2 ""3) =1
h1h2h3{!f
!u1#
!f
!u2"
!f
!u3} =
1
h1h2h3J
Note: J ! 0 $ "1,"2,"3 are not coplanar, and none are null.
P
u1
u2
u3
12P
C1P
2P
C3P
!1
!2!3
99
Any vector acan be resolved into components in directions
!1,!2,!3:
a ="1!1 +"2
!2 +"3!3
The components are the lengths of the sides of the
parellogram as in diagram.
(Note: For a general curvilinear coordinate system "1 ! a.!1)
a
!1
!2
!3
Components of Vector
100100
Volume of the elementary parallelopiped
with sides h1du1!1, h2du2!2, h3du3!3
dV = (h1du1 !1).(h2du2 !2 )!(h3du3 !3)
=
h1h1h1!
1.(!
2!!
3)du1du2du3
= J du1du2du3
"dV ="(f1, f2,f3)
"(u1, u2, u3)du1du2du3
The Element of Volume
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101
Volume Integrals(change of variable)
"(x)dVD
###
= "(x)dx1D
### dx2dx3
= "(x(u1,u2 ,u3)Du
### )$(x1,x2 ,x3)$(u1,u2 ,u3)
du1du2du3
= "(x(u1,u2 ,u3)Du
### )J du1du2du3
102102
If !1.(!2!!3) =1 curvilinear coordinate system
orthogonal (RH:+1, LH: -1).
Arc length:
dl2= h
1
2du
1
2+h
2
2du
2
2+h
3
2du
3
2
Jacobian: J= h1h2h3
Element of volume:
dV = h1h2h3du1du2du3
Volume integrals:
"V
""" (x)dV = "(x(u))Vu
""" h1h2h3du1du2du3
Orthogonal Curvilinear Coordinates
103
Example: Consider the ellipsoid
x2
a2+y2
b2+z2
c2=1
We can find the volume by changing variables
(x,y,z)" (u,v,w) via
x =au,y =bv,z =cw
In (u,v,w) space the ellipsoid is the sphere
u2+ v
2+w
2=1
J=#(x,y,z)
#(u,v,w)=
#x
#u
#x
#v
#x
#w. . .
. . .
=
a 0 0
0 b 0
0 0 c
=abc
V= dxdydzellipsoid
$$$ = abcdudvdwunit sphere
$$$ =4%
3abc
104
Spherical Polar Coordinates(orthogonal curvilinear)
Spherical Polars: u = (r,",#)
x =r sin"cos#e1 + r sin"sin#e2 + r cos"e3
$x
$r= (sin"cos#,sin"sin#,cos")
$x
$"= r(cos"cos#,cos"sin#,%sin")
$x
$#= (%sin"sin#,sin"cos#,0)
h1 =$x
$r=1
h2 =$x
$"=r
h3 =$x
$#=r sin"
Z
X
Y!
P " coordinatecurve
!coordinatecurve r coordinate
curve
" r
!
dV=r2sin"drd"d#
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105
Cylindrical Polar Coordinates (orthogonal curvilinear)
Cylindrical Polars: u = (R,",z)
x =R cos"e1 +Rsin"e2 +ze3
#x
#R= (cos",sin",0)
#x
#"=R($sin",cos",0),
#x
#z= (0,0,1)
h1 =
#x
#R=1
h2 =
#x
#"= R
h3 =
#x
#z=1
P
z
! R
X
Y
Z
!coordinatecurve
R coordinatecurve
z coordinatecurve
dV=RdRd"dz106
Exercise: Show that the volume of the region bounded above
by the sphere x2+ y
2+z
2= a
2and below by the cone
x2 + y2 = tan2! z2
is2"a3
3
(1! cos!)
Z
O
107
VECTOR CALCULUSPart 2: Vector differential operators
108
The vector differential operator "Del" or "Nabla" is defined by it effect on a
scalar or vector functions;
! = e1!
!x1+e2
!
!x2+ e3
!
!x3
!operates on a scalar field "(x) as follows:
Gradient:
grad " =!"(x)=
(e1!
!x1+e2
!
!x2+e3
!
!x3)"(x) =e1
!"
!x1+ e2
!"
!x2+ e3
!"
!x3
So when "del" operates on a scalar field "(x), the result is the
vector field grad ".
Grad, div and curl in Cartesians
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109
"del" operates on a vector field a(x) in two different ways, producing either
another scalar field, or another vector field:
Divergence: div a =!. a = (e1!
!x1
+e2
!
!x2
+e3
!
!x3
). (a1e1 +a2e2 + a3e3)
=!a1
!x1
+
!a2
!x2
+
!a3
!x3
Curl: curl a =!" a = (e1 !!x1
+e2 !!x2
+e3 !!x3
) " (a1e1 +a2e2 + a3e3)
=
e1 e2 e3
!
!x1
!
!x2
!
!x3
a1 a2 a3
= (!a3
!x2
#!a2
!x3
)e1 +..........
The above definitions of grad, div and curl use cartesian components. However,
there are coordinate free definitions which allow us to interpret these in a
geometrical way; 110
Example :
F=r=xi +y j+zk
F= x2+y
2+z
2=r
divF= ".F=#x
#x+
#y
#y+
#z
#z= 3
div measures divergence
Example :
F= $yi +x j (2D field)
F= x2+y
2=r (in 2D)
curl F=
i j k
#
#x
#
#y
#
#z
$y x 0
= 0i +0j+2k
= 2k
Curl measures rotation
X
Y
X
Y
111
"= "(x) - Scalar Field
Consider level (equipotential) surfaces
through neighbouring points P, #P
"(x)= "P, "(x)= "
P+$"
P
nis unit normal in direction of "increasing :
Rate of change of "in direction n(unit normal)
%"%n
= limNP&0
"N'"P
NP
Rate of change of "in arbitrary direction u(unit vector)
%"
%u= lim
#P P&0
"#P'"
P
#P P= lim
NP&0{"
N'"
P
NP(u.n)}
=%"
%n(u.n)
(maximum rate of change of "is%"
%n (when u.n =1)
Grad: a coordinate free
definition
P
P/
N
u
n
!= !P+ !P
!= !p
n, uunit vectors
112
Define vector field by
"#=$#
$nn (ndirection of #increasing)
Then$#
$u= u ."#
"#is a vector perpendicular to #= const
at every point on surface. It is in the direction of
#increasing
n."#gives maximum rate of change of #
P
P/
N
u
n
!= !P+ "!P
!= !p
n, uunit vectors
"#
"x1= e1.$#,
"#
"x2= e2 .$#,
"#
"x3= e3.$#
%$#="#
"x1e1 +
"#
"x2e2 +
"#
"x3e3 as before
!= const "#
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113
Ex : Calculate a unit normal to the surface
S : x2+ y
2+ z
2=1
Consider the scalar field "(x,y,z) = x 2 + y2 + z2 #1
The surface S corresponds to the equipotential "(x,y,z)= 0.
$The required direction is given by %".
%"= (2x,2y,2z)
n =%"
%"=
1
4x2+ 4y
2+ 4z
2(2x,2y,2z)=
1
r(x,y,z) =
r
r
nis in the radial direction, which is not surprising because S is a sphere.
Note: had we chosen "(x,y,z)=1- x2# y
2# z
2, we would have obtained
n = #r
r
n
r
114
O != const
n
P
P0
Tangent Plane
r0
r
Tangent plane to a surface Ex : Calculate equation of tangent plane to the surface x
2+ 2y
2+ xz = 6
at r0 = (1,1,3)
"(x, y,z) = x2+ 2y
2+ xz #6
$"= (2x + z, 4y,x)
$"(r0 )= (5, 4,1)
Equation of tangent plane is
(r # r0).$"(r0) = 0
5(x #1)+ 4(y #1)+1(z # 3)= 0
5x + 4y + z =12
115
n
a
P
SurfaceS
a.n
volumeV
S: Closed surface about P with outward
unt normal nsuch that x-xP
V : Interior of S
a(x) : Vector field (visualise as fluid velocity)
(a. n)dS= "flux" of aacross dS
a. n = "flux" per unit area
[div a]P= lim
!"0
a. n dS##vol V
= lim!"0
total outward flux of athrough S
vol V
$
%&'
()
If ais velocity, outward flux is volume rate of outflow of fluid.
Divergence: coordinate free definition
diva measuresdivergence of fluid flow per unit volume116
Demonstration for a rectangular volume elementQ,R belong to S, x1R = x1Q + !x1
"a1
"x1
!
"#
$
%&Q
="a1
"x1
!
"#
$
%&P
+o(1),"a1
"x1
!
"#
$
%&R
="a1
"x1
!
"#
$
%&P
+o(1)
'a1R( a1Q ="a1
"x1
!
"#
$
%&P
!x1 +o(!x1)
limrate of outflow
volume
)
*+,
-.
= lim
("a1
"x1)P!x1 +o(!x1)
/01
234!x2!x3 + two similar expressions
!x1!x2!x3
)
*
++++
,
-
.
.
.
.
="a1
"x1+...+
"a3
"x3
= div a
lim it independent of shape of boundary (later)
Q Ra1Ra1Q
!x1
!x2
!x3
!x2
!x3P
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117
Curl: coordinate free definition
a
!
a.!C
m
P
m : unit vector in some direction
C: boundary of a (small) plane surface
having area A normal to m and such that
x -xp < ", x #C
$ : unit tangent to Cwith direction related to m
by RH rule
Then
m. (curl a) = lim"%0
1
Aa.$
C
& dl (dlelement of length)
= lim"%0
{circulation of aabout closed loop C
area A}
Exercise: Show this for a rectangular areaCurl ameasures circulation per unit area
Curl
118
Example 1: v = (x,0,0)
div v = ".v =1, curl v = "# v = 0
Excess of flux leaving region ABCD
No rotation imparted to region.
Note: v = "(1
2x
2 ) - -can be derived from scalar potential
Example 2 : v = (y,0,0)
div v = ".v = 0, curl v = "# v = (0,0,$1)
Flux leaving region balances flux
entering region.
Rotation about - Z direction
No scalar potential exists for flow.
Example 3: v = ( x
x2+ y
2,
y
x2+ y
2,0)
div v = ".v = 0 (except at origin)
curl v = "#v = 0
Zero net flux and rotation
v = "[1
2ln(x2 + y2 )]
X
Y
X
Y
Y
X
119
! = e1!
!x1+e2
!
!x2+e3
!
!x3 (cartesians)
!behaves essentially like b in bf, b. a, b"a but since !is a differential
operator, the sequence in which the symbols are written matter. Thus
(a. b) c = c(a. b),
but (a.!) c # (!. a) c
(a.!) c = (a1!
!x1+ a2
!
!x2+ a3
!
!x3) (e1c1 +e2c2 + e3c3)
= (a1!c1
!x1+ a2
!c1
!x2+ a3
!c1
!x3)e1 + two similar terms
(!. a) c = (!a1
!x1+
!a2
!x2+
!a3
!x3) (e1c1 + e2c2 + e3c3)
Some properties of del
120
If "and#are scalar fields, aand bare vector fields and
$and %are scalars
&($"+ %#) =$&"+%
&. (a + b) =&. a +&. b
&'(a + b) =&' a +&' b
These, and other identities can be proved by
taking components
Differentiation of Sums
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121
div ("a) = "diva +grad". a #. ("a) = "#.a +#". a
curl ( "a) = "curla +grad"$a #$("a) = "#$a +#"$a
div (a$b) =b.curla - a.curl b #. (a$b) = (#$a).b -(#$b).a
curl (a$b) =a divb -b diva + (b.grad)a - (a.grad)b
grad (a.b) =a$curlb + b$ curla + (b.grad)a +(a.grad)b
#$(a$b) = (b.#)a% b(#.a)% (a.#)b+a(#.b)
#(a.b) =b$(#$a)+a$(#$b)+ (b.#)a+ (a.#)b
Differentiation of Products
122
Example of proof by components :
[curl "a]1 =#("a
3)
#x2$#("a
2)
#x3=
#"
#x2a3 +
"#a
3
#x2$#"
#x3a2 $"
#a2
#x3
= "(#a3
#x2$#a2
#x3)+(
#"
#x2a3 $
#"
#x3a2 )
= "[curl a]1 +[grad"%a]1
&curl "a = "curl a+grad"%a
123
Example:!
1
r=
d
dr (
1
r )r = "
1
r2 r
Example:
curl (1
r2r )=
1
r2
curlr +grad(1
r2)#r
=1
r2
0"2
r3
r
r#r = 0
Exercise: If f(r) =f(r) show that
!f =df
dr
r
r=
df
drr
where r is a unit vector in the radial rdirection.
124
Second Order Operators Starting with a vector field a(x) and a scalar field !(x)
we have constructed
First order operators: div a, curl aand grad!
We can differentiate again and construct
Second order operators: grad (diva
) div (curl a)
curl (curl a)
curl (grad !)
div (grad !)
Some operations like curl( div a) do not make sense
in vector calculus (may in Tensor calculus!)
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125
LAPLACIAN : !2
!2!= div grad != !.(!!) --- (coordinate free definition)
="
2!
"x12+
"2!
"x22 +
"2!
"x32
--- (in RC system)
!2= (
"2
"x12 +
"2
"x22+
"2
"x32
) when operating on a scalar field and is the "Laplacian".
When !2 operates on a vector field it is defined by its action as
!2a = grad (div a)-curl(curla) ----- (coordinate free definition)
Caution ...
!2a = (!
2a1,!
2a2,!
2a3) only in RC system. In any other system,
!2a = grad (div a )-curl(curl a), where in evaluating grad and div
one must take into consideration the fact that the unit vectors are
functions of positions in space. The expressions for grad, div and
curl in a general orthogonal curvilinear coordinate system are given
later .
The Laplacian
126
Proof (by taking components in RC system)
"#= ($#
$x1,$#
$x2,$#
$x3)
"%a =
e1 e2 e3
$
$x1
$
$x2
$
$x3a1 a2 a3
&["%"#]1 =$
$x2
$#
$x3
'
()
*
+,-
$
$x3
$#
$x2
'
()
*
+,= 0
similarly for other two components
&curl grad #="%"#= 0
curl (grad ") =#$#"= 0
div (curl a) =#.#$a = 0
Two important identities
127
Laplaces and Poissons equationoccurs in many branches of applied
mathematics. Electrostatics Gravitation theory Incompressible Fluid Flows where is a scalar potential function that describes the particular
vector field(e.g. gravitational potential, electrostatic potential etc.) Discussed in partial differential equations course.
Two partial differential equations of
mathematical physics involving the Laplacian "
2#(x,y,z) = 0 Laplace's Equation
"2#(x,y,z) =$(x,y,z) Poisson's Equation
where $(x,y,z) is a source term.
!(x,y,z)
128
Vector Integral Theorems In evaluating definite integrals, we effectively "take out" one dimension
df
dxa
b
! dx = f(b)" f(a)
"a 1-D integral along a line is reduced to evaluating a function at the
end points of the line
S!! #
C! """"
Stokes theorem
-a 2-D surface integral across an open surface Sis reduced to evaluating a 1-D line integral
over the bounding curve C
V
!!! #S
!! ----- Divergence theorem or Gauss's theorem
-a 3-D volume integral across a volume V is reduced to evaluating a 2-D surface integral
over the bounding surface S
We prove these theorems by first proving Green's theorem in the plane.
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129
Greens Theorem in the Plane C: Closed positively directed plane curve
S: Union of Cand its interior
Under suitable restrictions
!"
!x!!#
!y
"
#$
%
&'
S
(( dxdy = #dx +"dyC
!(
We prove the result for
(a) standard domain (convex) region with simple piecewise smooth
boundary curve Cwith #,") C1 on S(functions and first partial derivatives
continuous)
The result follows also
(b) for sum of standard domains Snwhere the boundary C
nis
simple piecewise smooth with #,") C1 on Sn, #,") C0 on S.
130
Proof of (a): Standard domain (convex: xsimple andysimple)
g1(x)" y " g2(x), f1(y)" x " f2(y) describe C
#dx = #dx + #dxC2
$C1
$C
$
= #(x,g1(x))dx +a
b
$ #(x,g2(x))dxb
a
$
= % #(x,g2(x) % #(x,g1(x)[ ]dxa
b
$
= %
&ydy
g1 (x )
g2 (x )
$'
())
*
+,,dx
a
b
$ = %
&ydxdy
S
$$
Similarly
-dy =C
$ +&-
&xdxdy
S
$$
Hence
#dx +-dy =C
$ &-
&x%
&y
'
()
*
+,dxdy
S
$$
C2
x=f1(y)
x=f2(y)
X
C1
c
d
2
y=g1(x)
y=g2(x)
Xa b
C1
131
Example : Evaluate
I= (y2 + sinx2 )dx+ (cosy2!! "x)dyover the square C: 0#x#1: 0#y#1
I=$
$x(cosy
2 "x)"$
$y(y
2+ sinx
2 )%
&'
(
)*
S
!! dxdy (using Green's Theorem)
= ("1" 2y)dxdy = "0
1
!0
1
! (1+2y)0
1
! dy = "2
Note: In this case area integral is easier to evaluate than
line integral!
132
(Proof of b):
For more complicates regions, divide into sub- regions each
of which is standard (convex).
S=S1 +S2 + ...Sn
We show result for two regions S1 and S2 .
"#
"x$"%
"y
&
'(
)
*+dxdy
S1 +S2
,, ="#
"x$"%
"y
&
'(
)
*+dxdy
S1
,, +"#
"x$"%
"y
&
'(
)
*+dxdy
S2
,,
= %dx +#dy +C1
, %dx +#dy +C3
, %dx +#dy +C4
, %dx +#dyC2
,
But contributions to line integrals from common boundaries
cancel (because %,#- C0). Thus
%dx +#dy = $ %dx +#dyC4
,C3
,
. "#
"x$"%
"y
&
'(
)
*+dxdy
S1 +S2
,, = %dx +#dy +C1
, %dx +#dyC2
,
Similarly for more than two regions.
S1
S2
C1
C2
C3
C4
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133
Example: Verify Green's theorem for y2dx+2x dy
C
!!
x2+y
2= 9.
y2dx +2x dy
C
!! =!
!x(2x)"
!
!y(y
2 )#
$%
&
'(
S
!! = 2"2y[ ]S
!! dxdy
To evaluate the area integral, use plane polar coordinates;
x = rcos", y = rsin", dA = rdrd"
RHS : 2"2y[ ]S
!! dxdy = (2" 2.3rsin")rdrd"0
3
! =0
2#
! (9 "18sin0
2#
! ")d"= 18#
To evaluate the line integral, parametrise circle by
x = 3cos t,y = 3sin t(0)t) 2#)
LHS: y2 dx+2x dyC
!! = 9sin2
t("3sin t)dt+0
2#
! 2.3cos t.3cos t dt= 0+18#0
2#
!
X3
3
S
C
134
Example: Show that for a closed plane curve C
-ydx+xdy
x2+y
2
C
!! = 2! -if C encircles origin
= 0 -otherwise
"=
-y
x2 + y2 ,#=
x
x2 + y2
""
"y=
y2 # x2
(x2+ y
2)
2,
"#
"x=
y2 # x2
(x2+ y
2)
2
First order partial derivatives are continuous except at (0,0).
Expect to be able to apply Green's theorem to any region
not containing origin.
Y
C
135
Case 1: C excludes origin;
!-ydx+xdy
x2+y
2
C!" =
= !dxC!" +!dy = (
#!
#xS"" $
#"
#y)dxdy
= [ y
2 $x2
(x2+y
2 )2S
"" $ y
2 $x2
(x2+y
2 )2]dxdy = 0
136
X
Y
C
C0L1 L2
Case 2 : C encloses origin. In this case construct a closed contour which excludes
the origin (see diagram). Here we take C0to be a small circle of radius a.
-ydx +xdy
x 2 +y2C+L1+L2+C0" = 0 becase C+L1 +L2 +C0 excludes O.
-ydx +xdy
x2+y
2C
" = #-ydx +xdy
x2+y
2C0
"
On circle set x =a cost, y =a sint (Note : t decreases from 2$ to 0)
-ydx +xdy
x2 +y2C" = #
#asint(#asint)+a cost(a cost)
a 22$
0
" dt
= 2$
Note: C+C0is traversed so that the S is to the "left " of C
(positively directed - with respect to enclosed area which has
a normal in the k direction).
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137
Divergence Theorem (3D)
S: Closed surface outward normal n
V : Volume of S and its interior
"#
"xkV$$$ dV= nk#
S
$$ dS (k=1,2, 3)
Under suitable restrictions :
[ Rule:"
"xk on V becomes nk on S]
Notation : As before
f(x)!Cm on domain Dif fhas continuous partial
derivatives (formed by values offon Donly) of orders 0,1,.....m
(e1,e1,e1)" (i,j,k)
(x1,x2,x3)" (x,y,z)
138
V : f(x,y)"z "g(x,y), f,gcontinuous on Sxy
+two similar expressions describe V
#$
#zV%%% dxdydz = dxdy
Sxy
%% #$
#zdz
f
g
%
= $(x,y,g)dxdySxy
%% & $(x,y,f)dxdySxy
%%
nzdS= dxdy on z =g(x,y)
nzdS= &dxdy on z = f(x,y)
' #$
#zV%%% dxdydz = nz$
S
%% dS
Similarly
#
#xV%%% ,
#
#yV%%%
X
Y
Z
n = n+ (nz > 0)
n = n- (nz< 0 )
z = g(x,y)
z = f(x,y)
Sxy
Standard Domain
139
Alternative form:
Set "=ak (k=1,2,3)
a = (a1,a
2,a
3) = (a
x,a
y,a
z)
#a1
#xV$$$ dxdydz = n1a1
S
$$ dS
#a2
#yV$$$ dxdydz = n2a2
S
$$ dS
#a3
#zV$$$ dxdydz = n3a3S$$ dS
and sum;
(#a
1
#x+#a
2
#y+#a
3
#zV$$$ )dxdydz = (a.n)
S
$$ dS
divaV
""" dV= (n.a)dSS
"" = a.dSS
""140
(a) Proved for standard domain V
(b) True for sum of standard domainsVn
:
Apply result to each Vn. Integrals along common boundaries cancel.
(c) Sufficient Restrictions:
Boundary Snof each V
n is orientable sum of simple smooth surfaces
!!C1 on each Vn
!!C0 on V
Finite number of standard domains
V1 V2
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141
Example : A surface Sencloses a volume V.
Show that the volume Vis given by
V=1
3(r.n)dS
S
""
where ris the position vector.
We use the divergence theorem:1
3(r.n)dS
S
"" =1
3r.dS
S
""
=
1
3div rdV
V"""
=
1
33 dV
V"""
= dV""" =V
142
X Y
!
z=1-x-y
y=1-x
Example: Verify the divergence theorem for the tetrahedron bounded
by the coordinate planes and the plane x+y+z=1 and the vector field
A=3x2i+xy j+zk
div A = 6x +x +1= 7x +1
(7x +1)V
!!! dV = (7x +1)dzdydx =11`
240
1"x"y
!0
1"x
!0
1
!
A.ndSSxy
!! = (3x2Sxy
!! i+xy j+ok).("k)dS= 0
A.ndSSyz
!! = (0Syz
!! i+xy j+zk).("i)dS= 0
A.ndSSzx
!! = (3x2Szx
!! i+0j+zk).("j)dS= 0
143
On inclined plane, we parametrise explicitly in terms of (x,y) :
Equation of plane is
r =xi +y j+(1-x -y)k
"r
"x= i # k,
"r
"y= j# k
"r
"x $
"r
"y = i+ j+ k
dS= (i +j+ k)dxdy
A.S
%% dS= (3x2
Sxy
%% i +xy j+zk).(i+ j+ k)dxdy
= (0
1#x
%0
1
% 3x2+xy+ (1#x #y))dydx =
11
24
144
2-D divergence theorem from Greens Theorem Consider a vector field a = i!!j"and a closed 2D curve C enclosing a surface S
divS
"" adxdy = aC
!" .!dl (Divergence Theorem-2D)
Now unit tangent " =dr
dl= i
dx
dl+j
dy
dl. Hence unit normal is
!= idy
dl!j
dx
dl ("#k= !)
and diva =
#$
#x!#%
#y
[#$
#x!#%
#y]
S
"" dxdy = [i$!j%]C
!" .[idy
dl!j
dx
dl]dl (2D divergence T)
[!"
!x!!#
!y]
S
"" dxdy = #dx+C
!" !dy (Green's Theorem in the Plane!)
X
y
C!
"
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145
2D Stokes Theorem from Greens theorem
X
y
C!
"
Let abe a 2-D vector field given by
a = i!(x,y)+j"(x,y) and let kbe unit vector in zdirection.
curla = (0,0,!#!
#y+
#"
#y)
k. curladxdyS
"" = (!#!
#y+
#"
#y)dxdy
S
""
= !dxC!" +!dy (using 2D Green's Theorem)
= (a.")C!" dl
= a.C
!" dl
For the enclosing area in the xyplane, dS= kdxdy
#k. curladxdyS
"" = curla. dSS
"" = a.C
!" dl
This is 2DStoke's Theorem which generalises to 3D as we now show. 146146
S: orientable smooth surface with a field of normals n
C: Edge of Spositively directed with respect to n
Under suitable restrictions:
(i) n!"!S## dS = !dx
C
!# = !"dlC
!#
(ii) ("!aS
## ).dS = a.!dlC
!# $$$$$ (standard form)
Stokess Theorem (3D)
(The Curl Theorem)
X
Y
Z
nSurface S
Boundary C
!
a
147
div (curl a) = 0 identically
! curlS
"" a.dS= div curlV
""" a dV= 0 - any closed S
Let S=S1#S2 : Ccommon boundary
curlS1
"" a. ndS+ curl a. ndSS2
"" = 0
curlS1
"" a. n1dS+ curl a. (-n2 )dSS2
"" = 0
curlS1"" a. n1dS= curl
S1"" a. n2dS
With n2 taken as the inward normal, Cis positively
directed with respect to both surfaces.
$ Surface integral over orientable surfaces with same boundary curve
must have same value!. Therefore the value can depend only on integral
over the boundary curve.
X
Y
Zn1
Surface S1
C
Surface S2
n2Volume V
A plausibility argument
148
X
Y
Zn1
Surface S1
C
Surface S2
n2Volume V
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Proof of 3D Stokes theorem Consider a simple smooth surface S
Let x = g(u, v) be a mapping which shows it to be so.
Then the unit "outward" normal is
n+
=xu !xv
xu !xv
and the element of area is
dS= xu !xv dudv
150
X
Y
Zn
Surface S
C
!
Suv
Cuv
u
v
Then
n+
!"!dS=xu!xvxu!xv
!"! xu !xv dudv
= [(xu."!)xv# (xv."!)xu]dudv
[n+
!"!]idS= [(xu."!)$x
i
$v # (xv."!)
$xi
$u ]dudv
="!
"xk
"xk
"u
"xi
"v#"!
"xk
"xk
"v
"xi
"u
%&'
()*
dudv (implicit summation over k)
=
"!
"u
"xi
"v#"!
"v
"xi
"u
="
"u(!
"xi
"v)#
"
"v(!
"xi
"u)
151151
n+
!"!dSS
##$%&
'()
i
=
"
"u(!
"xi
"v)*
"
"v(!
"xi
"u)
$%&
'()
Su
## dudv
= (Cu
!# !"x
i
"udu+!
"xi
"vdv) (by Green's Theorem for plane)
= !dxi
C
!#
+ n+
!"!dSS
## = !dxC
!#
(ii) For an orientable sum of simple smooth surfaces, apply the result to each,
integrals along common boundaries cancel.
Sufficient Restrictions: !, C1 on each Sn, !, Con S
152
(iii) Set != a1, take the 1st
component of above.
Now (n!"!)=
e1 e2 e3
n1 n2 n3
"a1
"x1
"a1
"x2
"a1
"x3
#
$
%%%%%%
&
'
((((((
) [n2"a1
"x3* n3
"a1
"x2]dS
S
++ = a1 dx1C
!+
!= a2, take the 2nd
component
[n3 "a2
"x1* n1 "a2
"x3]dS
S
++ = a2 dx2C
!+ and similarly
[n1"a3
"x2* n2
"a3
"x1]dS
S
++ = a3 dx3C
!+
Adding
[n1S
++ ("a3
"x2*"a2
"x3)+n2 (
"a1
"x3*"a3
"x1)+...]dS= (a1 dx1 + a2dx2
C
!+ +..)
n. curlaS
++ dS= a. dxC
!+ = aC
!+ .#dl
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153
The divergence theorem can be used to justify thecoordinate free definition of divergence (recall we previouslyused the special case of a rectangular volume element to demonstrateits plausibility)
Likewise, Stokess theorem can be used to justify thecoordinate free definition of curl.
154
Consider a surface S given by the part of the sphere
x2+y
2+z
2=1
bounded by the planes z = 0, z =1
2; a=zi +x j+yk
Unit outward normal (radial):
n = sin"cos#i +sin"sin#j+cos"k ($3%"% $
2,0 % #% 2$)
dS=&r
&"'&r
d"d#= sin"d"d#
((' aS
)) ).ndS= (i +j+ k).(sin"cos#i +sin"sin#j+cos"k)S"#
)) sin"d"d#
= (0
2$
)$ 3
$ 2
) sin2"cos#+sin
2"sin#+sin"cos")d"d#
= 2$ sin"cos"$ 3
$ 2
) d"=$
4
Z
X
Y
C1
C2 nn
!/3
Example: 3D Stokes theorem
155
Now consider the line integrals along C 1and C2.
On C1 :r = costi + sin t j+ 0k (t: 0" 2#)
a
C1
$ .dr = a0
2#
$ .dr
dtdt= (0i + cost j+ sintk).(
0
2#
$ %sin ti + cost j+ 0k)dt
= cos2
0
2#
$ tdt= #
On C2 :x2 +y 2 = ( 32
)2
r =3
2costi +
3
2sint j+
1
2k (t: 2#" 0)
a
C1
$ .dr = a2#
0
$ .dr
dtdt= %
3#
4 (check!)
Hence sum is#
4
Z
Y
C1
C2 nn
156
Now consider the line integrals along C 1and C2.
On C1 :r = costi + sin t j+ 0k (t: 0" 2#)
a
C1
$ .dr = a0
2#
$ .dr
dtdt= (0i + cost j+ sintk).(
0
2#
$ %sin ti + cost j+ 0k)dt
= cos2
0
2#
$ tdt= #
On C2 :x2+y 2 = ( 3
2)2
r =3
2costi +
3
2sint j+
1
2k (t: 2#" 0)
a
C1
$ .dr = a2#
0
$ .dr
dtdt= %
3#
4 (check!)
Hence sum is#
4
Z
Y
C1
C2 nn
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157
Greens Theorems (I and II)If !and "are scalar fields, the following theorems follow;
(!!2"V
""" +!!.!")dV = !#"
#nS
"" dS
where !# C1 piecewise on V, "# C2 piecewise on V
(!!2"V
""" $"!2!)dV = (!#"
#nS
"" $"#!
#n)dS
where !,"# C2 piecewise on V
158
("#2$V
%%% +#".#$)dV = "&$
&nS%% dS
Using
div("a)= "diva +#".a
"div(grad$) = div("grad$)'grad".grad$"div(grad$
V
%%% )dV= [div("grad$)' grad".grad$]dVV
%%%
= "grad$.dSS
%% ' grad".grad$dVV
%%%
( ("#2$V
%%% +#".#$)dV = "&$
&nS%% dS
Proof of first theorem
159
An application of integral IdentitiesEquation of continuity in fluid dynamics :
Rate of increase of mass of material within
a fixed surface S = rate of flow of mass (mass flux) intoS
i.e."
"t
#V
$$$ dV = "#
"tV$$$ dV= % (#v
S
$$ .n)dS
(V
$$$ "#
"t+div #v)dV= 0 applying divergence theorem.
But above is true for every volume V.
"#
"t+div #v = 0 %%%%% at all points of material.
n
V
160
Irrotational (or conservative) vector fieldsDefinition: F(x) is said to be irrotational (or curl free) in a domain D if curl F=0 in D
If F(x)("C1) is irrotational on a simply connected volume D, then
there exists a scalar potential #such that F=$#. Conversely,
if there exists #such that F=$#, then Fis irrotational.
(a) suppose $%F= 0 (that is, Fis irrotational)
Take a fixed reference point B " D, and define
#(xP )= F.dxB
P
&
#is independent of the path joining B and P because
F.d xBMP
& ' F.dxBNP
& = F.dxBMPNB
& = nS
&& .curlFdS (Using Stoke's Theorem)
= 0
B
P
S
N
M
Theorem
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161
B
P
Q
X1
X2
X3
!x1
curl F= 0 " F=#$
Now consider neighbouring points P and Q where PQ is
in the x1direction :
"#
"x1
$
%&
'
()P
= lim#Q*#
P
PQ
= lim+x1,0
1
+x1F
1(-,
x1P
x1P++x1
. x2P ,x3P )d-=F1(xP )
(using MVT)
Similarly for the other two components of/#
0F=/#
(b)Now suppose F=/#
Then we have, using the standard identity,
/1F=/1/#= 0
162
UniquenessSuppose "# is another potential for the vector field F.Then F=$ "# and F=$#. Writing U=#% "# we see that$U= 0
or
&U
&x=
&U
&y=
&U
&z= 0
'U=#% "# = a constant.The scalar potential is unique apart for an arbitrary constant.
163
An equivalent theoremFis irrotational on a simply connected domain D
if and only if the work integral
F.dxCAB
!
is independent of path connecting A and B,
for all pairs of points A,B in D.
Note: In a more general form of the above theorems,
F is allowed to be undefined at "exceptional points"
(a finite number). A scalar potential still exists, but it is
also not defined at this point (cf: force field of gravity)164
Irrotational fields (example)