ANU MATH2405 Vector Notes

8/12/2019 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

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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)

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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)

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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

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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) ]

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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

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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

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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!

!

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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

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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]

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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

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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

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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

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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


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


curves

y

r

!

drd!

dr rd!

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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"


12/54

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

#


13/54

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


14/54

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.


15/54

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


16/54

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.


17/54

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)


18/54

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


19/54

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


20/54

7777

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


21/54

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


22/54

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


23/54

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


24/54

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


25/54

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


26/54

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#


27/54

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


28/54

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 "#


29/54

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


30/54

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


31/54

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!)


32/54

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.


33/54

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).


35/54

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


36/54

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!

"


37/54

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


38/54

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


39/54

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


40/54

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


41/54

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)

ANU MATH2405 Vector Notes

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