EE2007: Engineering Mathematics II - NTULearn · EE2007: Engineering Mathematics II Vector Calculus ... Div, Grad, Curl and all that, an informal text on vector ... This makes vector

EE2007: Engineering Mathematics IIVector Calculus

Ling KV

School of EEE, NTU

[email protected]

Rm: S2-B2b-22

Ver 1.1: Ling KV, October 22, 2006

Ver 1.0: Ling KV, Jul 2005

EE2007/Ling KV/Aug 2006

EE2007: Engineering Mathematics IIVector Calculus

My part:

week 10 Introduction, Dot and Vector Products, Scalar Triple Productweek 10-11 Differential Calculusweek 11-12 Integral Calculusweek 13 Revision Classes

Text: Kreyszig, E. (1999). Advanced Engineering Mathematics, Chapters 8-9, 8thEd. John Wiley & Sons, Inc. [NTU Library, QA401.K92]

Reference:

1. H. M. Schey (1997). Div, Grad, Curl and all that, an informal text on vectorcalculus, W.W. Norton & Company, Inc. [QA433S328]

EE2007/Ling KV/Aug 2006 1

Overview

This module deals with vectors and vector functions in 3D-space and

extends the differential calculus to these vector functions. Forces,

velocities and various other quantities are vectors. This makes vector

calculus a natural instrument for engineers. In 3D space, geometrical

ideas become influential, and many geometrical quantities (tangents

and normals, for example) can be given by vectors.

We will first review the basic algebraic operation with vectors in 3D

space. Vector functions, which represent vector fields and have

various physical and geometrical applications, will be introduced

next. Three physically and geometrically important concepts related

to scalar and vector fields, namely, the gradient, divergence and curl

will be discussed. Integral theorems (Vector Integral Calculus)

involving these concepts will then follow.


Introduction

Length, temperature and

voltages are quantities

determined by its .

They are .

Force and velocity are quantities

determined by both and

. They are . A vector

is represented as an arrow or a

.


In printed work, we denote vectors by lowercase boldface letters, a,

v, etc; in handwritten work, we often denote vectors by arrow, ~a, ~v.

We sometimes also write the vector as an order pair of real numbers

and use boldface letter to emphasize that it represents a vector.

Thus r = [x, y, z].

The vector r has magnitude |r| = and the unit vector in the

direction of r is

r̂ =1|r|

r = .

A unit vector has magnitude , and can be used to indicate a

particular director. We can write r = |r|̂r .


Equality of Vectors

Two vectors a and b are equal, written a = b, if they have the same

and the same . Hence, a vector can be arbitrarily ,

that is, the initial point can be chosen arbitrarily. This definition is

practical in connection with forces and other applications.


Position Vector

A Cartesian coordinate system being given, the position vector, r of

a point A : (x, y, z) is the vector with the origin (0, 0, 0) as the

initial point and A as the terminal point.

Thus, a position vector can be

written as

r = xi + yj + zk, or r = [x, y, z]where i, j, k are standard

Cartesian unit vectors along the

x, y, z directions.


Examples

• Find the components of the vector v, given the initial point

P : (1, 0, 0) and the terminal point Q : (4, 2, 0). Find also |v|.

• If p = [1, 3,−1],q = [5, 0,−2], and u = [0,−1,−3] represent

forces. What is the resultant force?

• Determine the force p such that p, q = [0, 3,−4], and u = [1,−1, 0]are in equilibrium.

• If |p| = 6 and |q| = 4, what can be said about the magnitude and

direction of the resultant?


Geometrical Application

Using vectors, prove that the diagonals of a parallelogram bisect

each other.

The vector r from O to P is of

the form r = k(a + b). We also

have r = a + l(b − a). Thus,

ka + kb = (1 − l)a + lb. From

this, k = 1 − l, k = l, hence

k = 12, l = 1

2, and the proof is

complete.


Inner Product (Dot Product) of Vectors

The inner product or dot product, a · b of two vectors

a = [a1, a2, a3] and b = [b1, b2, b3] is defined as

a · b = |a||b| cos γ if a 6= 0, b 6= 0a · b = 0 if a = 0 or b = 0

where 0 ≤ γ ≤ π is the angle

between a and b (measured when

the vectors have their initial

points coinciding).

Note that the dot product is a .

In components1, a · b = a1b1 + a2b2 + a3b3 .1see text pp.410 for a derivation


Inner Product

• The inner product of two nonzero vectors is zero if and only if

these vectors are .

• [Length and angle in terms of inner product]a · a = |a|2, so that we have

|a| =√

a · a

We can also obtain the angle γ between two nonzero vectors

cos γ = a·b|a||b| = a·b

√a·a√

b·b


Component or Projection

The concept of the component or projection of a vector a in the

direction of another vector b(6= 0) is an important one.

From the figure,

p = |a| cos γ

Thus, p is the of the orthogonal projection of a on a straight

line l. Multiply p by |b|/|b|, we have

p = a·b|b| , (b 6= 0) since a · b = |a||b| cos γ.


General Properties of Inner Products

For any vectors, a,b, c and scalars q1, q2,

[q1a + q2b] · c = q1a · c + q2b · c (Linearity)a · b = b · a (Symmetry)

a · a ≥ 0a · a = 0 iff a = 0

}(Positive− definiteness)

(a + b) · c = a · c + b · c (Distributive)|a · b| ≤ |a||b| (Schwarzinequality)

|a + b| ≤ |a|+ |b| (Triangleinequality)|a + b|2 + |a− b|2 = 2(|a|2 + |b|2) (Parallelogramequality)


Dot Product Example

If a = [1, 2, 0] and b = [3, −2, 1], find the angle between the two

vectors.

[ 96.865o ]


Applications of Inner Products

• [Work done by a force] Find the work done by a force p = [2, 6, 6]acting on a body if the body is displaced from point A : (3, 4, 0)to point B : (5, 8, 0) along the straight line segment AB. Sketch

p and AB.

[ 28 ]


• [Component of a force in a given direction] What force in the

rope will hold a car of 5000kg in equilibrium?

[ 2113kg ]


Vector Product (Cross Product)

The vector product (cross product) v = a× b of two vectors

a = [a1, a2, a3] and b = [b1, b2, b3] is a vector

whose magnitude is given by

|v| = |a||b| sin γ

and the direction of v is

perpendicular to both a and band such that the three vectors

a, b, v, in this order, form a

right-handed triple.

Thus, if a and b have the same or opposite direction or if one of the

these vectors is the zero vector, then a× b = 0.


In component,

v = a× b =

∣∣∣∣∣∣∣i j k

a1 a2 a3

b1 b2 b3

∣∣∣∣∣∣∣Example: Let a = [4, 0,−1] and b = [−2, 1, 3]. Then

v = a× b =

∣∣∣∣∣∣∣i j k4 0 −1−2 1 3

∣∣∣∣∣∣∣ = i− 10j + 4k = [1,−10, 4].


General Properties of Vector Products

For any vectors a,b, c and scalar l,

(la)× b = l(a× b) = a× (lb)a× (b + c) = (a× b) + (a + c)(a + b)× c) = (a× c) + (b + c)

}(distributive)

b× a = −(a× b) (anti− communtative)a× (b× c) 6= (a× b)× c (not associative)


Typical Applications of Vector Products

Find the moment of a force p about a point Q.


Find the velocity of the rotating disk, whose angular speed is ω(> 0).

Let P be any point on the disk

and d its distance from the axis.

Then P has a speed ωd. Let r be

the position vector of P referred

to a coordinate system with the

origin on the axis of rotation.

If we define a vector w whose direction is that of the axis of rotation

and such that the rotation appears clockwise if one looks from the

initial point of w to its terminal point. Then length of w is equal to

the angular speed ω.


Then

ωd = |w||r| sin γ = |w× r|where γ is the angle between wand r.

Thus we see that the velocity vector v of P can be represented in

the form

v = w× r


Scalar Triple Product

The scalar triple product or mixed triple product of three vectors

a = [a1, a2, a3], b = [b1, b2, b3], c = [c1, c2, c3]

is denoted by (a b c) and is defined by

(a b c) = a · (b× c) =

∣∣∣∣∣∣∣a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣∣It can be shown that

(ka kb kc) = k(a b c) and a · (b× c) = c · (a× b)


Fact: a · (b× c) = c · (a× b) = b · (c× a)

Proof:

a · (b× c) =

∣∣∣∣∣∣∣a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣∣= −

∣∣∣∣∣∣∣c1 c2 c3

b1 b2 b3

a1 a2 a3

∣∣∣∣∣∣∣ = c · (a× b)

=

∣∣∣∣∣∣∣b1 b2 b3

c1 c2 c3

a1 a2 a3

∣∣∣∣∣∣∣ = b · (c× a)


Geometric Interpretation of Scalar Triple Product

Show that the absolute value of the scalar triple product is the

volume of the parallelepiped with a,b, c as edge vectors.


Parametric Representation of Curves

A curve in space is determined

by three equations of the form

x = x(t), y = y(t), z = z(t)Alternatively, a curve C in space

can be represented by a position vector

r(t) = [x(t), y(t), z(t)] = x(t)i + y(t)j + z(t)k

and t is a parameter specifying points (x(t),y(t),z(t)) along the

curve. The parameter t may be time or something else.

For example, a straight line is represented by

x(t) = 1 + 3t, y(t) = 2− 5t, z(t) = 3 + 3t


Examples: Parametric Representation of Curves

• Straight line. A straight line L through a point A with position

vector a in the direction of a constant vector b can be represented

in the form

r(t) = a + tb= [a1 + tb1, a2 + tb2, a3 + tb3].


• Ellipse and Circle. The vector function

r(t) = [a cos t, b sin t, 0]

represents an ellipse in the xy-plane with centre at the origin and

principal axes in the direction of the x and y axis. In fact, since

cos2 t + sin2 t = 1, we obtain

x2

a2+

y2

b2= 1, z = 0


• Circular Helix

The twisted curve C represented

by the vector function

r(t) = [a cos t, b sin t, ct], (c 6= 0)

is called a circular helix.


Application: Parametric Representation of Curves

We can use parametric equations to find the intersection of two curves.

Example: Two particles move through space with equationsr1(t) = [t, 1 + 2t, 3− 2t] and r2(t) = [−2− 2t, 1− 2t, 1 + t]. Do the particlescollide? Do their paths cross?

Solution: The particle will collide if .

⇒

Their paths cross if

⇒

⇒


Representation of Surfaces

A surface S in the xyz-space can be represented as

z = f(x, y) or g(x, y, z) = 0

For example, z = +√

a2 − x2 − y2 or x2 + y2 + z2 = 0, z ≥ 0represents a hemisphere of radius a and centre at the origin.

Just like the parametric

representation of a curve in

space, we could also represent a

surface parametrically. Surfaces

are two-dimensional, hence we

need two parameters, which we

denote as u and v.


Thus a parametric representation of a surface S in space is of the

form

r(u, v) = [x(u, v), y(u, v), z(u, v)], (u, v) in R.

where R is some region in the uv-plane.


Example: Parametric Representation of a Cylinder

x2 + y2 = a2, −1 ≤ z ≤ 1, represent a cylinder of radius a, height 2

in the z-direction.

A parametric representation of

this cylindrical surface is

r(u, v) = [a cos u, a sinu, v]where the parameters u, v vary

in the rectangle

R : 0 ≤ u ≤ 2π, −1 ≤ v ≤ 1


Example: Parametric Representation of a Sphere

A sphere xy + y2 + z2 = a2 can be represented in the form

r(u, v) = [a cos v cos u, a cos v sinu, a sin v]

where the parameters u, v vary in

the rectangle

R : 0 ≤ u ≤ 2π, −π/2 ≤ v ≤ π/2


Parametric Representations of Common Surfaces2

2Problem Set 9.5, p.495


Vector and Scalar Functions and Fields

Vector functions are functions whose values are vectors

v = v(P ) = [v1(P ), v2(P ), v3(P )]

depending on the point P in space.

A curve or surface in space which are described parametrically in

vector form, are examples of vector functions.

Scalar functions are functions whose values are scalars

f = f(P )

depending on P . An example of scalar functions is the spatial

distribution of temperature in a room.


A vector field is defined by a vector function within region of space

or surface in space or a curve in space.

Field of tangent vectors of a curve Field of normal vectors of a surface

Similarly, a scalar function defines a scalar field in a region on a

surface or curve. Examples of scalar field are temperature field in a

body and the pressure field of the air in the earth’s atmosphere.

Vector fields can be used to describe fluid flows, electromagnetism

and heat transfer in space. Vector and scalar functions may also

depend on time t or on other parameters.


Examples of Scalar and Vector Functions

• The distance f(P ) of any point P from a fixed point Po in space

is a scalar function

f(P ) = f(x, y, z) =√

(x− xo)2 + (y − yo)2 + (z − zo)2

•The velocity vector v(P ) of a

rotating body constitute a vector

field, the so-called velocity field

of the rotation.

v(x, y, z) = w× r = w× [x, y, z]


•According to Newton’s law of

gravitation, a particle of mass M

at a fixed point Po will have an

attractive force p on a particle in

space at point P . Its magnitude

is proportional to 1/r2 where r is

the distance between P and Po,

p = −GMm

r2r̂

where r̂ is the unit vector in the (outward) radial direction.


Examples of Vector Functions

F(x, y) = xi + yj G(x, y) =−yi + xj√

x2 + y2


Derivative of a Vector Function

The derivative of a vector function is defined as

v′(t) = lim∆t→0

v(t + ∆t)− v(t)∆t

In components,

v′(t) = [v′1(t), v′2(t), v

′3(t)]

A vector function is said to be differentiable at a point t if the

above limit exists.

If the vector function v(t) represents a curve C is space,

then v′(t) is a vector tangent to the curve C at the point

P : (x(t), y(t), z(t)), provided v(t) is differentiable and v′(t) 6= 0.


Note that the usual rules of differentiation carry over to vector

functions:

(u · v)′ =(u× v)′ =(u v w)′ =


Derivative of a Vector Function of Constant Length

Let v(t) be a vector function whose length is constant,

|v(t)| = c

Then

|v|2 =

and

(v · v)′ =

Thus, the derivative of a vector function v(t) of constant length is

either or is


Partial Derivatives of a Vector Function

Let r(t1, t2) = a cos t1i + a sin t1j + t2k. Then

∂r

∂t1= −a sin t1i + a cos t1j

∂r

∂t2= k


Tangent to a Curve

In slide 40, we mentioned that

if curve C is represented by r(t),then the tangent to the curve C

is r′(t).The unit tangent vector is then

given by u = 1|r′|r

′ .

The tangent to C at an arbitrary point P is given given by

q(w) = r + wr′ ,

where w is a scalar parameter.


Example: Tangent to an Ellipse

Find the tangent to the ellipse

14x2 + y2 = 1 at P : (

√2, 1/

√2).

Solution:

The ellipse can be represented as

r(t) =

Hence

r′(t) =

and P corresponds to t = . Thus, the tangent to the ellipse is

q(w) = .


Example, Helix, its Tangent Vector and Derivativeof Tangent


Velocity and Acceleration

Curves as just discussed from the viewpoint of geometry also play an

important role in mechanics paths of moving bodies. Consider a

path C given by r(t), where t is now time. Then, the derivative

r′(t) = ddtr(t)

is the velocity vector of the motion, and the acceleration vector is

a(t) = r′′(t).


Example: Centripetal Acceleration

The vector function r(t) = R cos ω(t)i + R sinωtj represents a circle

C of radius R with centre at the origin of the xy-plane and describes

a motion of a particle P in the counter-clockwise sense.

The velocity vectorv = r′ =

is to C, and its magnitude( ) is

|v| =√

r′ · r′ = is .

The angular speed is equal to . The acceleration vector is

a = v′ =

which is of a magnitude , with a direction . The

vector a is the .


Tangent Plane and Surface Normal

Recall that (in slide 40), we mentioned that if curve C is represented

by r(t), then the tangent to the curve C is the derivative dr(t)dt .

Thus, given the parametric representation of the surface S : r(u, v),we could obtain the surface normal as

N = ru × rv

where ru and rv are partial derivatives of r(u, v) wrt u and v

respectively.

The unit normal vector is

n =1|N|

N


Example: Normal Vector to a Spherical Surface

Since a parametric representation of the sphere is

r(u, v) = [a cos v cos u, a cos v sinu, a sin v]

so,

ru = [−a cos v sinu, a cos v cos u, 0]

rv = [−a sin v cos u,−a sin v sinu, a cos v]

and this gives

N = ru × rv = a2 cos v[cos v cos u, cos v sinu, sin v], and

|N| = a2 cos v


Hence

n =1|N|

N = [cos v cos u, cos v sinu, sin v] =1a[x, y, z]


Gradient of a Scalar Field

For a give scalar function f(x, y, z), the gradient of f is the vector

function3

grad f = ∂f∂xi +

∂f∂y j + ∂f

∂zk .

Introducing the differential operator (read nabla or del),

∇ = ∂∂xi +

∂∂yj + ∂

∂zk

we write

grad f = ∇f = ∂f∂xi +

∂f∂y j + ∂f

∂zk .

For example, if f = 2x2 + yz − 3y2, then ∇f = 2i + (z − 6y)j + yk.3assuming, of course, f is differentiable


∇f and the Directional Derivative

Gradient vector (∇f) arises when

we attempt to define a rate of

change for a scalar function of

several variables, f(x, y, z).

The rate of change of f(x, y, z) at a point P in the direction of a

vector b is called the directional derivative of f at P in the direction

of a vector b:

Dbf =∇f · b|b|

Geometrically, Dbf represents the slope of surface f(x, y, z) in the

direction of b.

Example: Given f = x2 + y2, then ∇f = 2xi + 2yj.


Example: Directional Derivative

Find the directional derivative of f(x, y, z) = 2x2 + 3y2 + z2 at the

point P : (2, 1, 3) in the direction of the vector b = i− 2k.

Solution: We obtain

∇f = [4x, 6y, 2z]

and at P , ∇f = [8, 6, 6]. Thus,

Dbf =[8, 6, 6] · [1, 0,−2]

|[1, 0,−2]|= − 4√

5

The minus sign indicates that f decreases at P in the direction of b.


Example: Gradient and Directional Derivative

See Problem Set 18.9. p.452

The force in an electrostatic field f(x, y, z) has the direction of

grad f = ∇f . Hence, if f(x, y) = x2 + 9y2, then the force at

P : (−2, 2) is ∇f = [2x, 18y] = [2(−2), 18(2)] = [−4, 36].

The flow of heat in a temperature field takes place in the direction

of maximum decrease of temperature T . Hence if the temperature

field is T (x, y) = x/y, then the heat will flow in the direction

−∇T = −[1/y, −x/y2].


∇f Characterizes Maximum Increase

Let f(P ) = f(x, y, z) be a scalar function having continuous first

partial derivaties. Then ∇f exists and its length and direction are

independent of the particular choice of Cartesian coordinates in

space. If a a point P , the gradient of f is not the zero vector, it has

the direction of maximum increase of f at P .

Proof: see text pp.448.

Example: If on a mountain the elevation above sea level is

z(x, y) = 1500− 3x2 − 5y2 (meters), what is the direction of

steepest ascent at P : (−0.2, 0.1)?

Solution: The direction of the steepest ascent is given by

∇z = [−6x, −10y] and at P , ∇z = [1.2, −1].


Gradient (∇f) as a Surface Normal Vector

Consider a surface S in space

given by

f(x, y, z) = c = const

Recall that a curve C in space can be represented by

r(t) = [x(t), y(t), z(t)]

If we want C to lie of S, then its components must satisfy

f(x(t), y(t), z(t)) = c

If we differentiate the above with respect to t, we get by chain rule,

∂f

∂xx′(t) +

∂f

∂yy′(t) +

∂f

∂zz′(t) = [

∂f

∂x,

∂f

∂y,

∂f

∂z] · r′(t) = 0


where r′(t) = [x′(t), y′(t), z′(t)], which is a tangent vector of C.

Since C lie of S, this vector is tangent to S. What this means is

that the vector

∇f = [∂f

∂x,

∂f

∂y,

∂f

∂z]

is perpendicular to tangent vector of S at P . Thus, we conclude:

Let f be a differentiable scalar function that represents a surface

S : f(x, y, z) = c. The gradient of f (i.e. ∇f) at a point P of S

is a normal vector of S at P is this vector is not the zero vector.


Example: Normal Vector to a Surface

Example: Find an equation of the plane tangent to the surface

x2 + 3y3 + 2z3 = 12 at the point (1,−1, 2).

Solution: A vector normal to the surface S : f(x, y, z) = c, where

f(x, y, z) = x2 + 3y3 + 2z3 is

∇f = [2x, 9y2, 6z2]

Recall that the equation of a plane has the form N · (r− a) = 0where N is a normal vector to the plane, and is a point on the plane.

Thus, the required equation is

[2x, 9y2, 6z2] · (r− [1,−1, 2]) = 0


Surface Normal of a Spherical Surface Re-visited

The sphere g(x, y, z) = x2 + y2 + z2 − a2 = 0 has a unit normal

vector given by

n = 1

|∇g|∇g

= 1

2√

x2+y2+z2[2x, 2y, 2z]

= 1a[x, y, z]

This agrees with the answer given in slide 50 where the surface

normal was computed using

n =1|N|

N where N = ru × rv


Example: Gradient (∇f) as Surface Normal Vector

Find a unit normal vector n of the cone of revolution

z2 = 4(x2 + y2) at the point P : (1, 0, 2).

Solution: The surface of the cone

can be represented by

f(x, y, z) = 4(x2 + y2)− z2 = 0Thus, ∇f = [8x, 8y,−2z] and

at P , ∇f = [8, 0,−4]. Hence, a

unit normal vector of the cone at

P is

n =1

|∇f |∇f =

1√5[2, 0,−1].

The other normal is -n.


Visualising Surface Normal, Contour, and Gradient


Conservative Vector Fields

Some vector fields can be obtained from scalar fields. Such a vector

field is given by a vector function v,

v = ∇f

The function f is called a potential function of v. Such a field is

called conservative because in such a vector field, energy is

conserved. Examples of conservative vector fields are gravitational

field and electrostatic field.


Example: Conservative Vector Field

Problem Set 8.9. p.452

Find f for a given v or state that v has no potential.

v = [yz, xz, xy] gives f = xyz + c where c is a constant.

v = [yex, ex, 1] gives f = yex + z + c where c is a constant.

v = [x, y]/(x2 + y2) gives f = 12 ln(x2 + y2) + c where c is a

constant.


A Physical Example: The Gradient Vector

If f(x, y, z) is a scalar-valued function describing, for example,

temperatures in a room, then Duf , the rate of change of f along a

fixed line whose direction is given by the unit vector u, is called the

directional directive of f in the direction of u. This derivative is a

spatial rate of change, not a temporal one. It gives the temperature

gradient, the change of temperature per unit length in a given

direction.

This gradient vector turns out to be orthogonal to the isotherms of

f , the surfaces along which the temperature described by f are

constant. The gradient vector points in the direction of increasing

temperatures, and the greatest rate of change in f is the length of

grad f .


Divergence of a Vector Field

Let v(x, y, z) = [v1, v2, v3] be a differentiable vector function, then

the function

div v = ∂v1∂x + ∂v2

∂y + ∂v3∂z

is called the divergence of v.

Another common notation for the divergence of v is ∇ · v,

div v = ∇ · v = ( ∂∂xi +

∂∂yj + ∂

∂zk) · (v)= ∂v1

∂x + ∂v2∂y + ∂v3

∂z

Example: If v = [3xz, 2xy,−yz2], then ∇ · v = .

Note that ∇ · v is a .


Physical Meaning of the Divergence

Roughly speaking, the divergence measures outflow minus inflow.

See also text pp.454, Examples 1 and 2. More on this later when we

talk about Divergence Theorem.

Example: Let v = [v1, v2, v3]represents the velocity of the

fluid flow.

Then the rate of fluid volume

flow across surface A is

v1(A)∆y∆z. ...


Visualizing the Divergence4

4(Problem Set 8.10, p.456) Plot the given velocity field v of a fluid flow in a square centered at

the origin. Recall that the divergence measures outflow minus inflow. By looking at the flow near

the sides of the square, can you see whether div v must be positive, or negative, or zero?


Curl of a Vector Field

Let v(x, y, z) = [v1, v2, v3] be a differentiable vector function. Then

the function

curl v = ∇× v =i j k∂∂x

∂∂y

∂∂z

v1 v2 v3

is called the curl of the vector function v.

Example: Let v = [yz, 3xz, z], then

∇× v =i j k∂∂x

∂∂y

∂∂z

yz 3xz z

= [−3x, y, 2z]


Example: Curl

We have seen earlier that the velocityfield of a rotating rigid body can berepresented in the form v = w×r wherewωk, and r = [x, y, z]. Thus,

v = w× r = [−ωy, ωx, 0]

Now, let compute

∇× v =i j k∂∂x

∂∂y

∂∂z

−ωy ωx 0=

Hence the curl of the velocity field of a rotating rigid body has the

direction in the axis of rotation, and its magnitude equals to twice

the angular speed of the rotation.


Summary, the ∇ Operator

∇ operates on field gives ∇f .

∇ operates on field gives ∇ · f or ∇× f.

If F = ∇f , then ∇× F = 0. Since the curl characterises the

rotation in a field, we say that F is irrotational. If F is not

associated with velocity, we usually say F is conservative.

Gradient vector (∇f) arises when we attempt to define a rate of

change for a scalar function of several variables (f(x, y, z)). This

vector is orthogonal to the level set of the scalar function (Surface

f(x, y, z) = c) and is tangent to the lines of flow of the gradient

field. The Gravitational and Electrostatic potentials are examples of

scalar fields for which the gradient field is physically significant.


Line Integrals

The concept of a line integral is a simple and natural generalisation

of a definite integral∫ b

af(x)dx.

If we represent the curve C by a parametric representation

r(t) = [x(t), y(t), z(t)], (a ≤ t ≤ b)

then the line integral of a vector function F(r) over a curve C is

defined by ∫C

F(r) · dr =∫ b

a

F(r(t)) · dr(t)dt

dt

In terms of components, with dr = [dx, dy, dz], the above becomes∫C

F(r) · dr =∫ b

a

[F1, F2, F3] · [dx

dt,dy

dt,dz

dt] dt


Example: Line Integral

• [pp.466] Find the value of the line integral when F(r) =[−y,−xy, 0] and C is the circular arc from A to B.

If F represents a force, then the line integral gives the by Fin the displacement along path C.


• [pp.466] Find the value of the line integral when F(r) = [z, x, y]and C is the helix r(t) = [cos t, sin t, 3t], (0 ≤ t ≤ 2π).


• [pp.467] Dependence of line integral on path


Line Integrals and Independent of Path

A line integral∫C

F(r) · dr =∫ b

a

[F1, F2, F3] · [dx

dt,dy

dt,dz

dt] dt

with continuous F1, F2, F3 in a domain D in space is independent of

path in D if and only if F is the gradient of some function f in D,

i.e. F = ∇f .

Proof: see text, pp.472.


Example: Independent of Path

Evaluate the integral∫C

F · dr =∫

C

[3x2, 2yz, y2] · [dx, dy, dz]

from A : (0, 1, 2) to B : (1,−1, 7) by showing that F is conservative.

Solution: F is conservative if there exists a scalar function f such

that F = ∇f . ......


Surface Integrals

Let the surface S be given by a parametric representation

r(u, v) = [x(u, v), y(u, v), z(u, v)], (u, v) in R

and S is piecewise smooth so that S has a normal vector

N = ru × rv and unit normal n =1|N|

N

Then, for a given vector function F, the surface integral over S is

defined as ∫ ∫S

F · ndA =∫ ∫

R

F(u, v) ·N(u, v) du dv

Note that the integrand is a scalar because we take dot product.

Indeed, F · n is the component of F normal to the surface. This


integrand arises naturally in flow problems where F represents the

velocity vector of the fluid and we sometimes call the surface

integral the flux integral.


Example: Flux Through a Surface

Compute the flux of water through the parabolic cylinder

S : y = x2, 0 ≤ x ≤ 2, 0 ≤ z ≤ 3

if the velocity vector is F = [3z2, 6, 6xz].


Solution: Let x = u and z = v, we have y = u2 and hence a

parametric representation of S is

S : r(u, v) = [u, u2, v], 0 ≤ u ≤ 2, 0 ≤ v ≤ 3

Thus,ru = [1, 2u, 0]rv = [0, 0, 1]N = ru × rv = [2u,−1, 0]F(u, v) = [3v2, 6, 6uv]

Hence∫ ∫SF · n dA =

∫ ∫R


=∫ 3

v=0

∫ 2

u=0[3v2, 6, 6uv] · [2u,−1, 0] du dv

= 72


Example: Surface Integral

If F = [x2, 0, 3y2] and S is the portion of the plane x + y + z = 1 in

the first octant. Evaluate the surface integral of F over S.


Solution: Let u = x and v = y,

we have z = 1− u− v. Hence a

parametric representation of S isS : r(u, v) = [u, v, 1− u− v],

0 ≤ u ≤ 1− v, 0 ≤ v ≤ 1

and

N = ru × rv = [1, 0,−1]× [0, 1,−1] = [1, 1, 1]

Hence∫ ∫SF · n dA =

∫ ∫R


=∫ 1

v=0

∫ 1−v

u=0[u2, 0, 3v2] · [1, 1, 1] du dv

= 13

Note that the normal vector can also be computed from the surface

function: g(x, y, z) = x + y + z − 1 = 0, i.e. N = ∇g = [1, 1, 1].


Divergence Theorem of Gauss

Let T be a closed bounded region in space whose boundary is a

piecewise smooth orientable surface S. Let F(x, y, z) be a vector

function that is continuous and has continuous first partial

derivatives in some domain containing T . Then∫ ∫ ∫T

∇ · (F ) dV =∫ ∫

S

F · n dA

where n is the outer unit normal vector of S.

Volume integral and surface integral are related through the

divergence theorem of Gauss. This is of practical interest because

one the two kinds of integral is often simpler than the other.


Evaluation of Surface Integral by the DivergenceTheorem

Evaluate∫ ∫

SF · n dA where F = [x3, x2y, x2z], S is the closed

surface consisting of the cylinder x2 + y2 = a2, 0 ≤ z ≤ b and the

circular disks at z = 0 and z = b.

Solution:


Stokes’s Theorem

Let S be a piecewise smooth oriented surface in space and let the

boundary of S be a piecewise smooth simple closed curve C. Let

F(x, y, z) be a continuous vector function that has continuous first

partial derivatives in a domain in space containing S. Then∫ ∫S

(∇× F) · n dA =∫

C

F · dr

where n is a unit normal vector

of S and, depending on n, the

integration around C is taken in

the sense shown in the figure.


Stokes’s Theorem and Path Independence

If ∇× F = 0, then by Stokes’s theorem,∫ ∫S

(∇× F) · n dA =∫

C

F · dr = 0

i.e., if F is conservative, the line integral is independent of path.


Verification of Stokes’s Theorem

Let F = [y, z, x] and S the

paraboloid

z = 1− (x2 + y2), z ≥ 0

The curve C is the circle r = [cos s, sin s, 0], 0 ≤ s ≤ 2π.

Consequently, the line integral is simply∫C

F · dr =∫ 2π

0

[sin s, 0, cos s] · [− sin s, cos, 0] ds = −π

On the other hand, ∇× F =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

y z x

∣∣∣∣∣∣∣ = [−1,−1,−1]


and

N = ∇(z − 1 + (x2 + y2)) = [2x, 2y, 1]

so that ∫ ∫S(∇× F) · n dA

=∫ ∫

R[−1,−1,−1] · [2x, 2y, 1] dx dy

=∫ ∫

R(−2x− 2y − 1) dx dy

=∫ 1

r=0

∫ 2π

θ=0(−2r cos θ − 2r sin θ − 1)r dθ dr

= −π


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EE2007: Engineering Mathematics II - NTULearn · EE2007: Engineering Mathematics II Vector Calculus ... Div, Grad, Curl and all that, an informal text on vector ... This makes vector