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ELEC 3600
TUTORIAL 2
VECTOR CALCULUS
Alwin Tam
[email protected]
Rm. 3121A
WHAT HAVE WE LEARNT SO FAR?
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Classification of vector & scalar fields
Differential length, area and volume
Line, surface and volume integrals
Del operator
Gradient of a scalar
Divergence of a vector
– Divergence theorem
Curl of a vector
– Stokes’ theorem
Laplacian of a scalar
SCALAR AND VECTOR FIELD

What is scalar field?


Quantities that can be completely described from its
magnitude and phase. i.e. weight, distance, speed,
voltage, impedance, current, energy
What is a vector field?
Quantities that can be completely described from its
magnitude, phase and LOCATION. i.e. force,
displacement, velocity, electric field, magnetic field
 Need some sense of direction i.e. up, down right and
left to specify

SCALAR AND VECTOR FIELD (CONT.)
Is temperature a scalar quantity?
A. Yes
B. No
Answer: A, because it can be completely described
by a number when someone ask how hot is today.

Is acceleration a scalar quantity?
A. Yes
B. No
Answer: B, because it requires both magnitude and
some sense of direction to describe i.e. is it
accelerating upward, downwards, left or right etc.

VECTOR CALCULUS

What is vector calculus?


Concern with vector differentiation and line, surface
and volume integral
So why do we need vector calculus??
To understand how the vector quantities i.e. electric
field, changes in space (vector differential)
 To determine the energy require for an object to
travel from one place to another through a
complicated path under a field that could be spatially
varying (line integral) i.e.

W=

𝑞𝑬 ∙ 𝑑𝒍
To pass ELEC 3600!! (vector differential and line
integral)
DIFFERENTIAL LENGTH, VOLUME AND
SURFACE (CARTESIAN COORDINATE)
Differential is infinitely small difference between 2 quantities

Differential length


Differential volume


A vector whose magnitude is
close to zero i.e. dx, dy and dz
→0
An object whose volume
approaches zero i.e. dv =
dxdydz → 0 (scalar)
Differential surface
A vector whose direction is
pointing normal to its surface
area
 Its surface area |dS| approach
zero i.e. shaded area ~ 0
 Calculated by cross product
of two differential vector
component

DIFFERENTIAL LENGTH, VOLUME AND
SURFACE (CYLINDRICAL COORDINATE)
All vector components MUST have
spatial units i.e. meters, cm, inch etc.
DIFFERENTIAL LENGTH, VOLUME AND
SURFACE (SPHERICAL COORDINATE)
z
All vector components MUST have
spatial units i.e. meters, cm, inch etc.
y
x
LINE INTEGRAL

Line integral: Integral of the tangential component of
vector field A along curve L.
2 vectors are involve inside the integral
 Result from line integral is a scalar

Line integral
Definite integral
Diagram
Maths description
Result
Information
required
A measure of the total
effect of a given field
along a given path
Area under the curve
1. Vector field
expression A
2. Path expression
1. Function f(x)
2. Integral limits
Integral limits depends on path
SURFACE & VOLUME INTEGRAL

Surface integral: Integral of the normal
component of vector field A along curve L.



Two vectors involve inside the integral
Result of surface integral is a scalar
Volume integral: Integral of a function f i.e.
inside a given volume V.


Two scalars involve inside the integral
Result of volume integral is a scalar
SURFACE & VOLUME INTEGRAL (CONT.)
Surface integral
Volume integral
Diagram
Maths description
Result
Information
required
A measure of the total
flux from vector field
passing through a given
surface
1. Vector field
expression A
2. Surface expression
Integral limits depends on surface
A measure of the total
effect of a scalar function
i.e. temperature, inside a
given volume
1. Scalar Function rv
2. Volume expression
Integral limits depends on volume
PROBLEM 1
2
2
F

x
a

xz
a

y
az , calculate the
 Given that
x
y
circulation of F around the (closed) path shown in
the following figure.
Solution:
𝐿
𝐹 ∙ 𝑑𝑙 = −1/6
DEL OPERATOR



Vector differential operator
Must operate on a quantity (i.e. function or
vector) to have a meaning
Mathematical form:
Cartesian
Cylindrical
Spherical
SUMMARY OF GRAD, DIV & CURL
Gradient
𝛻 must operate
on
Scalar f(x,y)
Divergence
Vector A
Curl
Vector A
Expression
(Cartesian)
Expression
(Cylindrical)
Expression
(Spherical)
Result
Vector
Scalar
Vector
SUMMARY OF GRAD, DIV & CURL
Gradient
Divergence
A vector that gives
direction of the
maximum rate of
change
of
a
quantity i.e. temp
A
scalar
that
measures
the
magnitude
of
a
source or sink at a
given point
Physical
meaning
𝛻∙𝑨<0
Sink
i.e. Flux out
< flux in
𝛻∙𝑨<0
Source
Curl
A vector operator
that describes the
rotation/ununiformity
of a vector field
𝛻𝑋𝑨 > 0
RHC rotation
i.e. Flux out
> flux in
𝛻𝑋𝑨 < 0
LHC rotation
𝛻∙𝑨=0
Incompressible
Flux out = flux in
𝛻𝑋𝑨 = 0
Irrotational
DIVERGENCE THEOREM

Divergence theorem:

Total outward flux of a vector field A through a
closed surface S is the same as the volume integral of
divA. i.e. Transformation of volume integral
involving divA to surface integral involving A

Equation:

Physical meaning: The total flux from field A
passing through a volume V is equivalent to
summing all the flux at the surface of V.
PROBLEM 2 (MIDTERM EXAM 2013)
Verify the divergence theorem for the vector r2ar
within the semisphere.
STOKE’S THEOREM

Stoke’s Theorem:

The line integral of field A at the boundary of a closed
surface S is the same as the total rotation of field A at
the surface. i.e. Transformation of surface integral
involving curlA to line integral of A

Equation:

Physical meaning: The total effect of field A along a
closed path is equivalent to summing all the rotational
component of the field inside the surface of which the
path enclose.
LAPLACIAN OF A SCALAR FUNCTION
2
2
2

U

U

U
2
 U    U   2  2  2
x
y
z
 2 2 2 
  2 , 2 , 2 
 x y z 
2
U is a scalar function of x, y, z (i.e. temperature)
 Laplacian of a scalar = Divergence of a Gradient
of scalar function.
 Important operator when working with
MAXWELL’S EQUATION!!

PROBLEM 3
Given that F  x yax  yay , find
(a)  F  dl Where L is shown in the following
L
figure
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(b)
L
   F dS
S
Where S is the area bounded by
(c) Is Stokes’s theorem satisfied?
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