B.Sc. Physics
Vector Operations
B.Sc. PHYSICS (PAPER-A)
VECTOR OPERATIONS
1. SCALARS
The physical quantities which are completely described by magnitude with proper unit are called
scalars. Mass, length, time density, energy, work, temperature and charge are the examples of scalars.
Scalars can be added, multiplied and subtracted by ordinary rules of algebra.
2. VECTORS
The physical quantities which are completely described by magnitude, with proper unit and direction
are called vectors. Force velocity, acceleration, momentum, torque, electric field intensity and magnetic field
induction are the examples of vectors.
Vectors are added, multiplied and subtracted by vector algebra. However, parallel and antiparallel
vectors are added by ordinary algebra.
3. UNIT VECTOR
A vector having unit magnitude is called unit vector. It is used to describe the direction of any vector.
If we have a vector ⃗ , then a unit vector in the direction of ⃗ is written as:
̂
⃗
| |
Where ̂ is the unit vector in the direction of ⃗ and | | is its magnitude.
4. VECTOR ADDITION
The process in which two or more than two vectors are added to
get a single vector is called vector addition.
The vectors can be added graphically by head to tail rule. According
to this rule, the addition of two vectors ⃗ and ⃗ consists of following
steps:
(i) Place the tail of vector ⃗ on the head of vector ⃗ .
(ii) Draw a vector from the tail of vector ⃗ to the head of vector ⃗ , called the resultant vector ⃗⃗ .
Important Note:
The vector sum ⃗
⃗ and ⃗
⃗ has the same resultant ⃗⃗ , as
shown in the figure. Therefore we can write:
⃗
⃗
⃗
⃗
So vector addition is commutative.
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5. RESULTANT VECTOR
Vector Operations
It is the sum of two or more vectors which along has the same effect as the
combined effect of all the vectors to be added.
The vector ⃗⃗ has the same effect as the combined effect of vectors ⃗ , ⃗ ,
and ⃗ . So ⃗⃗ is the resultant vector of the vectors ⃗ , ⃗ ,
and ⃗ .
6. RECTANGULAR COMPONENTS
Consider a vector ⃗ represented by ̅̅̅̅ as shown in the figure. This vector can be decomposed into
three mutual perpendicular components along x, y and z-axis. Let these components are denoted by ⃗ , ⃗
and ⃗ . These components form three sides of a rectangular
parallelepiped as shown.
Draw perpendicular ̅̅̅̅ on
̅̅̅̅
̅̅̅̅
, then from figure:
̅̅̅̅
The vector ̅̅̅̅ is along z-axis and is denoted by ⃗ . Therefore:
⃗
̅̅̅̅
⃗
( )
Now draw perpendicular ̅̅̅̅ on x-axis, then:
̅̅̅̅
̅̅̅̅
̅̅̅̅
̅̅̅̅ is along x-axis and is denoted by ⃗ . ̅̅̅̅ is along y-axis and is denoted by ⃗ . Therefore,
̅̅̅̅
⃗
⃗
Putting values in (1), we have:
⃗
⃗
⃗
⃗
( )
̂ are the unit vectors along x, y and z-axis respectively, then:
If ̂ ̂
⃗
,̂ ⃗
̂
,̂ ⃗
Equation (2) can be written as:
⃗
̂
̂
̂
The magnitude of this vector will be:
|⃗ |
√
7. DIRECTION COSINES
Figure shows a vector ⃗ in space. Let this vector makes angles , ,
with x,
y, z-axis respectively. This
⃗
̂
̂
̂
Now the unit vector in the direction of ⃗ is written as:
̂
⃗
| |
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Where ̂ is the unit vector in the direction of ⃗ and | | is its magnitude.
̂
̂
̂
̂
̂
| |
̂
| |
Now | |
Vector Operations
| |
̂
,| |
| |
̂
( )
,| |
Equation (1) becomes:
̂
̂
̂
̂
are called direction cosines of vector ⃗ and are denoted as:
Where
So the above equation becomes:
̂
̂
̂
̂
Taking magnitude on both sides:
|̂|
√
√
Squaring both side, we have:
So the sum of the squares of direction cosines is equal to unity.
Problem: Find the length (i.e., the magnitude) of the vector ⃗
̂
̂ . Also, calculate the angles
̂
which this vector makes with the axes x, y and z.
Solution: The magnitude of vector ⃗ is described by formula:
|⃗ |
( )
√
Here
Putting values in (1), we have:
√( )
( )
Let ⃗ makes angles , ,
( )
√
with x, y, z-axis respectively, then:
( )
| |
( )
(
)
| |
( )
( )
| |
( )
| |
| |
| |
√
Problem: Find the resultant of the vectors ⃗
̂
̂
̂ ⃗
̂
̂
̂ ⃗
̂
̂
̂ . Also,
calculate the angles which resultant vector makes with the axes x, y and z.
Solution: If ⃗ the resultant of vectors ⃗ ⃗
⃗ and
are the rectangular components of resultant,
then:
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Vector Operations
The magnitude of resultant is given by:
|⃗ |
√
√( )
√
√
Let ⃗ makes angles , ,
( )
(
)
with x, y, z-axis respectively, then:
| |
( )
| |
(
)
| |
( )
| |
(
)
| |
( )
| |
(
)
Problem: Find the angle between the direction of the vector given by the difference of the following two
vectors and the z-axis. ⃗
√ ̂
̂ ⃗
√ ̂
√ ̂
⃗
Solution: Suppose the difference of the vectors: ⃗
√
√
√
√
√
√
√ ̂
̂
⃗
The magnitude of resultant is given by:
|⃗ |
√
√( √ )
√
√
Let ⃗ makes angle
( √ )
( )
with z-axis, then:
(
| |
)
| |
(
)
8. SPHERICAL POLAR COORDINATES
Consider a vector ⃗ in three dimensions having components
,
,
along x, y, z-axis respectively.
Draw projection of ⃗ on xy-plane.

The angle
between ⃗ and z-axis is called polar angle.

The angle
between x-axis projection of ⃗ in xy-plane is
called Azimuthal angle.
From the figure, it is clear that:
Here (
) are called spherical polar coordinates.
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9. APPLICATIONS OF SPHERICAL POLAR COORDINATES
Vector Operations
The spherical polar coordinates are superior than Cartesian Coordinates for the study of Physical
problems. For example, the gravitational force of the earth on distant objects has the symmetry of sphere and
its properties can be described in an easy way by the help of spherical polar coordinates.
10. VECTOR DERIVATIVE
Consider a particle moving along a curve as shown. Its position at
any time is given by a vector ( ). With the passage of time, the direction
and magnitude of this vector may change.
Let ( ) and ( ) denote the positions of the particle at time
and
, respectively. Then, the displacement
between two points
and
is described as:
( )
( )
is also a vector collinear with
and hord
written as
. As
, point
becomes the tangent at
and is called derivative of
approaches point
. When
, the ratio
is
with respect to time . But by definition
is the velocity
of the
particle. Thus:
The acceleration
being the rate of change of velocity is given by:
(
)
̂
In Cartesian coordinates:
̂
̂
Now the position vector of the moving particle can also be written as:
( )
( ) ̂( )
Then the derivative of ( ) is defined as:
(
)
(
By Taylor’s Theorem: (
( )
) ̂(
)
* ( )
( ) ̂( )
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( )
( ) ̂( )
and ̂ (
̂
+ [ ̂( )
( )
̂
]
)
̂( )
̂
( ) ̂( )
̂( )
̂
(
)
( ) ̂( )
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Vector Operations
̂
[ ( )
̂ ( )]
̂
[ ( )
̂
( )
̂ ( )]
̂
[ ( )
̂
(
(
)
̂
)
̂
̂ ( )]
̂( )
( )
This is of the form:
( ⃗)
⃗
⃗
In equation (1),
⃗
gives change in magnitude of
and
̂
gives change in direction of
11. CIRCULAR MOTION
Consider an object is revolving along a circular path with constant angular velocity
. The position of the
body revolving in a circle is given by:
( )
̂( )
Suppose that the center of the circle is at origin O. Now the magnitude of
remains constant and the
unit vector ̂ ( ) rotates at a constant rate. A circular motion is an example of a motion in two dimension i.e.,
in a plane. So ̂ ( ) can be written as:
̂( )
̂
̂
̂( )
where
̂
( )
̂
is the angular velocity which is constant.
Velocity
The linear velocity of a body is then given as:
[
̂ ( )]
̂( )
[
[
]̂
̂
( ) ̂]
( )̂
̂
̂
Magnitude of velocity is given by:
| |
)
√(
(
)
√
√
(
√
( )
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Acceleration
Vector Operations
The acceleration of the particle moving in a circle can be obtained by differentiating its instantaneous
velocity with respect to time:
[
̂]
̂
[
̂]
̂
[
( ) ̂]
( )̂
[
̂]
̂
̂( )
[
̂ ( )]
( )
The magnitude of acceleration will be:
| |
|
|
| |
| |
( )
The negative sign in equation (2) shows that acceleration is directed towards the origin i.e., towards
the center of the circle and the equation (3) describes that the magnitude of acceleration is proportional to the
distance from the center (origin).
As, for the case of circular motion:
Equation (3) becomes:
( )
(
)
This is called centripetal acceleration.
12. SCALAR FIELD
Consider a scalar U e.g., temperature or density of a medium, which changes from point to point. The region
in which the scalar varies from point to point is called scalar field.
13. VECTOR FIELD
Consider a vector ⃗ e.g., electric field intensity which changes from point to point. The region in which the
vector varies from point to point is called vector field.
14. LINE INTEGRAL
Consider a vector ⃗ at point P of the curve AN of length l as shown in the figure. Take
a small element of length ‘⃗⃗⃗ ’, then the dot product of ⃗ and ⃗⃗⃗ is given by:
⃗ ⃗⃗⃗
|⃗ ||⃗⃗⃗ |
⃗ ⃗⃗⃗
⃗ ̂
Or
Where ̂ is the unit vector in the direction of ⃗⃗⃗ .
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Integrating the above expression over the entire length ‘l’, we have
∫ ⃗ ⃗⃗⃗
∫
∫ ⃗ ⃗⃗⃗
∫⃗ ̂
Vector Operations
Or
This integral is called line integral of vector along the curve AB. The common example of line integral is the
definition of work.
15. SURFACE INTEGRAL
Consider a vector ⃗ at point P of a surface area . Take a small element of the surface
⃗ ⃗⃗⃗⃗
Where
|⃗ ||⃗⃗⃗⃗ |
is the angle between ⃗ and outward drawn normal to
⃗ ⃗⃗⃗⃗
. Then
. Or
⃗ ̂
Where ̂ is the unit vector normal to surface
.
Integrating the above expression over whole surface, we have
∫ ⃗ ⃗⃗⃗⃗
∫
∫ ⃗ ⃗⃗⃗⃗
∫⃗ ̂
Or
It is called the surface integral of vector ⃗ over the whole surface.

If ⃗ is the velocity of fluid at any point, then ∫ ⃗ ⃗⃗⃗⃗ is called the rate of flow.

If ⃗ is the electric field strength (or magnetic field strength) at any point, then ∫ ⃗ ⃗⃗⃗⃗ is gives the
electric flux (or magnetic flux).
16. VOLUME INTEGRAL
Consider a close surface having uniform volume charge density ( ). Take a small element of volume
having mass
, then
Integrating over the whole volume, we get:
∫
And ∫
∫
∭
∫
∫
is called volume integral. Therefore,
∭
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17. OPERATORS
Vector Operations
These are the quantities whose operation on a function gives a new function. Operators always
operate on something placed after them. There are many types of operators e.g.,

Number Operator (

Differential Operator (

Integral Operator (∫

Logarithmic Operators (
)
)
)
∫
)
But now we shall introduce a vector differential operator called del , denoted by ⃗ , which is described as:
⃗
̂
̂
̂
It can operate both on scalar and vector.
18. OPERATIONS OF ⃗
(i) Gradient Operation
(ii) Divergence Operation
(iii) Curl Operation
Gradient Operation
The operation of ⃗ on a scalar function is called gradient operation. If we have a scalar function U, then
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗
Note that ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗
is a vector. In Cartesian coordinates, ⃗
(
(
̂
̂)
̂
̂
is written as:
̂)
̂
Divergence Operation
The operation of ⃗ on a vector function through dot product is called divergence operation. If we have a
vector function ⃗ , then:
⃗ ⃗
⃗
⃗ is a scalar.
Note that
⃗ ⃗
(
̂
̂
̂) (
̂
̂
̂)
⃗ ⃗
Curl Operation
The operation of ⃗ on a vector function through cross product is called curl operation. If we have a vector
function ⃗ , then:
⃗
⃗
⃗
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⃗ is a vector. In Cartesian Coordinates,
Note that
̂
⃗
⃗ is described as:
̂
̂
||
⃗
Vector Operations
||
) ̂
(
(
) ̂
(
) ̂
19. GRADIENT OF SCALAR
The maximum rate of increase of a scalar function with respect to space in a particular direction is
called gradient of a scalar function.
20. LEVEL SURFACES (DEFINITION):
The surface in which all the point of a scalar field has same value is called level surface. An entire scalar field
can mapped out by level surfaces, each associated with a constant value of a scalar function.
21. Show that ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗ , where U is a scalar point function.
Consider a scalar U e.g., temperature or density of a medium which changes from point to point. The
region in which the scalar varies from point to point is called scalar field. In such a field, U has definite value
at each point.
Let two level surfaces
constant values
and
and
associated with
of the scalar function,
respectively.
Consider a point A on the surface
vector ⃗⃗⃗⃗⃗
with position
with respect to origin O. Let ⃗⃗⃗⃗⃗
the position vectors of any point B taken on surface
be
as
shown in the figure. By head to tail rule, we have:
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
The shortest distance between the level surfaces
and
is taken along the outward drawn normal
⃗⃗⃗⃗⃗ specified by the unit normal vector ̂. We can follow different paths lengths between the level surfaces
and
. Let us consider on the path lengths ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ , where ⃗⃗⃗⃗⃗ is the shortest path length.
⃗⃗⃗⃗⃗
Now
⃗⃗⃗⃗⃗
and
We have:
Thus the rate of variation of
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Taking dot product of
takes place in the direction of directed normal ̂ and is called gradient of U:
̂
on both sides, we have:
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Vector Operations
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
̂
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Since U is the function of one variable, so we can replace partial differential coefficient by ordinary
differential coefficient. Therefore,
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
( )
In this scalar field U has a definite U is value at every point. So U is a function of coordinates of
point (
) i.e.,
(
)
Now, let the two points A and B has the coordinates (
The displacement
) in the scalar field.
between P and Q is described as:
̂
The change in
) and (
̂
̂
as we move from A to B (Total Differential
(
̂
̂) (
̂
̂
̂
) is described (by Calculus) as:
̂)
̂
̂)
As the body moves from high P.E. to low P.E., the change in P.E. is equal to –
. So,
(
̂
̂
̂)
( )
(
̂
⃗
So, equation (2) becomes:
⃗
( )
Comparing equation (1) and (3), we have:
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗
Hence Proved.
Problem: Show that
, where
is the P.E. of the body.
Solution: As we know that for a scalar V,
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
( )
( )
Comparing equation (1) and (2), we get:
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Hence Proved
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Vector Operations
Problem: The P.E. of a body of mass m held at a height h above the surface of the earth is mgh. Find the
force of gravity (or weight) of the body.
Solution:
( )
Given
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
As
(
̂
(
Put
̂)
̂
̂
̂)
̂
( )
in equation (1), we have:
Thus equation (2) becomes:
(
(
)̂
(
(
(
)̂
)̂
(
(
̂
) ̂)
(
)̂
(
) ̂)
)
̂
This expression shows that force of gravity acts in the downward direction and the magnitude of force of
gravity is mg.
Problem: Show that if a vector is the gradient of a scalar function, then its line integral around a closed path
is zero.
Solution: Let
be a vector which is equal to gradient of a scalar function
i.e.,
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Taking dot product of
on both sides:
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Taking line integral from point A to point B on both sides:
∫
∫ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
(
( )
As
̂
̂
̂) (
̂
̂
̂)
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Expression (1) becomes:
∫
∫
| |
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Vector Operations
∫
This shows that line integral depends upon the values of
at point A and B. If we have a close path then the
line integral around the close path is given by:
∫
| |
∫
∫
∫
Hence proved that if a vector is the gradient of a scalar function, then its line integral around a closed path is
zero.
, then show that ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Problem: If a scalar function
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ( )
̂
̂
̂
Solution: As
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Also
̂
̂
̂
̂
( )
̂
̂
| |
√
( )
( )
( )
From differentiating (partially) both sides of equation (2) w.r.t x, we have
( )
(
)
Equation (3) becomes:
( )
Putting values in equation (1), we have:
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
̂
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̂
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Vector Operations
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
( ̂
̂)
̂
( )
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Hence Proved.
Problem: Given that ⃗
Solution: As ⃗ ⃗
̂
̂
(
⃗ ⃗
(
̂) (
̂
)( )
̂ and ⃗
̂
(
)(
̂
)
̂
̂
̂ . Find ⃗⃗⃗⃗⃗⃗⃗⃗⃗ (⃗ ⃗ ).
̂)
̂
( )( )
⃗ ⃗
Now ⃗⃗⃗⃗⃗⃗⃗⃗⃗ (⃗ ⃗ )
⃗ (⃗ ⃗ )
⃗⃗⃗⃗⃗⃗⃗⃗⃗ (⃗ ⃗ )
(
̂
(⃗ ⃗ )
⃗⃗⃗⃗⃗⃗⃗⃗⃗ (⃗ ⃗ )
̂ ) (⃗ ⃗ )
̂
(⃗ ⃗ )
̂
(⃗ ⃗ )
̂
(
⃗⃗⃗⃗⃗⃗⃗⃗⃗ (⃗ ⃗ )
⃗⃗⃗⃗⃗⃗⃗⃗⃗ (⃗ ⃗ )
)
(
)̂
̂
̂
(
(
)
)̂
̂
(
(
)
̂
)̂
22. Flux of a Vector Field
Consider a vector ⃗ e.g. velocity of a fluid which changes from point to point. The region in which the
vector varies from point to point is called vector field. In such a field, ⃗ changes from point to point. So ⃗ is a
function of position coordinates x,y,z and is written as:
⃗
⃗(
)
If we consider an element of area ⃗ in the vector field, then the scalar product of ⃗ and ⃗ is called flux of
vector field. Thus:
⃗
If
⃗
⃗ , then:
is the angle between ⃗
|⃗ || ⃗ |
If the plane of area ⃗ is perpendicular to ⃗ , then
⃗ ⃗
⃗
23. Show that
, so
Consider a parallelepiped having side
in a vector field. Let ⃗ be
the value of vector at the center of parallelepiped, where
⃗
̂
̂
̂
We consider two faces 1 and 2 of parallelepiped perpendicular to the x-axis
each of area

.
The value of x-component of ⃗ at the center of face-2
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This may be taken as the value for the whole face-2 because the face is very small.

(
)
(
)
The flux entering the parallelepiped through face-1
Vector Operations
Similarly,

The value of x-component of ⃗ at the center of face-2

The flux leaving the parallelepiped through face-2
(
)
(
(
)
)
(
(
)
)
Similarly, the net outward flux through parallelepiped along y-axis and along z-axis is given as:
(
)
(
)
(
)
(
(
)
(
)
Here
)
(
(
)
(
)
)
is the volume of parallelepiped.
(
(
)
)
(
)
Now by the definition, the net flux per unit volume is called divergence of vector ⃗ . It is denoted by
⃗.
Thus:
⃗
This can be written as:
⃗
(
⃗
̂
̂
̂) (
̂
̂
̂)
⃗ ⃗
Note: If flux entering a volume element is equal to flux leaving the volume element, then
⃗
Problem: Given a position vector
̂
̂
̂ . Evaluate
.
⃗
Solution: As
⃗
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Vector Operations
(
̂
̂
Problem: If
⃗
Solution: As
̂) ( ̂
̂
̂)
̂
̂ . Find divergence at point (
̂
)
⃗ ⃗
⃗
(
⃗
̂
̂) (
̂
(
)
(
̂
̂)
̂
)
(
)
⃗
Now
⃗
(
)
⃗
(
⃗
24. Prove that
( )
( ) ( )
(
)( )
)
⃗
⃗
Consider a fluid flowing with constant velocity ⃗ at any point (
) with
as its rectangular
components.

Initially, we consider flow of fluid in
plane,
along y-axis as shown in the figure. The
successive layers will move alike. However, if
is a function of z, then different layers in the
-
plane will slide over or move relative to one
another. The velocity of different layers of fluid
goes on increasing along z-axis. Let
be the
velocity of central layer along y-axis, then the
velocity of the layer just above it will be [
(
[
)
]. The factor
(
)
(
)
] and velocity of layer just below it will be
gives the rotation of the fluid in clockwise direction around x-axis from
z to y.

Similarly, we consider the flow of fluid in yz-plane along z-axis as shown in the figure. The velocity of
different layers of fluid goes on increasing along y-axis. Let
axis, then the velocity of the layer just above it will be *
will be *
(
)
+. The factor
(
)
be the velocity of central layer along z(
)
+ and velocity of layer just below it
gives the rotation of the fluid in anticlockwise direction around x-
axis from y to z.
Taking counter-clockwise direction positive, the net rotation about x-axis must be proportional to
(
) ̂
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Similarly, Net rotation around y-axis will be:
(
Vector Operations
) ̂
And net rotation around z-axis is:
) ̂
(
Now the total rotation of the fluid gives the Curl of vector ⃗ . Hence,
⃗
) ̂
(
̂)
⃗
(
) ̂
) ̂
(
( )
As,
⃗
(
̂
̂
̂
̂
̂
̂
̂
̂
Now, ⃗
⃗
|
|
⃗
⃗
(
) ̂
(
) ̂
) ̂
(
( )
Comparing (1) and (2), we have:
⃗
⃗
⃗
Hence proved
Problem: If ⃗
̂
̂
Solution:
̂ , find
̂
̂
̂|
̂(
̂(
|
(
̂
̂
||
⃗
⃗
)
)
)
̂
̂
⃗
⃗ at the point (
||
̂|
|
̂|
(
|
(
))
̂(
(̂
)
̂(
)
(
)̂
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)
(
))
̂
̂(
(
)
(
))
̂
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25. GAUSS’S DIVERGENCE THEOREM
Vector Operations
The surface normal integral of vector taken over a closed surface is equal to the volume integral of the
divergence of a vector over the volume enclosed by the surface. Mathematically, it is described as:
∫ ⃗ ⃗⃗⃗⃗
⃗
∫
∫⃗ ̂
∫
⃗
Proof
Consider small volume element having volume
⃗
⃗ ⃗
⃗
(
⃗
(
Where
̂
̂) (
̂
enclosed in surface S. then by definition:
̂
̂)
̂
)
are the components of A along x, y, z-axis respectively. Multiplying both sides
by
, we get:
⃗
(
)
Integrating both sides, we get:
⃗
∭
∭(
∫
⃗
∭
∫
⃗
∫
Since
)
∭
∬
∭
∫
∬
∫
∬
are functions of only one variable, so we can change partial derivative in total derivative:
⃗
∫
∫
∬
∫
⃗
∬
∫
⃗
∬
∫
⃗
∬
∫
⃗
∬(
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∫
∬
∬
∬
∬
∬
∬
∫
∬
∬
)
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Vector Operations
∫
⃗
∬ ⃗ ⃗⃗⃗⃗
∫
⃗
∫ ⃗ ⃗⃗⃗⃗
This theorem enable us to transform a surface integral into volume integral and vice versa.
̂
Problem. Evaluate ∫
Where S is the close surface.
Solution.
By Gauss’s Divergence theorem:
∫
∫⃗
̂
∫
̂
∫(
∫
̂
∫(
̂
̂) ( ̂
̂
̂
̂)
)
∫(
)
)
∫(
∫
∫
STOKE’S THEOREM
The line integral of a vector function around the closed curve (boundary edge) of a surface is equal to
the surface normal integral of the curl of vector function over that
surface.
If ⃗ is a vector function, then mathematically:
∮ ⃗ ⃗⃗⃗
∫
⃗ ⃗⃗⃗⃗
Proof
Consider a surface enclosed by a curve ABCD. We divide it into
large number of small meshes. Let the area of a mesh be ⃗⃗⃗⃗ .
As
is the line integral per unit area. So
⃗⃗⃗⃗ ]
[
⃗ ⃗⃗⃗⃗
Suppose that we take the line integral of all the meshes within the curve ABCD. If we add all these
line integrals, the line integral along each side of the mesh where the line integral is taken twice, will cancel
away and we are left with line integral only along the curve ABCD.
Hence the line integral ∮ ⃗ ⃗⃗⃗ taken over the curve ABCD and the surface integral ∫
⃗ ⃗⃗⃗⃗ taken
over the surface ABCD are equal. Therefore,
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∮ ⃗ ⃗⃗⃗
Vector Operations
⃗ ⃗⃗⃗⃗
∫
This is called Stoke’s theorem. This theorem enables us to transform a line integral into surface
integral and vice versa.
Prove the following vector identities:
a)
⃗
b)
Proof:
a)
̂
⃗
(⃗ )
̂
̂
̂
̂
̂
|
|
|
|
|
|
|
|
⃗
b)
⃗
⃗
⃗ ⃗
|
|
|
|
Problem: Prove that
⃗
⃗
⃗(
⃗
⃗
⃗ (⃗ ⃗ )
⃗)
(
(⃗ ⃗ )⃗
⃗
⃗)
⃗
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