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17. Vector Calculus with Applications
17.1 INTRODUCTION
In vector calculus, we deal with two types of functions: Scalar Functions (or Scalar Field) and
Vector Functions (or Vector Field).
Scalar Point Function
A scalar function πΉ(π₯, π¦, π§)defined over some region R of space is a function which associates, to
each point π(π₯, π¦, π§)in R, a scalar value πΉ π = πΉ(π₯, π¦, π§). And the set of all scalars πΉ(π)for all
values of P in R is called the scalar field over R.
Precisely, we can say that scalar function defines a scalar field in a region or on a space or a curve.
Examples are the temperature field in a body, pressure field in the air in earthβs atmosphere.
Moreover, if the position vector of the point P is π , then we may also write the scalar field as
πΉ π = πΉ(π ). This notation emphasizes the fact that the scalar value πΉ(π ) is associated with the
position vector π in the region R.
E.g. 1) The distance πΉ π of any point π(π₯, π¦, π§) from a fixed point πβ²(π₯ β², π¦β², π§β²) in the space is a
scalar function whose domain of definition is the whole space and is given by
πΉ π = (π₯ β π₯ β²)2 + (π¦ β π¦ β²)2 + (π§ β π§ β²)2. Also πΉ π defines a scalar field in space.
E.g. 2) The function πΉ π₯, π¦, π§ = π₯π¦2 + π¦π§ + π₯2 for the point (π₯, π¦, π§) inside the unit sphere
π₯2 + π¦2 + π§2 = 1 is a scalar function and also is defines a scalar field throughout the sphere.
Note: In the physical problems, the scalar function F depends on time variable t in addition to the
point P and then we write it as πΉ π, π‘ = πΉ π , π‘ = πΉ(π₯, π¦, π§, π‘) . The example of such a time
dependent scalar function is the temperature distribution throughout a block of metal heated in such
a way that its temperature varies with time.
Vector Point Function
A vector functions πΉ π₯, π¦, π§ defined over some region R of space is a function which associates, to
each point π(π₯, π¦, π§) in R, a vector value πΉ π = πΉ (π₯, π¦, π§) and the set of all vectors πΉ π for all
points P in R is called the vector field over R
Moreover, if the position vector of the point P is π , then we may write the vector field as
πΉ π = πΉ (π ). This notation emphasizes the fact that the vector value πΉ (π ) is associated with the
position vector π in the region R. Also the general form (component form) of the vector function is
πΉ π = πΉ1 π π + πΉ2 π π + πΉ3 π π , where the components πΉ1 π , πΉ2 π and πΉ3 π are the scalar
functions.
E.g. 1) The function πΉ π₯, π¦, π§ = 2π₯π¦π + sin π₯ π + 3π§2π for point P( x, y, z ) inside an ellipsoid
π₯2
9+
π¦2
16+
π§2
4 = 1 is a vector function and defines a vector field throughout the ellipsoid.
E.g. 2) The force field given by πΉ π₯, π¦, π§ = π₯π + 2π¦π + π§2π is a vector field.
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Note: Like the time dependent scalar field, time dependent vector field also exists. Such a field depends on
time variable t in addition to the point in the region R and may be expressed as
πΉ π , π‘ = πΉ1 π , π‘ π + πΉ2 π , π‘ π + πΉ3 π , π‘ π , where πΉ1 , πΉ2 and πΉ3 are scalar functions. An example of time
dependent vector field is the fluid velocity vector in the unsteady flow of water around a bridge support
column, because this velocity depends on the position vector π in the water and the time variable t and is
given as π (π , π‘).
Vector Function of Single Variable
A vector function πΉ of single variable t is a function which assigns a vector value πΉ (π‘) to each
scalar value t in interval π β€ π‘ β€ π . In the component form, it may be written as
πΉ π‘ = πΉ1 π‘ π + πΉ2 π‘ π + πΉ3 π‘ π where πΉ1, πΉ2 and πΉ3 are called components and are scalar functions
of the same single variable t.
For example, the functions given byπΉ π‘ = π‘ π + sin(π‘ β 2) π + cos 3π‘ π and πΊ π‘ = π‘2π + ππ‘π + log π‘ π
are vector functions of a single variable t.
Limit of a Vector Function of Single Variable
A vector function πΉ π‘ = πΉ1 π‘ π + πΉ2 π‘ π + πΉ3(π‘)π of single variable t is said to have a limit
πΏ = πΏ1π +πΏ2π +πΏ3π as π‘ β π‘0 , if πΉ π‘ is defined in the neighborhood of t0 and Ltπ‘βπ‘0 πΉ π‘ β πΏ = 0
or Ltπ‘βπ‘0
πΉ1 π‘ β πΏ1 = Ltπ‘βπ‘0 πΉ2 π‘ β πΏ2 = Ltπ‘βπ‘0
πΉ3 π‘ β πΏ3 = 0, then we write it as Ltπ‘βπ‘0πΉ (π‘) = πΏ .
Continuity of a Vector Function of Single Variable
A vector function πΉ π‘ = πΉ1 π‘ π + πΉ2 π‘ π + πΉ3(π‘)π of a single variable t is said to be continuous at
t = t0, if it is defined in some neighborhood of t0 and Ltπ‘βπ‘0πΉ π‘ = πΉ π‘0 .
Moreover, πΉ π‘ is said to be continuous at π‘ = π‘0 if and only if its three components F1, F2 and F3
are continuous as π‘ = π‘0.
DIFFERENTIAL VECTOR CALCULUS
17.2 DIFFERENTIATION OF VECTORES
Differentiability of a Vector Function of Single Variable
A vector function πΉ π‘ = πΉ1 π‘ π + πΉ2 π‘ π + πΉ3(π‘)π of a single variable t defined over the interval
π β€ π‘ β€ π is said to be differentiable at t = t0 if the following limit exists.
Ltπ‘βπ‘0
πΉ π‘ βπΉ (π‘0)
π‘βπ‘0 = πΉ β²(π‘0)
And πΉ β²(π‘0) is called the derivative of πΉ π‘ at t = t0.
Also πΉ π‘ is said to be differentiable over the interval π β€ π‘ β€ π, if it is differentiable at each of the
points of the interval. In component form, πΉ π‘ is said to be differentiable at t = t0 if and only if its
three components are differentiable at t = t0. In general, the derivative of πΉ (π‘) is given by
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πΉ β² π‘ = Ltπ‘βπ‘0
πΉ π‘+βπ‘ βπΉ (π‘)
βπ‘, provided the limit exists and in terms of components
πΉ β² π‘ = πΉ1β² π‘ π + πΉ2
β² π‘ π + πΉ3β² (π‘)π or
ππΉ
ππ‘=
ππΉ1
ππ‘π +
ππΉ2
ππ‘π +
ππΉ3
ππ‘π .
In the similar manner, π2πΉ
ππ‘ 2=
π
ππ‘ ππΉ
ππ‘ ,
π3πΉ
ππ‘ 3=
π
ππ‘
π2πΉ
ππ‘ 2 =π2
ππ‘ 2 ππΉ
ππ‘ .
Rules for Differentiation of Vector Functions
If πΉ π‘ , πΊ π‘ & π» π‘ are the vector functions and π(π‘) is a scalar function of single variable π‘ defined
over the interval π β€ π‘ β€ π, then
1. ππΆ
ππ‘ = 0 , where π is a constant vector.
2. π πΆ πΉ π‘
ππ‘ = πΆ
ππΉ
ππ‘, where C is a constant.
3. π πΉ π‘ Β±πΊ (π‘)
ππ‘=
ππΉ
ππ‘Β±
ππΊ
ππ‘
4. π π π‘ πΉ π‘
ππ‘ = π π‘
ππΉ
ππ‘+
ππ
ππ‘ πΉ (π‘)
5. π πΉ π‘ β πΊ (π‘)
ππ‘ =
ππΉ
ππ‘ β πΊ (π‘) + πΉ (π‘) β
ππΊ
ππ‘
6. π πΉ π‘ ΓπΊ (π‘)
ππ‘ =
ππΉ
ππ‘ Γ πΊ (π‘) + πΉ (π‘) Γ
ππΊ
ππ‘
7. π
ππ‘ πΉ π‘ , πΊ π‘ , π» (π‘) =
ππΉ
ππ‘, πΊ π‘ , π» (π‘) + πΉ π‘ ,
ππΊ
ππ‘, π» (π‘) + πΉ π‘ , πΊ π‘ ,
ππ»
ππ‘
8. π
ππ‘ πΉ π‘ Γ πΊ π‘ Γ π» π‘ =
ππΉ
ππ‘Γ πΊ π‘ Γ π» (π‘) + πΉ π‘ Γ
ππΊ
ππ‘Γ π» (π‘) + πΉ π‘ Γ πΊ π‘ Γ
ππ»
ππ‘
9. If πΉ π‘ is differentiable function of π‘ and π‘ = π‘(π ) is differentiable function then
ππΉ
ππ =
ππΉ
ππ‘ ππ‘
ππ .
Observations:
(i) If πΉ (π‘) has a constant magnitude, then πΉ β ππΉ
ππ‘ = 0. For πΉ π‘ β πΉ π‘ = πΉ (π‘)
2= ππππ π‘πππ‘,
implying πΉ β ππΉ
ππ‘ = 0 or πΉ β₯
ππΉ
ππ‘ .
(ii) If πΉ (π‘) has a constant (fixed) direction, then πΉ Γ ππΉ
ππ‘ = 0 .
Let πΉ π‘ = π(π‘)πΊ (π‘), where πΊ (π‘) is a unit vector in the direction of πΉ (π‘).
β΄ dF
dt=
π π π‘ πΊ π‘
ππ‘ = π π‘
ππΊ
ππ‘+
ππ
ππ‘ πΊ π‘ =
ππ
ππ‘ πΊ π‘ π ππππ, πΊ ππ π ππππ π‘πππ‘, π π
ππΊ
ππ‘= 0
and πΉ Γ ππΉ
ππ‘= π(π‘)πΊ (π‘) Γ
ππ
ππ‘ πΊ π‘ = π π‘
ππ
ππ‘ πΊ π‘ Γ πΊ π‘ = 0 π ππππ, πΊ Γ πΊ = 0
Theorem 1: Derivative of a constant vector is a zero vector. A vector is said to be constant if
both its magnitude and direction are constant (fixed).
Proof: Let π = π be a constant vector, then π + πΏπ = π .
On subtraction, πΏπ = 0 .Which further implies that πΏπ
πΏπ‘ = 0 .
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Implying,πΏπ‘
πΏπ‘ β 0 πΏπ
πΏπ‘ = 0 i.e.
ππ
ππ‘ = 0 .
Theorem 2: The necessary and sufficient condition for the vector function π of a single
variable t to have constant magnitude is π β π π
π π = π.
Proof:
Necessary condition: Suppose πΉ has constant magnitude, so πΉ π‘ β πΉ π‘ = πΉ (π‘) 2
= ππππ π‘πππ‘.
=> π
ππ‘ πΉ β πΉ = 0 i.e. πΉ β
ππΉ
ππ‘ +
ππΉ
ππ‘ β πΉ = 0
=> 2πΉ β ππΉ
ππ‘ = 0 i.e. πΉ β
ππΉ
ππ‘ = 0.
Sufficient condition: Suppose πΉ β ππΉ
ππ‘ = 0 => 2πΉ β
ππΉ
ππ‘ = 0
=> πΉ β ππΉ
ππ‘ +
ππΉ
ππ‘ β πΉ = 0 =>
π
ππ‘ πΉ β πΉ = 0
=> πΉ β πΉ = ππππ π‘πππ‘ => πΉ 2
= ππππ π‘πππ‘
Therefore πΉ has a constant magnitude.
Theorem 3: The necessary and sufficient condition for the vector function π of a single
variable t to have a constant direction is π Γ π π
π π = π .
Proof: Suppose that π is a unit vector in the direction of πΉ and πΉ = πΉ , then π = πΉ
πΉ i.e.
πΉ = πΉπ β¦ (1)
And ππΉ
ππ‘ = πΉ
ππ
ππ‘+
ππΉ
ππ‘ π β¦ (2)
Thus πΉ Γ ππΉ
ππ‘ = πΉπ Γ (πΉ
ππ
ππ‘+
ππΉ
ππ‘ π ) (using (1) and (2))
= πΉ2π Γ ππ
ππ‘ +πΉ
ππΉ
ππ‘ (π Γ π )
= πΉ2π Γ ππ
ππ‘ (since π Γ π = 0 ) β¦ (3)
Necessary condition: Suppose πΉ has a constant direction, then π has a constant direction and
constant magnitude. So ππ
ππ‘ = 0 . Thus from (3), πΉ Γ
ππΉ
ππ‘ = 0 .
Sufficient condition : Suppose that πΉ Γ ππΉ
ππ‘ = 0 .
Then by (3), πΉ2π Γ ππ
ππ‘ = 0 i.e. π Γ
ππ
ππ‘ = 0 β¦ (4)
Since π has a constant magnitude, so, by theorem 2, π β ππ
ππ‘ = 0 β¦ (5)
Form (4) and (5), ππ
ππ‘ = 0.
Which implies π is a constant vector i.e. π has a constant direction. Hence πΉ has a constant
direction.
Example 1: Show that if π = π π¬π’π§ππ + π ππ¨π¬ ππ where π , π and π are constants, then
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π ππ
π ππ = βπππ and π Γ
π π
π π = βπ(π Γ π) .
Solution: Given π = π sin ππ‘ + π cos ππ‘
Differentiating w. r. to t, ππ
ππ‘ = π π cos ππ‘ β π π sin ππ‘
Again differentiating w. r. to t, π2π
ππ‘ 2 = βπ π2 sin ππ‘ β π π2 cos ππ‘
= β π2 π sin ππ‘ + π cos ππ‘ = βπ2π
Also π Γ ππ
ππ‘ = π sin ππ‘ + π cos ππ‘ Γ (π π cos ππ‘ β π π sin ππ‘)
= π Γ π π sin ππ‘ cos ππ‘ + π Γ π π cos2 ππ‘ β π Γ π π sin2 ππ‘ β π Γ π π sin ππ‘ cos ππ‘
= β π Γ π π(cos2 ππ‘ + sin2 ππ‘) (since π Γ π = π Γ π = 0 )
= β π Γ π π = βπ(π Γ π ).
Example 2: If π = ππππ π β ππππ π + πππ π and π = ππ π + π π β ππ π , find ππ
ππππ π Γ π at
(1, 0, -2).
Solution: Here π Γ π = π₯2π¦π§ π β 2π₯π§3 π + π₯π§2 π Γ 2π§ π + π¦ π β π₯2 π
= π₯2π¦2π§ π + π₯4π¦π§ π + 4π₯π§4 π + 2π₯3π§3 π + 2π₯π§3 π β π₯π¦π§2 π
(β΅ π Γ π = π Γ π = π Γ π = 0 πππ π Γ π = π , π Γ π = βπ ππ‘π.)
= 2π₯3π§3 β π₯π¦π§2 π + π₯4π¦π§ + 2π₯π§3 π + π₯2π¦2π§ + 4π₯π§4 π
Now π2
ππ₯ππ¦ π Γ π =
π2
ππ₯ππ¦ 2π₯3π§3 β π₯π¦π§2 π + π₯4π¦π§ + 2π₯π§3 π + π₯2π¦2π§ + 4π₯π§4 π
= π
ππ₯
π
ππ¦ 2π₯3π§3 β π₯π¦π§2 π + π₯4π¦π§ + 2π₯π§3 π + π₯2π¦2π§ + 4π₯π§4 π
= π
ππ₯ βπ₯π§2 π + π₯4π§ π + 2π₯2π¦π§ π = βπ§2 π + 4π₯3π§ π + 4π₯π¦π§ π
At the point (1, 0, -2) π2
ππ₯ππ¦ π Γ π = β4π β 8π
Example 3: If π· = ππππ + πππ β ππ and πΈ = π π¬π’π§ π π β ππ¨π¬ π π + πππ , then find (a) π
π π π· β πΈ
(b) π
π π π· Γ πΈ .
Solution: Consider π = 5π‘2π + π‘3π β π‘π and π = 2 sin π‘ π β cos π‘ π + 5π‘π
So π π
π π = 10π‘ π + 3π‘2π β π and
π π
π π = 2 cos π‘ π + sin π‘ π + 5π
a) π
ππ‘ π β π =
π π
π π β π + π β
π π
π π
= 10π‘ π + 3π‘2π β π β 2 sin π‘ π β cos π‘ π + 5π‘π
+ 5π‘2π + π‘3π β π‘π β 2 cos π‘ π + sin π‘ π + 5π
= 20 π‘ sin π‘ β 3π‘2 cos π‘ β 5π‘ + 10 π‘2 cos π‘ + π‘3 sin π‘ β 5π‘
= π‘3 sin π‘ + 7π‘2 cos π‘ + 20 π‘ sin π‘ β 10 π‘
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b) π
ππ‘ π Γ π =
π π
π π Γ π + π Γ
π π
π π
= 10π‘ π + 3π‘2π β π Γ 2 sin π‘ π β cos π‘ π + 5π‘π
+ 5π‘2π + π‘3π β π‘π Γ 2 cos π‘ π + sin π‘ π + 5π
= π π π
10 π‘ 3π‘2 β12 sin π‘ β cos π‘ 5π‘
+ π π π
5π‘2 π‘3 βπ‘2 cos π‘ sin π‘ 5
= π 15π‘3 β cos π‘ + π β2 sin π‘ β 50 π‘2 + π β10π‘ cos π‘ β 6π‘2 sin π‘
+ π 5π‘3 + π‘ sin π‘ + π β2π‘ cos π‘ β 25 π‘2 + π 5π‘2 sin π‘ β 2π‘3 cos π‘
= π 20π‘3 + π‘ sin π‘ β cos π‘ β π 2π‘ cos π‘ + 2 sin π‘ + 75 π‘2
βπ 2π‘3 cos π‘ + 10π‘ cos π‘ + π‘2 sin π‘
Example 4: If π πΌ
π π = πΎ Γ πΌ and
π π½
π π = πΎ Γ π½ , then prove that
π
π π πΌ Γ π½ = πΎ Γ (πΌ Γ π½ ).
Solution: Given ππ
ππ‘ = π Γ π and
ππ
ππ‘ = π Γ π β¦ (1)
Consider π
ππ‘ π Γ π =
ππ
ππ‘ Γ π + π Γ
ππ
ππ‘ = π Γ π Γ π + π Γ π Γ π
= π β π π β π β π π + π β π π β π β π π = π β π π β π β π π
= π Γ (π Γ π )
π’π πππ π Γ π Γ π = π β π π β π β π π πππ π Γ π Γ π = π β π π β π β π π
7.3 CURVES IN SPACE
1. Tangent Vector:
Let π π‘ = π₯ π‘ π + π¦ π‘ π + π§(π‘)π be the position vector
of a point P. Then for different values of the scalar
parameter t, point P traces the curve in space (Fig.
17.1). For neighboring point Q with position vector
π (π‘ + πΏπ‘) , πΏπ = π π‘ + πΏπ‘ β π π‘ implying πΏπ
πΏπ‘=
π π‘+πΏπ‘ βπ (π‘)
πΏπ‘ is directed along the chord PQ.
As πΏπ‘ β 0, πΏπ
πΏπ‘ becomes the tangent to the space curve
at P provided there exists a non zero limit.
Thus a vector ππ
ππ‘= π β² is a tangent to the space curve π = πΉ (π‘).
Let P0 be a fixed point on the space curve corresponding to t=t0, and the arc length π0π = π , then
πΏπ
πΏπ‘=
πΏπ
πΏπ
πΏπ
πΏπ‘=
πππ ππ
πππππ ππ πΏπ
πΏπ‘ . β¦ (1)
As π β π along the curve QP, i.e. πΏπ‘ β 0, then the πππ ππ
πππππ ππ β 1 and
ππ
ππ‘=
ππ
ππ‘ = π β²(π‘) β¦ (2)
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If ππ
ππ‘ is continuous, then π =
ππ
ππ‘ ππ‘
π‘
π‘0 = π₯β² 2+ π¦β² 2+ π§β² 2ππ‘
π‘
π‘0 β¦ (3)
Further, if we take s as the parameter in place of t, then the magnitude of the tangent vector i.e.
ππ
ππ = 1. Thus denoting the unit tangent vector by π , we have π =
ππ
ππ β¦(4)
Example 5: Find the unit tangent vector at any point on the curve π = ππ + π, π = ππ β π,
π = πππ β ππ where t is variable. Also determine the unit tangent vector at t = 2.
Solution: Let π be the position vector of any point π₯, π¦, π§ on the given curve,
then π = π₯π + π¦π + π§π
β΄ π = π‘2 + 2 π + 4π‘ β 5 π + 2π‘2 β 6π‘ π
The vector tangent to the curve at any point π₯, π¦, π§ is ππ
ππ‘ = 2π‘ π + 4 π + 4π‘ β 6 π
Now ππ
ππ‘ = 2π‘ 2 + 4 2 + 4π‘ β 6 2 = 2 5π‘2 β 12π‘ + 13
Therefore unit tangent vector at π₯, π¦, π§ =ππ
ππ‘ ππ
ππ‘ =
2π‘ π + 4 π + 4π‘β6 π
2 5π‘2β12π‘+13
And the unit tangent vector at t = 2 is ππ
ππ‘ ππ
ππ‘
π‘=2 =
2π‘ π + 4 π + 4π‘β6 π
2 5π‘2β12π‘+13 π‘ = 2
=2π +2π +π
3.
2. Principal Normal: Since π is a unit vector, so ππ
ππ β π = 0 i.e. either
ππ
ππ is perpendicular to π or
ππ
ππ = 0, in which case π is a constant vector w.r.t. the arc length s and so has a fixed direction i.e.
the curve is a straight line. Now, if we denote the unit normal vector to the curve at P by π , then ππ
ππ
is in the direction of π which is known as the principal normal to the curve at P. The plane of π
and π is called osculating plane of the curve at P.
3. Binormal: A unit vector π΅ = π Γ π is called the binormal at P. As π and π both are unit vectors,
so π΅ is also a unit vector normal to both π and π
i.e. to the osculating plane of π and π .
Thus at each point P on the curve C, there are
three mutually perpendicular unit vectors π , π
and π΅ , which form a moving trihedral such that
π΅ = π Γ π , π = π΅ Γ π , π = π Γ π΅ . β¦ (1)
This moving trihedral determines three
fundamental planes at each point of the curve C.
(i) The osculating plane of π and π .
(ii) The normal plane of π and π΅ .
(iii) The rectifying plane of π΅ and π .
4. Curvture: The arc rate of turning of the tangent viz. ππ
ππ is called the curvature of the curve
and is denoted by . As ππ
ππ is in the direction of the principal normal π , therefore
ππ
ππ = π β¦ (2)
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5. Torsion: As the binormal π΅ is a unit vector, so ππ΅
ππ β π΅ = 0. Also π΅ β π = 0, therefore
ππ΅
ππ β π + π΅ β
ππ
ππ = 0 or
ππ΅
ππ β π + π΅ β π = 0 or
ππ΅
ππ β π = 0. Hence,
ππ΅
ππ is perpendicular to both π΅ and π and
is, therefore, parallel to π . The arc rate of turning of the binormal viz. ππ΅
ππ is called torsion of the
curve and is denoted by π. So, we can write it as
ππ΅
ππ = βππ (the negative sign indicates that for π > 0,
ππ΅
ππ has a direction of βπ ) β¦(3)
Further, we know that π = π΅ Γ π , which on differentiation gives ,
ππ
ππ =
ππ΅
ππ Γ π + π΅ Γ
ππ
ππ = βππ Γ π + π΅ Γ π
ππ
ππ = ππ΅ β π (using (1)) β¦ (4)
The relations in (1), (2) and (3) constitutes the well known Frenet Formulas for the curve C.
Observations:
(i) π =1
is called the radius of curvature.
(ii) π =1
π is called the radius of torsion.
(iii) π = 0 for a plane curve.
Example 6: Find π΅ (π) and π΅ (π) for the curve represented by π π = ππ π + ππππ .
Solution: For given π π‘ , we have π β² π‘ = 3π + 4π‘π and π β² π‘ = 9 + 16π‘2
Which implies that the unit tangent vector π π‘ = π β² π‘
π β² π‘ =
3π +4π‘π
9+16π‘2 β¦ (1)
Differentiating π π‘ w. r. to t, π β² π‘ = 1
9+16π‘2 4π β
16π‘
9+16π‘2 32
(3π + 4π‘π ) = 12(β4π‘π +3π )
9+16π‘2 3/2 β¦(2)
And π β² π‘ = 12 9+16π‘2
9+16π‘2 3=
12
9+16π‘2 β¦ (3)
Therefore, the principal unit normal vector is π π‘ = π β²(π‘)
π β²(π‘) =
β4π‘π +3π
9+16π‘2 β¦ (4)
But at π‘ = 1, the principal unit normal vector is π 1 = 1
5 (β4π + 3π ).
Example 7: Find the angle between the tangents to the curve π = ππ π + ππ π β ππ π at the point
t = Β± 1.
Solution: Differentiating the given curve w. r. to t, we get
ππ
ππ‘ = 2π‘ π + 2 π β 3π‘2 π which is the tangent vector to the curve at any point t.
Let π1 & π2
are the tangent vectors to the curve at t = 1 and t = -1 respectively, then
π1 = 2 π + 2 π β 3 π and π2
= β2 π + 2 π β 3 π
Let π be the angle between the tangents π1 & π2
, then
πΆππ π = π1 β π2
π1 π2 =
2 π +2 π β3 π β β2 π +2 π β3 π
2 π +2 π β3 π β2 π +2 π β3 π =
β4 + 4 + 9
17 17=
9
17
9
β΄ π = πππ β1 9
17
Example 8: Find the curvature and torsion of the curve π = π πππ π, π = π πππ π, π = ππ. (This curve is drawn on a circular cylinder cutting its generators at a constant angle and is known as a circular helix)
Solution: Equation of the given curve in vector form is
π = π cos π‘ π + π sin π‘ π + ππ‘ π
Differentiating w. r. to t,
ππ
ππ‘ = βπ sin π‘ π + π cos π‘ π + π π
Now, the arc length of the curve from P0 (t = 0) to any
point P (t) is given by
π = ππ
ππ‘ ππ‘
π‘
0 = π2 + π2 π‘
β΄ ππ
ππ‘ = π2 + π2
Now, the unit tangent vector,
π =ππ
ππ =
ππ ππ‘
ππ ππ‘ =
βπ sin π‘ π + π cos π‘ π + π π
π2+π2
So ππ
ππ =
ππ ππ‘
ππ ππ‘ =
βπ cos π‘ π β π sin π‘ π
π2+π2
β΄ = ππ
ππ =
π
π2+π2 is the curvature of the given curve.
Also, the unit normal vector is π = β(cos π‘ π + sin π‘ π ) and π΅ = π Γ π = (b sin π‘ π βπ cos π‘π +ππ )
π2+π2
So ππ΅
ππ =
ππ΅ ππ‘
ππ ππ‘ =
π(cos π‘ π +sin π‘ π)
π2+π2 = βππ = π(cos π‘ π + sin π‘ π )
Hence π = π
π2+π2 .
Example 9: A circular helix is given by the equation π = π ππ¨π¬ π π + π π¬π’π§ π π + π . Find the
curvature and torsion of the curve at any point and show that they are constant.
Solution: Equation of the given curve in vector form is
π = 2 cos π‘ π + 2 sin π‘ π + π
Differentiating w. r. to t, ππ
ππ‘ = β2 sin π‘ π + 2 cos π‘ π + 0 π
Now, the arc length of the curve from P0 (t = 0) to any point P (t) is given by
π = ππ
ππ‘ ππ‘
π‘
0 = 2 π‘ implying
ππ
ππ‘ = 2
Now, the unit tangent vector, π =ππ
ππ =
ππ ππ‘
ππ ππ‘ =
β2 sin π‘ π + 2 cos π‘ π + 0 π
2
So ππ
ππ =
ππ ππ‘
ππ ππ‘ =
β cos π‘ π β sin π‘ π
2
β΄ = ππ
ππ =
1
2 is the curvature of the given curve and is a constant.
Also, the unit normal vector is π = β(cos π‘ π + sin π‘ π ) and
π΅ = π Γ π = β sin π‘ π + cos π‘ π Γ β cos π‘ π β sin π‘ π = π
So ππ΅
ππ =
ππ΅ ππ‘
ππ ππ‘ = 0 = βππ = π(cos π‘ π + sin π‘ π )
10
Hence π = 0 is the torsion of the given curve and is constant.
Example 10: Show that for the curve π = π ππ β ππ π + ππππ π + π(ππ + ππ)π , the curvature
equals torsion.
Solution: Given curve is π = π 3π‘ β π‘3 π + 3ππ‘2 π + π(3π‘ + π‘3)π
Differentiating w. r. to t, ππ
ππ‘= π 3 β 3π‘2 π + 6ππ‘ π + π 3 + 3π‘2 π
Now, the arc length of the curve P0 (t = 0) to any point P (t) is given by
π = ππ
ππ‘ ππ‘
π‘
0= π 3 β 3π‘2
2 + 6ππ‘ 2 + π 3 + 3π‘2
2 ππ‘
π‘
0
= 3π 2 π‘2 + 1 ππ‘π‘
0= 3π 2
π‘3
3+ π‘
β΄ ππ
ππ‘ = 3π 2 π‘2 + 1
Now, the unit tangent vector,
π = ππ
ππ =
ππ ππ‘
ππ ππ‘ =
π 3β3π‘2 π +6ππ‘ π +π 3+3π‘2 π
3π 2 π‘2+1 =
1βπ‘2 π +2π‘ π + 1+π‘2 π
2 π‘2+1
So ππ
ππ =
ππ ππ‘
ππ ππ‘ =
β2π‘ π + 1βπ‘2 π
3π 1+π‘2 3
β΄ = ππ
ππ =
1
3π 1+π‘2 2 is the curvature of the given curve.
Also, the unit normal vector is π =β2π‘ π + 1βπ‘2
π
1+π‘2
and
π΅ = π Γ π = 1βπ‘2 π +2π‘ π + 1+π‘2 π
2 1+π‘2
So ππ΅
ππ =
ππ΅ ππ‘
ππ ππ‘ = β
β2π‘ π + 1βπ‘2 π
3π 1+π‘2 3 = β
1
3π 1+π‘2 2 .
β2π‘ π + 1βπ‘2 π
1+π‘2 = βππ
β΄ π =1
3π 1+π‘2
2 is the torsion of the given curve.
Hence curvature equals torsion for the given curve.
ASSIGNMENT 1
1. If π = π₯2π¦π§ π β 2π₯π§3π + π₯π§2π and π = 2π§ π + π¦ π β π₯2π , find π2
ππ₯ππ¦ π Γ π at (1, 0, -2).
2. Given π = π‘π π΄ + π‘π π΅ , where π΄ and π΅ are constant vectors, show that, if π and π2π
ππ‘2 are parallel
vectors, then π + π = 1, unless π = π.
3. Find the equation of tangent line to the curve π₯ = π cos π , π¦ = π sin π , π§ = ππ tan πΌ at = π
4.
11
4. Find the unit tangent vector at any point on the curve π₯ = π‘2 + 2, π¦ = 4π‘ β 5, π§ = 2π‘2 β 6π‘, where t
is any variable. Also determine the unit tangent vector at the point π‘ = 2.
5. If π = π cos π‘ π + π sin π‘ π + ππ‘ tan πΌ π , find the value of (a) ππ
ππ‘Γ
π2π
ππ‘2 (b)
ππ
ππ‘
π2π
ππ‘2
π3π
ππ‘3 .
Also find the unit tangent vector at any point t on the curve.
6. Find the equation of the osculating plane and binormal to the curve
(a) π₯ = ππ cos π , π¦ = ππ sin π , π§ = ππ at π = 0 (b) π₯ = 2 coshπ
2, π¦ = 2 sinh
π
2, π§ = 2π at π = 0
7. Find the curvature of the (a) ellipse π = π cos π‘ π + π sin π‘ π (b) parabola π = 2π‘ π + π‘2 π at point
π‘ = 1.
17.4 VELOCITY AND ACCELERATION
1. Velocity: Let the position of particle P at a time t on the curve C is π π‘ and it comes to point Q
at time π‘ + πΏπ‘ having position π π‘ + πΏπ‘ , then πΏπ = π π‘ + πΏπ‘ β π π‘ i.e. πΏπ
πΏπ‘ is directed along PQ.
As π β π along C, the line PQ becomes tangent at P to the curve C.
So π£ π‘ = ππ
ππ‘ = LtπΏπ‘β0
πΏπ
πΏπ‘ is the tangent vector to C at point P which is the velocity vector π£ (π‘) of
the motion and its magnitude gives the speed π£ = ππ
ππ‘, where s is the arc length of P from a fixed
point P0 (s=0) on C.
2. Acceleration: Acceleration vector π π‘ of a particle is the derivative of the velocity vector π£ (π‘),
and it is given by π π‘ = ππ£
ππ‘=
π2π
ππ‘2 . It is an interesting fact that the magnitude of acceleration is not
always the rate of change of π£ = π£ , as π π‘ is not always tangential to the curve C. There are two
components of acceleration, which are given as: (i) Tangential Acceleration (ii) Normal
Acceleration
Observation: The acceleration is the time rate of change of π£ π‘ = ππ
ππ‘, if and only if the normal acceleration
is zero, for then π = π2π
ππ‘2 ππ
ππ =
π2π
ππ‘2 .
3. Relative Velocity and Acceleration: Let two particles P and
Q moving along the curves C1 and C2 have position vectors π1 (π‘)
and π2 (π‘) at time t, so that π π‘ = ππ = π2 π‘ β π1 (π‘)
Differentiating w. r. to t, ππ
ππ‘=
ππ 2
ππ‘β
ππ 1
ππ‘ β¦ (1)
This defines the relative velocity of Q w. r. t. P and states that the
velocity of Q relative to P = Velocity vector of Q β Velocity
vector of P.
Again differentiating (1) w. r. to t., we have
π2π
ππ‘2=
π2π 2
ππ‘2β
π2π 1
ππ‘ 2
This defines the relative acceleration of Q w. r. t. and states that Acceleration of Q relative to P =
Acceleration of Q β Acceleration of P.
12
Example 11: Find the tangential and normal acceleration of a particle moving in a plane
curve in Cartesian coordinates.
Solution: Let π be the position vector the point P, a function of a scalar t. In particular, if the scalar
variable t is taken as an arc length s along the curve C measured from some fixed point, that is,
π₯ = π₯ π , π¦ = π¦ π , π§ = π§(π ) then π = π₯ π π + π¦ π π + π§(π )π
So that ππ
ππ =
ππ₯
ππ π +
ππ¦
ππ π +
ππ§
ππ π β¦ (1)
And ππ
ππ
2
= ππ₯
ππ
2+
ππ¦
ππ
2+
ππ§
ππ
2 β¦ (2)
For two dimension curves we have in calculus
ππ 2 = ππ₯ 2 + ππ¦ 2
which when extended to the space, becomes
ππ 2 = ππ₯ 2 + ππ¦ 2 + ππ§ 2
Or ππ₯
ππ
2+
ππ¦
ππ
2+
ππ§
ππ
2= 1
Therefore (2) gives ππ
ππ
2
= 1
That means, ππ
ππ is a unit vector along the tangent and (1) represents a unit tangent vector along the
curve C in space.
Therefore, Velocity π£ of the particle at any point of the curve is given by
π£ =ππ
ππ‘= ππ
ππ ππ ππ‘
= π£ π β¦ (3)
where π£ =ππ
ππ‘ and π =
ππ
ππ is the unit vector along the tangent.
Thus π£ =ππ
ππ‘ is the tangential component of the velocity and the normal component of the velocity
is zero.
Next, acceleration π =ππ£
ππ‘= π(π£π)
ππ‘= ππ£
ππ‘ π + π£
ππ
ππ‘
or π =π2π
ππ‘2 π + ππ
ππ‘
ππ
ππ
ππ
ππ‘=
ππ£
ππ‘ π + π£2
ππ
ππ
ππ
ππ
=ππ£
ππ‘ π +
π£2
π ππ
ππ (since radius of curvature, π =
ππ
ππ ) β¦ (4)
From the adjoining figure, π = ππ is along the tangent at P to the curve C and π is the unit vector
along the normal to P.
β΄ π = cos π π + sin π π and π = cos π
2+ π π + sin
π
2+ π π = βsin π π + cos π π
Now ππ
ππ = βsin π π + cos π π = π
Therefore, equation (4) becomes π =ππ£
ππ‘ π +
π£2
ππ
Which shows that tangential and normal components of acceleration at the point P are ππ£
ππ‘ and
π£2
π .
Since ππ£
ππ‘=
ππ£
ππ
ππ
ππ‘ = v
ππ£
ππ , so the tangential component of acceleration is also written as v
ππ£
ππ .
13
Example 12: Find the radial and transverse acceleration of a particle moving in a plane curve
in Polar coordinates.
Solution: Let the position vector of a moving particle π(π, π) be π so that
π = π π = π (cos π π + sin π π ) at any time t.
Then the velocity of the particle is π£ = ππ
ππ‘=
ππ
ππ‘ π + π
ππ
ππ‘
As π = (cos π π + sin π π ) so ππ
ππ‘= (βsinπ π + cos π π )
ππ
ππ‘
Therefore, ππ
ππ‘ is perpendicular to π and
ππ
ππ‘ =
ππ
ππ‘ i.e. if π’
is a unit vector perpendicular to π , then ππ
ππ‘=
ππ
ππ‘ π’
And thus, π£ = ππ
ππ‘=
ππ
ππ‘ π + π
ππ
ππ‘ π’
So the radial and transverse components of the velocity are ππ
ππ‘ and
ππ
ππ‘.
Also π = ππ£
ππ‘=
π2π
ππ‘ 2 π +
ππ
ππ‘
ππ
ππ‘ +
ππ
ππ‘
ππ
ππ‘ π’ +π
π2π
ππ‘ 2 π’ + π
ππ
ππ‘
ππ’
ππ‘
= π2π
ππ‘ 2β π
ππ
ππ‘
2
π + 2ππ
ππ‘
ππ
ππ‘+ π
π2π
ππ‘2 π’
π ππππ π’ = β sin π π + cos π π πππ£ππ ππ’
ππ‘= β
ππ
ππ‘π
Thus radial and transverse components of the acceleration are π2π
ππ‘ 2β π
ππ
ππ‘
2
and
2ππ
ππ‘
ππ
ππ‘+ π
π2π
ππ‘ 2 .
Example 13: A particle moves along the curve π = ππ β ππ π + ππ + ππ π + πππ β πππ π
where t denotes the time. Find the magnitudes of acceleration along the tangent and normal at
time t = 2.
Solution: The velocity of the particle is π£ = ππ
ππ‘ = 3π‘2 β 4 π + 2π‘ + 4 π + 16π‘ β 9π‘2 π
And the acceleration is π = π2π
ππ‘ 2 = 6π‘ π + 2 π + 16 β 18π‘ π
At t = 2, π£ = 8 π + 8 π β 4 π and π = 12 π + 2 π β 20 π
Since the velocity vector is also the tangent vector to the curve, so the magnitude of acceleration
along the tangent at t = 2 is = π β π£
π£ = 12 π + 2 π β 20 π β
8 π +8 π β4 π
64+64+16
= 12 8 + 2 8 + β20 (β4)
12 = 16
And, the magnitude of acceleration along the normal at t = 2 is
π β πΆππππππππ‘ ππ π along the tangent at π‘ = 2 = 12 π + 2 π β 20 π β 16 8 π +8 π β4 π
12
= 4 π β26 π β44 π
3 = 2 73
Example 14: A person going east wards with a velocity of 4 km per hour, finds that the wind
appears to blow directly from the north. He doubles his speed and the wind seems to come
from north-east. Find the actual velocity of the wind.
14
Solution: Let the actual velocity of the wind is π£ = π₯π + π¦π , where π and π represent velocities of 1
km per hour towards the east and north respectively. As the person is going eastwards with a
velocity of 4 km per hour, his actual velocity is 4π .
Then the velocity of the wind relative to the man is π₯π + π¦π β 4π , which is parallel to βπ , as it
appears to blow from the north. Hence x = 4.
When the velocity of the person becomes 8π , the velocity of the wind relative to a man is π₯π + π¦π β
8π . But this is parallel to β π + π .
β΄ π₯β8
π¦ = 1 which gives π¦ = β4 (using (1))
Hence the actual velocity of the wind is 4π β 4π i.e. 4 2 km per hour towards south east.
Example 15: A particle moves along the curve π = ππ + π, π = ππ, π = ππ + π where t is the
time. Find the components of velocity and acceleration at t = 1in the direction of the vector
π + π + ππ .
Solution: Let π be the position vector of the particle at any time t,
then π = π‘3 + 1 π + π‘2 π + 2π‘ + 3 π
So the velocity is π£ = ππ
ππ‘ = 3π‘2 π + 2π‘ π + 2 π
And the acceleration is π = π2π
ππ‘ 2 = 6π‘ π + 2 π + 0 π
At t =1, π£ = 3 π + 2 π + 2 π and π = 6 π + 2 π + 0 π
Also the unit vector in the direction of the given vector π + π + 3π is = π +π +3π
π +π +3π =
π +π +3π
11
Now the component of velocity at t = 1, in the direction of the vector π + π + 3π =
3 π +2 π +2 π β π +π +3π
11=
3+2+6
11 = 11
And the component of acceleration at t = 1, in the direction of the vector π + π + 3π =
6 π +2 π +0 π β π +π +3π
11=
6+2+0
11=
8
11
ASSIGNMENT 2
1. The particle moves along a curve π₯ = πβπ‘ , π¦ = 2 cos 3π‘ , π§ = 2 sin 3π‘, where t is the time variable.
Determine its velocity and acceleration vectors and also the magnitudes of velocity and
acceleration at π‘ = 0.
2. A particle (with position vector π ) is moving in a circle with constant angular velocity π. Show
by vector methods, that the acceleration is equal to βπ2π .
3. A particle moves on the curve π₯ = 2π‘2 , π¦ = π‘2 β 4π‘, π§ = 3π‘ β 5, where t is the time. Find the
components of velocity and acceleration at time t = 1 in the direction π β 3π + 2π .
4. The position vector of a particle at time t is π = cos(π‘ β 1) π + sinh π‘ β 1 π + ππ‘3π . Find the
condition imposed on a by requiring that at time t = 1, the acceleration is normal to the position
vector.
15
5. A particle moves so that its position vector is given by π = cos ππ‘ π + sin ππ‘ π . Show that the
velocity π£ of the particle is perpendicular to π and π Γ π£ is a constant vector.
6. A particle moves along a catenary π = π tan π. The direction of acceleration at any point makes
equal angles with the tangent and normal to the path at that point. If the speed at vertex (π = 0)
be π£0, show that the magnitude of velocity and acceleration at any point are given by π£0 ππ and
2
ππ£0
2π2π cos2 π respectively.
7. The position vector of a moving particle at a time t is π = π‘2 π β π‘3 π + π‘4 π . Find the tangential
and normal components of acceleration at t = 1.
8. A vessel A is a sailing with a velocity of 11 knots per hour in the direction south-east and a
second vessel B is sailing with a velocity of 13 knots per hour in a direction 300 of north. Find
the velocity of A relative to B.
9. The velocity of a boat relative to water is represented by 3π + 4π and that of water relative to
earth is π β 3π . What is the velocity of the boat relative to earth if π and π represent one KM an
hour east and north respectively?
10. A person travelling towards the north-east with a velocity of 6 KM per hour finds that the wind
appears to blow from the north, but when he doubles his speed it seems to come from a
direction inclined at an angle tanβ1 2 to the north of east. Show that the actual velocity of the
wind is 3 2 KM per hour towards the east.
17.5 DEL APPLIED TO SCALAR POINT FUNCTIONS: GRADIENT [KUK 2009]
Del Operator: Del operator is a vector differential operator and is written as
π = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§
Gradient of a Scalar Function: Let β (π₯, π¦, π§) be a scalar function of three variables defined over a
region R of space. Then gradient of is a vector function defined as
β = π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§,
wherever the partial derivatives exists. It may also be denoted as ππππ(β ). The del operator is also
called the gradient operator.
Level Surface: Let β (π₯, π¦, π§) be a scalar valued function and C is a constant. The surface given by
β π₯, π¦, π§ = πΆ through a point π(π ) is such that at each point on it the function has same value, is
called the level surface of β π₯, π¦, π§ through P, e.g. equi-potential or isothermal surfaces.
In other words, locus of the point π(π ) satisfying β π = πΆ form a surface through P. This surface is
called the level surface through P.
Gradient as Normal or Geometrical Interpretation of Gradient
16
Let β π = β π₯, π¦, π§ is a scalar function where π = π₯π + π¦π + π§π . And β π₯, π¦, π§ = πΆ is the level
surface of β through π(π ). Let π(π + πΏπ ) be a point on neighboring level surface β + πΏβ , then
πβ β Ξ΄r = π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§ β πΏπ₯π + πΏπ¦π + πΏπ§π
= πβ
ππ₯ πΏπ₯ +
πβ
ππ¦ πΏπ¦ +
πβ
ππ§ πΏπ§ = πΏβ β¦ (1)
Now if P and Q lie on same level surface i.e. β & β + πΏβ are
same, then πΏβ = 0.
Implies πβ β Ξ΄r = 0 (using (1))
Therefore, πβ is perpendicular (normal) to every Ξ΄r lying on
this surface.
Hence, πβ is normal to the surface β π₯, π¦, π§ = πΆ and we can write πβ = πβ n , where π is unit
normal vector to the surface.
See the Fig. 17.7, if the perpendicular distance PM between the surfaces through P and Q be πΏπ, then
rate of change of β along the normal to the surface through P is
πβ
ππ= πΏπ‘πΏπβ0
πΏβ
πΏπ = πΏπ‘πΏπβ0
πβ βΞ΄r
πΏπ (using (1))
= LtπΏπβ0 πβ n βΞ΄r
πΏπ = πβ LtπΏπβ0
n βΞ΄r
πΏπ β¦ (2)
= πβ LtπΏπβ0Ξ΄n
πΏπ = πβ (π΄π π β πΏπ = πΏπ cos π = πΏπ)
Hence the magnitude of πβ i.e. πβ = πβ
ππ β¦ (3)
Thus ππππ β is normal vector to the level surface β π₯, π¦, π§ = πΆ and its magnitude represents the rate
of change of β along this normal.
Directional Derivative: If πΏπ denotes the length PQ and π be the unit vector in the direction of PQ,
the limiting value of πΏπ
πΏπ as πΏπ β 0 (i.e.
ππ
ππ) is known as the directional derivative of f along the
direction PQ.
Since πΏπ =πΏπ
cos πΌ=
πΏπ
π βπ’ , therefore
ππ
ππ= LtπΏπβ0 π β π’
πΏπ
πΏπ = π’ β
πΏπ
πΏπ π = π’ β ππ
Thus the directional derivative of f in the direction of π’ is the resolved part of ππ in the direction of π’ .
Since ππ β π’ = ππ cos πΌ β€ ππ
It follows that ππ gives the maximum rate of change of f.
Properties of Gradient Operator: Let β (π₯, π¦, π§) & Ο(π₯, π¦, π§) are two differentiable scalar functions
defined over some region R. then the gradient operator has following properties:
17
(i) Gradient of a constant multiple of scalar function β
ππππ πΆβ = πΆ ππππ(β ) or β πΆβ = πΆ ββ
Proof: Consider ππππ πΆβ = β πΆβ
= π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ πΆβ
= π π
ππ₯(πΆβ ) + π
π
ππ¦(πΆβ ) + π
π
ππ§(πΆβ )
= πΆ π πβ
ππ₯+ πΆ π
πβ
ππ¦+ πΆ π
πβ
ππ§
= πΆ π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§
= πΆ π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β
= πΆ ββ = πΆ ππππ β
Hence ππππ πΆβ = πΆ ππππ(β ).
(ii) Gradient of sum or difference of two scalar functions
ππππ β Β± π = ππππ β Β± ππππ(π) or β β Β± π = ββ Β± βπ
Proof: Consider ππππ β Β± π = β β Β± π
= π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β Β± π
= π π
ππ₯ β Β± π + π
π
ππ¦ β Β± π + π
π
ππ§ β Β± π
= π πβ
ππ₯Β±
ππ
ππ₯ + π
πβ
ππ¦Β±
ππ
ππ¦ + π
πβ
ππ§Β±
ππ
ππ§
= π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§ Β± π
ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§
= π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β Β± π
π
ππ₯+ π
π
ππ¦+ π
π
ππ§ π
= ββ Β± βπ = ππππ β Β± ππππ(π).
Hence ππππ β Β± π = ππππ β Β± ππππ(π).
(iii) Gradient of product of two scalar functions
ππππ β π = β ππππ π + π ππππ(β ) or β β π = β βπ + π ββ
Proof: Consider ππππ β π = β β π
= π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β π
= π π
ππ₯ β π + π
π
ππ¦ β π + π
π
ππ§ β π
18
= π β ππ
ππ₯+ π
πβ
ππ₯ + π β
ππ
ππ¦+ π
πβ
ππ¦ + π β
ππ
ππ§+ π
πβ
ππ§
= β π ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§ + π π
πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§
= β βπ + π ββ = β ππππ π Β± π ππππ(β )
Hence ππππ β π = β ππππ π Β± π ππππ(β ).
(iv) Gradient of quotient of two scalar functions
ππππ β
π =
π ππππ β ββ ππππ (π)
π 2 or β
β
π =
π β β ββ β(π)
π 2, provided π β 0.
Proof: Consider ππππ β
π = β
β
π
= π π
ππ₯+ π
π
ππ¦+ π
π
ππ§
β
π
= π π
ππ₯
β
π + π
π
ππ¦
β
π + π
π
ππ§
β
π
= π π
πβ
ππ₯ β β
ππ
ππ₯
π2 + π π
πβ
ππ¦ β β
ππ
ππ¦
π2 + π π
πβ
ππ§ β β
ππ
ππ§
π2
=1
π2 π π
πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§ β β π
ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§
=1
π2 π ππππ(β ) β β ππππ(π) =
π ππππ β ββ ππππ (π)
π 2
=π β β ββ β(π)
π 2
Hence ππππ β
π =
π ππππ β ββ ππππ (π)
π 2 .
Example 16: If β = ππππ β ππππ , then find ππππ β at (1, -2, -1).
Solution: ππππ β = β 3π₯2π¦ β π¦3π§2
= π π
ππ₯ 3π₯2π¦ β π¦3π§2 + π
π
ππ¦ 3π₯2π¦ β π¦3π§2 + π
π
ππ§ 3π₯2π¦ β π¦3π§2
= π 6π₯π¦ + π 3π₯2 β 3π¦2π§2 + π β2π¦3π§
At (1, -2, -1), ππππ β = π 6 1 β2 + π 3 1 2 β 3 β2 2 β1 2 + π β2 β2 3 β1
= β12π β 9π β 16π
Example 17: Prove that π ππ = π ππβπ π , where π = ππ + ππ + ππ .
Solution: Here π = π₯π + π¦π + π§π and π2 = π₯2 + π¦2 + π§2
So differentiating partially w. r. t. x, ππ
ππ₯=
π₯
π, Similarly,
ππ
ππ¦=
π¦
π and
ππ
ππ§=
π§
π β¦ (1)
19
Consider β ππ = π π
ππ₯ ππ + π
π
ππ¦ ππ + π
π
ππ§ ππ
= π π ππβ1 ππ
ππ₯ + π π ππβ1 ππ
ππ¦ + π π ππβ1 ππ
ππ§
= π π ππβ1 π₯
π + π π ππβ1 π¦
π + π π ππβ1 π§
π
= π π ππβ2 π₯ + π π ππβ2 π¦ + π π ππβ2 π§
= π ππβ2 π₯π + π¦π + π§π = π ππβ2 π
Hence β ππ = π ππβ2 π
Example 18: Find the unit vector normal to the surface ππ + ππ + ππππ = π at
(i) the point (1, 2, -1) * (ii) the point (1, 3, -1)** [KUK *2006, **2011]
Solution: We know that a vector normal to a surface is given by its gradient, so if π is the vector
normal to the given surface then
π = β π₯3 + π¦3 + 3π₯π¦π§ β 3 = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ π₯3 + π¦3 + 3π₯π¦π§ β 3
= 3π₯2 + 3π¦π§ π + 3π¦2 + 3π₯π§ π + 3π₯π¦ π
(i) At point (1, 2, -1), π = 3 1 2 + 3 2 β1 π + 3 2 2 + 3 1 β1 π + 3 1 2 π
= βππ + 9π + 6π
Also π = β3 2 + 9 2 + 6 2 = 126
Therefore the unit normal vector to the given surface at a point (1, 2, -1) is
π =1
126 βππ + 9π + 6π .
(ii) At point (1, 3, -1), π = 3 1 2 + 3 3 β1 π + 3 3 2 + 3 1 β1 π + 3 1 3 π
= β6π + 24π + 9π
Also π = β6 2 + 24 2 + 9 2 = 693
Therefore the unit normal vector to the given surface at a point (1, 3, -1) is
π =1
693 β 6 π + 24 π + 9 π .
Example 19: Show that ππππ π(ππ+ππ+ππ) = ππππ, π°π‘ππ«π ππ = π π = ππ + ππ + ππ.
Solution: Here π2 = π₯2 + π¦2 + π§2
So differentiating w. r. t. x, ππ
ππ₯=
π₯
π, Similarly,
ππ
ππ¦=
π¦
π and
ππ
ππ§=
π§
π.
Now ππππ π π₯2+π¦2+π§2 = β ππ2= π
π
ππ₯ ππ2
+ π π
ππ¦ ππ2
+ π π
ππ§ ππ2
= π ππ2 2π
ππ
ππ₯ + π ππ2
2πππ
ππ¦ + π ππ2
2πππ
ππ§
20
= ππ2 2π π
ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§ = ππ2
2π π π₯
π+ π
π¦
π+ π
π§
π
= 2ππ2π
Example 20: If π = π + π + π, π = ππ + ππ + ππ ππ§π π = ππ + ππ + ππ , then show that
ππππ π, ππππ π ππ§π ππππ π are coplanar.
Solution: Consider ππππ π’ = π ππ’
ππ₯+ π
ππ’
ππ¦+ π
ππ’
ππ§ = π 1 + π 1 + π 1 = π + π + π
ππππ π£ = π ππ£
ππ₯+ π
ππ£
ππ¦+ π
ππ£
ππ§= π 2π₯ + π 2π¦ + π 2π§ = 2π₯ π + 2π¦ π + 2π§ π
ππππ π€ = π ππ€
ππ₯+ π
ππ€
ππ¦+ π
ππ€
ππ§= π¦ + π§ π + π§ + π₯ π + π₯ + π¦ π
We know that three vectors π , π πππ π are coplanar if their scalar triple product is zero i.e.
π π π = 0
Consider ππππ π’ ππππ π£ ππππ π€ = 1 1 1
2π₯ 2π¦ 2π§π¦ + π§ π§ + π₯ π₯ + π¦
= 2 1 1 1π₯ π¦ π§
π¦ + π§ π§ + π₯ π₯ + π¦ (taking common 2 from R2)
= 2 1 1 1
π₯ + π¦ + π§ π₯ + π¦ + π§ π₯ + π¦ + π§π¦ + π§ π§ + π₯ π₯ + π¦
(adding R2 and R3)
= 2(π₯ + π¦ + π§) 1 1 11 1 1
π¦ + π§ π§ + π₯ π₯ + π¦
(taking common (x + y + z) from R2)
= 2 π₯ + π¦ + π§ 0 = 0
Hence ππππ π’, ππππ π£ and ππππ π€ are coplanar.
Example 21: Show that ππππ π π Γ π = π .
Solution: Here ππππ π π = β π(π) = π π
ππ₯ π(π) + π
π
ππ¦ π(π) + π
π
ππ§ π(π)
= π π β²(π) ππ
ππ₯ + π π β²(π)
ππ
ππ¦ + π π β²(π)
ππ
ππ§
= π β²(π) π ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§ = π β²(π) π
π₯
π+ π
π¦
π+ π
π§
π
= π β²(π) π
π
21
Now ππππ π π Γ π = π β²(π) π
πΓ π = π β²(π)
1
π π Γ π = 0
π ππππ π Γ π = 0
Example 22: Find the directional derivative of π π, π, π = ππππππ at the point (1, 1, -1) in the
direction of the tangent to the curve π = ππ, π = π πππ π + π, π = π β πππ π at t = 0.
Solution: Consider β π π₯, π¦, π§ = β π₯2π¦2π§2 = 2π₯π¦2π§2 π + 2π¦π₯2π§2 π + 2π§π₯2π¦2 π
At (1, 1, -1), β π π₯, π¦, π§ = 2 π + 2 π β 2 π
Now π = π₯π + π¦π + π§π = ππ‘ π + 2 sin π‘ + 1 π + (π‘ β cos π‘)π
So tangent to the curve is ππ
ππ‘ = ππ‘ π + 2 cos π‘ π + (1 + sin π‘)π
At t = 0, ππ
ππ‘ = π + 2π + π
And the unit tangent vector is ππ
ππ‘/
ππ
ππ‘ =
π +2π +π
6
So the required directional derivative in the direction of the tangent is βπ π₯, π¦, π§ β ππ
ππ‘/
ππ
ππ‘
= 2 π + 2 π β 2 π β π + 2π + π / 6 =4
6=
2 3
3
Example 23: If the directional derivative β = π πππ + π πππ + π πππ at the point (1, 1, 1) has
maximum magnitude 15 in the direction parallel to the line πβπ
π=
πβπ
βπ=
π
π , find the values of
a, b and c. [Madrass 2004]
Solution: Consider β β = β (π π₯2π¦ + π π¦2π§ + π π§2π₯)
= 2π π₯π¦ + π π§2 π + π π₯2 + 2π π¦π§ π + (π π¦2 + 2π π§π₯)π
At (1, 1, 1), β β = 2π + π π + π + 2π π + (π + 2π)π
We know that directional derivative of β is maximum in the direction of its normal vector β β , but it
is given to be maximum in the direction of the line π₯β1
2=
π¦β3
β2=
π§
1 .
Therefore, the line and normal vector are parallel to each other, which results as:
2π+π
2=
π+2π
β2=
π+2π
1 β¦ (1)
Taking first two members of (1), 3π + 2π + π = 0
and by last two members of (1), π + 4π + 4π = 0
Solving the two obtained equations, π
4=
π
β11=
π
10 = π πΏππ‘
=> π = 4π, π = β11π πππ π = 10π β¦ (2)
Also given that maximum magnitude of directional derivative is 15 units i.e. ββ = 15
So , 2π + π 2 + π + 2π 2 + π + 2π 2 = 15 2 β¦(3)
Putting the values of a, b and c from (2),
22
8π + 10π 2 + 4π β 22π 2 + β11π + 20π 2 = 15 2 => π = Β± 5
9
Hence π = Β± 20
9 , π = β
55
9 , π = Β±
50
9.
Example 24: In what direction from (3, 1, -2) is the directional derivative of β = ππππππ
maximum? Find also the magnitude of this maximum. [KUK 2010, 2007, 2006]
Solution: The vector normal to the given surface is
π = β β = β π₯2π¦2π§4 = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ π₯2π¦2π§4
= 2π₯π¦2π§4 π + 2π₯2π¦π§4 π + 4π₯2π¦2π§3 π
At point (3, 1, -2) π = 2 3 1 2 β2 4 π + 2 3 2 1 β2 4 π + 4 3 2 1 2 β2 3 π
= 96 π + 288 π β 288 π
Also π = 96 2 + 288 2 + β288 2 = 96 19
So the directional derivative of given surface will be maximum in the direction of
96 π + 288 π β 288 π and the magnitude of this maximum is 96 19.
Example 25: Find the angle between the surfaces ππ + ππ + ππ = π and π = ππ + ππ β π at
(2, -1, 2). [KUK 2008]
Solution: Given surfaces are
β 1 = π₯2 + π¦2 + π§2 β 9 = 0 β¦ (1)
and β 2 = π₯2 + π¦2 β π§ β 3 = 0 β¦ (2)
We know that gradient of a surface gives the vector normal to the surface. Let π 1 and π 2 are the
vectors normal to the surfaces (1) and (2) respectively.
Now ββ 1 = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β π₯2 + π¦2 + π§2 = 9 = 2π₯ π + 2π¦ π + 2π§ π β¦ (3)
ββ 2 = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β π₯2 + π¦2 β π§ β 3 = 2π₯ π + 2π¦ π β π β¦ (4)
So, π 1 = ββ 1 at (2,β1,2) = 4 π β 2 π + 4 π and π 2 = ββ 2 at (2,β1,2) = 4 π β 2 π β π
Let π be the angle between the given surfaces at point (2, -1, 2), then π will also be an angle between
their normals π 1 and π 2.
Therefore, cos π = π 1βπ 2
π 1 π 2 =
4 π β2 π +4 π β 4 π β2 π β π
4 π β2 π +4 π 4 π β2 π β π =
16+4β4
36 21=
8
3 21.
Example 26: Find the constants a and b so that the surface πππ β πππ = π + π π is orthogonal
to the surface ππππ + ππ = π at the point (1, -1, 2).
Solution: Given surfaces are
π = ππ₯2 β ππ¦π§ β π + 2 π₯ = 0 β¦ (1)
and π = 4π₯2π¦ + π§3 β 4 = 0 β¦(2)
23
Let π1 and π2 are the vectors normal to the surfaces (1) and (2) at (1, -1, 2), respectively.
Consider βπ = π ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§ = π 2ππ₯ β π + 2 + π (βππ§) + π (βππ¦) β¦ (3)
And βπ = π ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§ = π 8π₯π¦ + π (4π₯2) + π (3π§2) β¦ (4)
So, π1 = βπ (1, β1, 2) = π π β 2 + π β2π + π π = π β 2 π β 2ππ + ππ
π2 = βπ (1, β1, 2) = π β8 + π 4 + π 12 = β8π + 4π + 12π
Given that surfaces (1) and (2) cut orthogonally at (1, -1, 2), so their normal vectors i.e. π1 and π2
should also be orthogonal to each other.
Therefore, π1 . π2 = 0
=> π β 2 π β 2ππ + ππ . β8π + 4π + 12π = 0
=> β8 π β 2 β 8π + 12π = 0
=> 2π β π = 4 β¦ (5)
Also the point (1, -2, 1) lies on the surface (1), so we have
π + 2π β π + 2 = 0 ππ 2π β 2 = 0 ππ π = π
Putting value of b in (3), we get 2π β 1 = 4 ππ π =π
π
Example 27: Show that the components of a vector π along and normal (perpendicular) to a
vector π , in the plane of π and π , are π βπ
π π π and
π Γ π Γπ
π π.
Solution: Let ππ΄ = π and ππ΅ = π and ππ be the projection
of π on π (Fig. 17.8)
β΄ Component of π along π = OM (unit vector along π )
= π β π π = π βπ
π
π
π =
π βπ
π 2 π
Also the component of π normal to π = ππ΅ = ππ΅ β ππ
= π β π βπ
π 2 π =
π βπ π β π βπ π
π 2=
a Γ π Γπ
π 2
Example 28: If f and π are point functions, prove that the components of the latter normal and
tangential to the surface f = 0 are π βππ ππ
ππ π and
ππΓ π Γππ
ππ π.
Solution: We know that for the given surface f = 0, the vector normal to the surface is given by the
gradient i.e. βπ.
Now the component of πΉ normal to the given surface is = πΉ β βπ
βπ
βπ
βπ =
πΉ ββπ βπ
βπ 2=
πΉ ββπ βπ
βπ 2
And the component of πΉ tangential to the given surface is = πΉ β π‘ππ ππππππ πππππππππ‘ ππ πΉ
24
= πΉ β πΉ ββπ βπ
βπ 2=
πΉ βπ 2β πΉ ββπ βπ
βπ 2=
πΉ βπββπ β πΉ ββπ βπ
βπ 2=
βπΓ πΉ Γβπ
βπ 2
ASSIGNMENT 3
1. Find ββ , if β = log π₯2 + π¦2 + π§2 .
2. Show that 1
π= β
π
π3.
3. What is the directional derivative of β = π₯π¦2 + π¦π§3 at the point (2, -1, 1) in the direction of the
normal to the surface π₯ log π§ β π¦2 = β4 at (-1, 2, 1)? [JNTU 2005; VTU 2004]
4. What is the directional derivative of β = π₯2π¦π§ + 4π₯π§2at the point (1, -2, 1) in the direction of the
vector 2π β π β 2π . [VTU 2007; UP Tech, JNTU 2006]
5. The temperature of points in a space is given by π π₯, π¦, π§ = π₯2 + π¦2 β π§. A mosquito located at
(1, 1, 2) desires to fly in such a direction that it will get warm as soon as possible. In what
direction should it move?
6. What is the greatest rate of increase of π’ = π₯2 + π¦π§2 at the point (1, -1, 3)?
7. Find the angle between the tangent planes to the surfaces π₯ log π§ = π¦2 β 1 and π₯2π¦ = 2 β π§ at the
point (1, 1, 1). [JNTU 2003]
8. Calculate the angle between the normals to the surface π₯π¦ = π§2 at the points (4, 1, 2) and
(3, 3, -3).
9. Find the angle between the surfaces π₯2 + π¦2 + π§2 = 9 and π§ = π₯2 + π¦2 β 3 at (2, -1, 2).
10. Find the values of π and π so that the surface ππ₯2π¦ + ππ§3 = 4 may cut the surface
5π₯2 = 2π¦π§ + 9π₯ orthogonally at (1, -1, 2)
17.6 DEL APPLIED TO VECTOR POINT FUNCTIONS (Divergence & Curl)
1. Divergence: Let π (π₯, π¦, π§) = π1(π₯, π¦, π§)π + π2(π₯, π¦, π§)π + π3(π₯, π¦, π§)π be a continuously differentiable
vector point function. Divergence π (π₯, π¦, π§) of is a scalar which is denoted by β β π and is defined as
β β π = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β π1 π + π2 π + π3π =
ππ1
ππ₯+
ππ2
ππ¦+
ππ3
ππ§.
t is also denoted by πππ£ π .
Physical interpretation of Divergence
Consider the case of fluid flow.
Let π£ = π£π₯ π + π£π¦ π + π£π§π be the velocity of the fluid at a
point π(π₯, π¦, π§) . Consider a small parallelopiped with
edges πΏπ₯, πΏπ¦ and πΏπ§ parallel to the x, y and z axis
25
respectively in the mass of fluid, with one of its corner at point π.
So, the mass of fluid flowing in through the face πππ π per unit time = π£π¦ πΏπ§ πΏπ₯
and the mass of fluid flowing out of the face πβ²πβ²π β²πβ² per unit time
= π£π¦+πΏπ¦ πΏπ§ πΏπ₯ = π£π¦ +ππ£π¦
ππ¦ πΏπ¦ πΏπ§ πΏπ₯
β΄ The net decrease in fluid mass in the parallelopiped corresponding to flow along y-axis
= π£π¦ +ππ£π¦
ππ¦ πΏπ¦ πΏπ§ πΏπ₯ β π£π¦ πΏπ§ πΏπ₯ =
ππ£π¦
ππ¦ πΏπ₯ πΏπ¦ πΏπ§
Similarly, the net decrease in fluid mass in the parallelopiped corresponding to the flow along x-axis
and z-axis is ππ£π₯
ππ₯ πΏπ₯ πΏπ¦ πΏπ§ and
ππ£π§
ππ§ πΏπ₯ πΏπ¦ πΏπ§ respectively.
So, total decrease in mass of fluid mass in the parallelopiped per unit time
= ππ£π₯
ππ₯+
ππ£π¦
ππ¦+
ππ£π§
ππ§ πΏπ₯ πΏπ¦ πΏπ§
Thus, the rate of loss of fluid per unit volume =ππ£π₯
ππ₯+
ππ£π¦
ππ¦+
ππ£π§
ππ§
= π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β π£π₯ π + π£π¦ π + π£π§π
= β β π£ = πππ£ π£
Hence, πππ£ π£ gives the rate at which fluid is originating or diminishing at a point per unit volume.
If the fluid is incompressible, there can be no loss or gain in the volume element i.e. πππ£ π£ = 0.
Observations:
(i) if π£ represent the electric flux, then πππ£ π£ is the amount of flux which diverges per unit volume in unit
time.
(ii) if π£ represent the heat flux, then πππ£ π£ is the rate at which the heat is issuing from a point per unit
volume.
(iii) If the flux entering any element of space is the same as that leaving it i.e.πππ£ π£ = 0 everywhere, then
such a vector point function is called Solenoidal.
Example 29: If π = ππ + ππ + ππ then show that π ππ π = π.
Solution: πππ£ π = β β π = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β π₯π + π¦π + π§π
=π
ππ₯ π₯ +
π
ππ¦ π¦ +
π
ππ§ π§ = 1 + 1 + 1 = 3
Example 30: Evaluate π ππ π where π = ππππ π β ππππ π + ππππ π at (1, 1, 1).
Solution: πππ£ π = π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§ β 2π₯2π§ π β π₯π¦2π§ π + 3π¦2π₯ π
26
=π
ππ₯ 2π₯2π§ +
π
ππ¦ βπ₯π¦2π§ +
π
ππ§ 3π¦2π₯
= 4π₯π§ β 2π₯π¦π§ + 0
At (1, 1, 1), πππ£ π = 4 1 1 β 2 1 1 1 = 2
Example 31: Determine the constant a so that the vector π = π + ππ π + π β ππ π + (π + ππ) π
is solenoidal.
Solution: Given that the vector π is solenoidal, so πππ£ π = 0
π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β π₯ + 3π¦ π + π¦ β 2π§ π + (π₯ + ππ§) π = 0
π
ππ₯ π₯ + 3π¦ +
π
ππ¦ π¦ β 2π§ +
π
ππ§ π₯ + ππ§ = 0
1 + 1 + π = 0 => π = β2
2. Curl: Let π (π₯, π¦, π§) = π1(π₯, π¦, π§)π + π2(π₯, π¦, π§)π + π3(π₯, π¦, π§)π be a continuously differentiable vector
point function. Curl of π (π₯, π¦, π§) of is a vector which is denoted by β Γ π and is defined as
β Γ π = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ Γ π = π Γ
ππ
ππ₯ +π Γ
ππ
ππ¦ +π Γ
ππ
ππ§
Also in component form, curl of π (π₯, π¦, π§) is
β Γ π = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ Γ π1 π + π2 π + π3π
=
π π π
π
ππ₯
π
ππ¦
π
ππ§
π1 π2 π3
= π ππ3
ππ¦β
ππ2
ππ§ + π
ππ1
ππ§β
ππ3
ππ₯ + π
ππ2
ππ₯β
ππ1
ππ¦
Physical Interpretation of Curl
Consider the motion of a rigid body rotating with
angular velocity π about an axis OA, where O is a
fixed point in the body. Let π be the position vector of
any point P of the body. The point P describing a circle
whose center is M and radius is PM = π sin π where π is
the angle between π and π , then the velocity of P is
ππ sin π. This velocity is normal to the plane POM i.e.
normal to the plane of π and π .
So, if π£ is the linear velocity of P, then π£ =
π π sin π π = π Γ π
27
(π πππππ ππππππ π‘π π πππ π )
Now, if π = π1 π + π2 π + π3π and π = π₯ π + π¦ π + π§ π , then
πΆπ’ππ π£ = π Γ π = π π π
π1 π2 π3
π₯ π¦ π§ = π π2π§ β π3π¦ + π π3π₯ β π1π§ + π π1π¦ β π2π₯
And
πΆπ’ππ π£ =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π2π§ β π3π¦ π3π₯ β π1π§ π1π¦ β π2π₯
= π π1 + π1 + π π2 + π2 + π π3 + π3
= 2π1 π + 2π2 π + 2π3 π = 2 π
Hence π =1
2πΆπ’ππ π£
Thus the angular velocity of rotation at any point is equal to half the curl of the velocity vector.
Observations:
(i) The curl of a vector point function gives the measure of the angular velocity at a point.
(ii) If the curl of a vector point function becomes zero i.e. β Γ π = 0, then π is called an irrotational
vector.
Example 32: If π = ππ + ππ + ππ then show that ππππ π = π .
Solution: ππ’ππ π = β Γ π =
π π π π
ππ₯
π
ππ¦
π
ππ§
π₯ π¦ π§
= π ππ§
ππ¦β
ππ¦
ππ§ + π
ππ₯
ππ§β
ππ§
ππ₯ + π
ππ¦
ππ₯β
ππ₯
ππ¦
= π 0 β 0 + π 0 β 0 + π 0 β 0 = 0 .
Example 33: Find a so that the vector π = πππ β ππ π + π β π ππ π + π β π πππ π is
irrotational.
Solution: Given that π is irrotational, therefore ππ’ππ π = 0 β¦ (1)
But ππ’ππ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
ππ₯π¦ β π§3 π β 2 π₯2 1 β π π₯π§2
= π 0 β 0 + π β3π§2 β 1 β π π§2 + π π β 2 2π₯ β ππ₯
= 0 π + β4 + π π§2 π + β4 + π π₯ π
28
Using (1), 0 π + β4 + π π§2 π + β4 + π π₯ π = 0 π + 0 π + 0 π
Comparing the corresponding components both sides,
β4 + π = 0 => π = 4.
Example 34: If π = πππ π + πππππ π β ππππ π , find the ππππ π at the point (1, -1, 1).
Solution: ππ’ππ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π₯π¦2 2π₯2π¦π§ β3π¦π§2
= π β3π§2 β 2π₯2π¦ + π 0 β 0 + π 4π₯π¦π§ β 2π₯π¦
At (1, -1, 1),
ππ’ππ π = π β3(1)2 β 2 1 2(β1) + π 0 β 0 + π 4 1 β1 (1) β 2(1)(β1)
= βπ β 2 π .
17.7 DEL APPLIED TO THE PRODUCT OF POINT FUNCTIONS
Let π, π are two scalar point functions and π , π are two vector point functions, then
1. β Ο Ο = Ο βΟ + Ο βΟ
2. β β Ο π = βΟ β π + Ο β β π
3. β Γ Ο π = βΟ Γ π + Ο β Γ π
4. β π β π = π β β π + π β β π + π Γ β Γ π + π Γ β Γ π [KUK 2007]
5. β β π Γ π = π β β Γ π β π β β Γ π
6. β Γ π Γ π = π β β π β π β β π + π β β π β π β β π [KUK 2011]
Proof 1: Consider β Ο Ο = π π
ππ₯ Ο Ο = π Ο
πΟ
ππ₯+ Ο
πΟ
ππ₯ = π Ο
πΟ
ππ₯ + π Ο
πΟ
ππ₯
= Ο π πΟ
ππ₯ + Ο π
πΟ
ππ₯ = Ο βΟ + Ο βΟ
Proof 2: Consider β β Ο π = π β π
ππ₯ Ο π = π β
πΟ
ππ₯π + Ο
ππ
ππ₯ = π β
πΟ
ππ₯π + π β Ο
ππ
ππ₯
= π πΟ
ππ₯ β π + Ο π β
ππ
ππ₯ = βΟ β π + Ο β β π
Proof 3: Consider β Γ Ο π = π Γ π
ππ₯ Ο π = π Γ
πΟ
ππ₯π + Ο
ππ
ππ₯ = π Γ
πΟ
ππ₯π + π Γ Ο
ππ
ππ₯
= π πΟ
ππ₯ Γ π + Ο π Γ
ππ
ππ₯ = βΟ Γ π + Ο β Γ π
Proof 4: Consider β π β π = π π
ππ₯ π β π = π
ππ
ππ₯β π + π β
ππ
ππ₯ = π
ππ
ππ₯ β π + π π β
ππ
ππ₯
β¦ (1)
29
But, π Γ π Γππ
ππ₯ = π β
ππ
ππ₯ π β π β π
ππ
ππ₯
or π βππ
ππ₯ π = π Γ π Γ
ππ
ππ₯ + π β π
ππ
ππ₯
So, π βππ
ππ₯ π = π Γ π Γ
ππ
ππ₯ + π β π
ππ
ππ₯= π Γ β Γ π + π β β π β¦ (2)
Interchanging π and π in (2)
π βππ
ππ₯ π = π Γ π Γ
ππ
ππ₯ + π β π
ππ
ππ₯= π Γ β Γ π + π β β π β¦ (3)
Using (2) and (3) in (1), we get
β π β π = π β β π + π β β π + π Γ β Γ π + π Γ β Γ π
Proof 5: Consider β β π Γ π = π β π
ππ₯ π Γ π = π β
ππ
ππ₯Γ π + π Γ
ππ
ππ₯
= π β ππ
ππ₯Γ π + π β π Γ
ππ
ππ₯ = π β π Γ
ππ
ππ₯ β π β π Γ
ππ
ππ₯
= π β β Γ π β π β β Γ π
[using π β π Γ π = π β π Γ π = βπ β π Γ π ]
Proof 6: Consider β Γ π Γ π = π Γ π
ππ₯ π Γ π = π Γ
ππ
ππ₯Γ π + π Γ
ππ
ππ₯
= π Γ ππ
ππ₯Γ π + π Γ π Γ
ππ
ππ₯
= π β π ππ
ππ₯β π β
ππ
ππ₯ π + π β
ππ
ππ₯ π β π β π
ππ
ππ₯
[using π Γ π Γ π = π β π π β π β π π ]
= π β π ππ
ππ₯β π π β
ππ
ππ₯ + π π β
ππ
ππ₯ β π β π
ππ
ππ₯
= π β β π β π β β π + π β β π β π β β π
= π β β π β π β β π + π β β π β π β β π
17.8 DEL APPLIED TWICE TO POINT FUNCTIONS
Let β be a scalar point function and π be a vector point function, then ββ and β Γ π being the vector
point functions, we can find their divergence and curl; whereas β β π being the scalar point function,
we can find its gradient only. Thus we have following formulae:
1. πππ£ ππππ β = β β ββ = β2β =π2β
ππ₯2 +π2β
ππ¦2 +π2β
ππ§2
2. ππ’ππ ππππ β = β Γ ββ = 0
3. πππ£ ππ’ππ π = β β β Γ π = 0
4. ππ’ππ ππ’ππ π = β Γ β Γ π = β β β π β β2π = ππππ πππ£ π β β2π [KUK 2006]
5. ππππ πππ£ π = β β β π = ππ’ππ ππ’ππ π + β2π = β Γ β Γ π + β2π
30
Proof 1: β2β = β β ββ = β β π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§
=π
ππ₯ πβ
ππ₯ +
π
ππ¦
πβ
ππ¦ +
π
ππ§
πβ
ππ§ =
π2β
ππ₯2+
π2β
ππ¦2+
π2β
ππ§2
Here β2 is called the Laplacian Operator and β2β = 0 is called the Laplaceβs Equation.
Proof 2: ππ’ππ ππππ β = β Γ ββ = β Γ π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§
=
π π π
π
ππ₯
π
ππ¦
π
ππ§
πβ
ππ₯
πβ
ππ¦
πβ
ππ§
= π
π2β
ππ¦ππ§β
π2β
ππ§ππ¦ = 0
Proof 3: πππ£ ππ’ππ π = β β β Γ π = π π
ππ₯ β π Γ
ππ
ππ₯+ π Γ
ππ
ππ¦+ π Γ
ππ
ππ§
= π β π Γπ2π
ππ₯2+ π Γ
π2π
ππ₯ππ¦+ π Γ
π2π
ππ₯ππ§
= π Γ π βπ2π
ππ₯2 + π Γ π βπ2π
ππ₯ππ¦+ π Γ π β
π2π
ππ₯ππ§
= π βπ2π
ππ₯ππ¦β π β
π2π
ππ₯ππ§ = 0
Proof 4: ππ’ππ ππ’ππ π = β Γ β Γ π = π π
ππ₯ Γ π Γ
ππ
ππ₯+ π Γ
ππ
ππ¦+ π Γ
ππ
ππ§
= π Γ π Γπ2π
ππ₯2 + π Γπ2π
ππ₯ππ¦+ π Γ
π2π
ππ₯ππ§
= π Γ π Γπ2π
ππ₯2 + π Γ π Γπ2π
ππ₯ππ¦ + π Γ π Γ
π2π
ππ₯ππ§
= π βπ2π
ππ₯2 π β π β π π2π
ππ₯2 + π βπ2π
ππ₯ππ¦ π β π β π
π2π
ππ₯ππ¦ + π β
π2π
ππ₯ππ§ π β π β π
π2π
ππ₯ππ§
= π βπ2π
ππ₯2 π + π βπ2π
ππ₯ππ¦ π + π β
π2π
ππ₯ππ§ π β
π2π
ππ₯2
= π π
ππ₯ π β
ππ
ππ₯+ π β
ππ
ππ¦+ π β
ππ
ππ§ β
π2π
ππ₯2 = β β β π β β2π
Proof 5: To get this formula we are to re-arrange the terms in last proof.
Example 35: Show that ππ ππ = π(π + π)ππβπ.
Solution: We know that π2 = π₯2 + π¦2 + π§2
On differentiation w. r. t. x, ππ
ππ₯=
π₯
π
Similarly ππ
ππ¦=
π¦
π and
ππ
ππ§=
π§
π
Now β2 ππ = π2
ππ₯2+
π2
ππ¦2+
π2
ππ§2 ππ =
π
ππ₯
πππ
ππ₯ +
π
ππ¦
πππ
ππ¦ +
π
ππ§
πππ
ππ§ β¦ (1)
And π
ππ₯
πππ
ππ₯ =
π
ππ₯ π ππβ1 ππ
ππ₯ =
π
ππ₯ π ππβ1 π₯
π = π
π
ππ₯ ππβ2π₯
31
= π ππβ2 + π₯ (π β 2)ππβ3 ππ
ππ₯ = π ππβ2 + π₯2(π β 2)ππβ4
Similarly π
ππ¦ πππ
ππ¦ = π ππβ2 + π¦2(π β 2)ππβ4
π
ππ§
πππ
ππ§ = π ππβ2 + π§2(π β 2)ππβ4
Using all these values in (1),
β2 ππ = π ππβ2 + π₯2(π β 2)ππβ4 ) + π ππβ2 + (π β 2)ππβ4 + π ππβ2 + π§2(π β 2)ππβ4
= 3π ππβ2 + π π β 2 ππβ4 π₯2 + π¦2 + π§2
= 3π ππβ2 + π π β 2 ππβ2 = π π + 1 ππβ2
Example 36: Show that *(i) πππ π = π β²β² π +π
π π β² (π) (ii) π β π ππ β π ππ = π πππ β π πππ
*[KUK 2008]
Solution: (i) β2π π = π2
ππ₯2 +π2
ππ¦2 +π2
ππ§2 π π =π
ππ₯
π
ππ₯π π +
π
ππ¦
π
ππ¦π π +
π
ππ§
π
ππ§π π β¦(1)
And π
ππ₯
π
ππ₯π π =
π
ππ₯ π β²(π)
ππ
ππ₯ =
π
ππ₯ π β²(π)
π₯
π = π β² (π)
π
ππ₯ π₯
π +
π₯
π π β²β²(π)
ππ
ππ₯
= π β² (π) 1
πβ
π₯
π2
ππ
ππ₯ +
π₯
π π β²β²(π)
ππ
ππ₯
= π β² π
πβ π₯2 π β² π
π3+ π₯2 π β²β² (π)
π2
Similarly π
ππ¦
π
ππ¦π π =
π β² π
πβ π¦2 π β² π
π3+ π¦2 π β²β² (π)
π2
π
ππ§
π
ππ§π π =
π β² π
πβ π§2 π β² π
π3+ π§2 π β²β² (π)
π2
Using all these values in (1)
β2π π = π β² π
πβ π₯2 π β² π
π3+ π₯2 π β²β²
π
π2 +
π β² π
πβ π¦2 π β² π
π3+ π¦2 π β²β²
π
π2
+ π β² π
πβ π§2 π β² π
π3+ π§2 π β²β² (π)
π2
=3 π β² π
πβ π₯2 + π¦2 + π§2
π β² π
π3+
π β²β² π
π2 π₯2 + π¦2 + π§2
=3 π β² (π)
πβ
π β² (π)
π+ π β²β² (π)
=2 π β² (π)
π+ π β²β² (π)
Hence β2π π = π β²β² π +2
π π β² (π) .
(ii) Consider β β π βπ β π βπ = β β π βπ β β β π βπ
= βπ β βπ + π β β βπ β βπ β βπ + π β β βπ
= βπ β βπ + π β2π β βπ β βπ β π β2π
= π β2π β π β2π
32
Hence β β π βπ β π βπ = π β2π β π β2π.
Example 37: Find the value of n for which the vector ππ π is solenoidal, where
π = ππ + ππ + ππ .
Solution: Consider πππ£ ππ π = β β ππ π = βππ β π + ππ β β π β¦ (1)
But βππ = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ ππ = π ππβ1 π
ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§
= π ππβ1 π₯
ππ +
π¦
ππ +
π§
ππ = π ππβ2 π β¦ (2)
And β β π = 3 β¦ (3)
Therefore, πππ£ ππ π = π ππβ2 π β π + 3 ππ = π ππβ2 π β π + 3ππ
= π ππβ2 π2 + 3ππ = (π + 3)ππ β¦ (4)
As given the vector ππ π is solenoidal, so πππ£ ππ π = 0
So using (4), π + 3 ππ = 0 implies that π = β3 (since π β 0)
Example 38: If π and π are irrotational, prove that π Γ π is solenoidal.
Solution: Given π and π are irrotational, so β Γ π = 0 = β Γ π β¦ (1)
Consider π·ππ£ π Γ π = π β β Γ π β π β β Γ π = π β 0 β π β 0 = 0 [using (1)]
Thus π Γ π is solenoidal.
Example 39: Show that the vector field π = ππ + ππ + ππ π + ππ + ππ + π π + (π + πππ)π is
irrotational but not solenoidal. Also obtain a scalar function β such that πβ = π .
Solution: Consider πΆπ’ππ π = β Γ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π§2 + 2π₯ + 3π¦ 3π₯ + 2π¦ + π§ π¦ + 2π§π₯
= π 1 β 1 β π 2π§ β 2π§ + π 3 β 3 = 0
So π is irrotational vector field.
Also consider πππ£ π = β β π = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ β π = 2 + 2 + 2π₯ = 2(π₯ + 2) β 0
So π is not solenoidal vector field.
Now πβ =πβ
ππ₯ ππ₯ +
πβ
ππ¦ ππ¦ +
πβ
ππ§ ππ§ = π
πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§ β ππ₯ π + ππ¦ π + ππ§ π
= ππππ β β ππ = π β ππ (as π = ππππ β )
= π§2 + 2π₯ + 3π¦ π + 3π₯ + 2π¦ + π§ π + (π¦ + 2π§π₯)π β ππ₯ π + ππ¦ π + ππ§ π
= π§2 + 2π₯ + 3π¦ ππ₯ + 3π₯ + 2π¦ + π§ ππ¦ + (π¦ + 2π§π₯)ππ§
= π§2ππ₯ + 2π§π₯ ππ§ + 3π¦ ππ₯ + 3π₯ ππ¦ + π§ ππ¦ + π¦ ππ§ + 2π₯ ππ₯ + 2π¦ ππ¦
= π π₯π§2 + 3π π₯π¦ + π π¦π§ + π π₯2 + π(π¦2)
Integrating both sides, we get
33
β = π₯π§2 + 3π₯π¦ + π¦π§ + π₯2 + π¦2 + π
Example 40: If π½ π ππ§π π½ π be the vectors joining the fixed points ππ, ππ, ππ and ππ, ππ, ππ
respectively to a variable point (π, π, π), prove that *[KUK 2010]
(i) π ππ π½ π Γ π½ π = π *(ii) ππππ π½ π Γ π½ π = π(π½ π β π½ π) (iii) ππππ π½ π β π½ π = π½ π + π½ π.
Solution: Here, π 1 = π₯ β π₯1 π + π¦ β π¦1 π + π§ β π§1 π and π 2 = π₯ β π₯2 π + π¦ β π¦2 π + π§ β π§2 π
(i) π 1 Γ π 2 = π π π
π₯ β π₯1 π¦ β π¦1 π§ β π§1
π₯ β π₯2 π¦ β π¦2 π§ β π§2
= π¦ β π¦1 π§ β π§2 β π¦ β π¦2 π§ β π§1 π + π₯ β π₯2 π§ β π§1 β π₯ β π₯1 π§ β π§2 π
+ π₯ β π₯1 π¦ β π¦2 β π₯ β π₯2 π¦ β π¦1 π
So πππ£ π 1 Γ π 2 =π
ππ₯ π¦ β π¦1 π§ β π§2 β π¦ β π¦2 π§ β π§1
+π
ππ¦ π₯ β π₯2 π§ β π§1 β π₯ β π₯1 π§ β π§2
+π
ππ§ π₯ β π₯1 π¦ β π¦2 β π₯ β π₯2 π¦ β π¦1 = 0
(ii) ππ’ππ π 1 Γ π 2 =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π¦ β π¦1 π§ β π§2 β π¦ β π¦2 π§ β π§1 π₯ β π₯2 π§ β π§1 β π₯ β π₯1 π§ β π§2 π₯ β π₯1 π¦ β π¦2 β π₯ β π₯2 π¦ β π¦1
= π₯ β π₯1 β π₯ β π₯2 β π₯ β π₯2 + π₯ β π₯1 π
+[ π¦ β π¦1 β π¦ β π¦2 β π¦ β π¦2 + π¦ β π¦1 ]π
+[ π§ β π§1 β π§ β π§2 β π§ β π§2 + π§ β π§1 ]π
= 2 π₯ β π₯1 π + (π¦ β π¦1)π + π§ β π§1 π β 2[ π₯ β π₯2 π + π¦ β π¦2 π + π§ β π§2 π ]
= 2(π 1 β π 2)
(iii) π 1 β π 2 = π₯ β π₯1 π₯ β π₯2 + π¦ β π¦1 π¦ β π¦2 + π§ β π§1 π§ β π§2
So ππππ π 1 β π 2 = π π
ππ₯ π₯ β π₯1 π₯ β π₯2 + π
π
ππ¦ π¦ β π¦1 π¦ β π¦2 + π
π
ππ§ π§ β π§1 π§ β π§2
= π π₯ β π₯1 + π₯ β π₯2 + π π¦ β π¦1 + π¦ β π¦2 + π π§ β π§1 + π§ β π§2
= π₯ β π₯1 π + π¦ β π¦1 π + π§ β π§1 π + π₯ β π₯2 π + π¦ β π¦2 π + π§ β π§2 π
= π 1 + π 2
34
Example 41: If π is a constant vector and π = π π + π π + π π , prove that [KUK 2009]
(i) ππππ π β π = π (ii) π ππ π Γ π = π (iii) ) ππππ π Γ π = ππ (iv) ππππ π β π π = π Γ π
Solution: Let π = π1π + π2π + π3 π is the constant vector.
(i) π β π = π1π + π2 π + π3 π β π₯ π + π¦ π + π§ π = π1π₯ + π2π¦ + π3π§
So ππππ π β π = π π
ππ₯ π1π₯ + π2π¦ + π3π§ + π
π
ππ¦ π1π₯ + π2π¦ + π3π§ + π
π
ππ§ π1π₯ + π2π¦ + π3π§
= π1π + π2 π + π3 π = π
(ii) π Γ π = π π π
π1 π2 π3
π₯ π¦ π§ = π π2π§ β π3π¦ + π π3π₯ β π1π§ + π π1π¦ β π2π₯
πππ£ π Γ π =π
ππ₯ π2π§ β π3π¦ +
π
ππ¦ π3π₯ β π1π§ +
π
ππ§ π1π¦ β π2π₯ = 0
(iii) ππ’ππ π Γ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π2π§ β π3π¦ π3π₯ β π1π§ π1π¦ β π2π₯
= π π1 + π1 + π π2 + π2 + π π3 + π3 = 2 π1π + π2π + π3 π = 2π
(iv) π β π π = π1π₯ + π2π¦ + π3π§ π₯π + π1π₯ + π2π¦ + π3π§ π¦π + π1π₯ + π2π¦ + π3π§ π§ π
πΆπ’ππ π β π π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π1π₯ + π2π¦ + π3π§ π₯ π1π₯ + π2π¦ + π3π§ π¦ π1π₯ + π2π¦ + π3π§ π§
= π π2π§ β π3π¦ + π π3π₯ β π1π§ + π π1π¦ β π2π₯ = π Γ π [using part (ii)]
Example 42: Find π Γ π Γ π at the point (1, -1, 2),
if π = πππ π + ππ π β πππ π , π = πππ π + πππ π β ππ π .
Solution: β Γ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
3π₯π§ 2π¦π§ βπ§2
= π 0 β 2π¦ + π 3π₯ β 0 + π 0 β 0 = β2π¦ π + 3π₯ π + 0 π
Now π Γ β Γ π = π π π
π₯π§2 2π¦ β3π₯π§β2π¦ 3π₯ 0
= 9π₯2π§ π + 6π₯π¦π§ π + 3π₯2π§2 + 4π¦2 π
At (1, -1, 2), π Γ β Γ π = 9 1 2(2) π + 6 1 β1 (2) π + 3(1)2(2)2 + 4(β1)2 π
= 18 π β 12 π + 16 π
Example 43: If π = πππ π β ππππ π + ππππ π , π = ππ π + ππ π β ππ π and β = πππ; find
(i) π Γ πβ (ii) π Γ π β (iii) π Γ π Γ π (iv) π β π Γ π
Solution: ββ = π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§ = π¦π§ π + π§π₯ π + π₯π¦ π
35
(i) π Γ ββ = π π π
π¦π§2 β3π₯π§2 2π₯π¦π§π¦π§ π₯π§ π₯π¦
= β5π₯2π¦π§2 π + π₯π¦2π§2 π + 4π₯π¦π§3 π
(ii) π Γ β=
π π π
π¦π§2 β3π₯π§2 2π₯π¦π§π
ππ₯
π
ππ¦
π
ππ§
= π β3π₯π§2 π
ππ§β 2π₯π¦π§
π
ππ¦ + π 2π₯π¦π§
π
ππ₯β π¦π§2 π
ππ§ + π π¦π§2 π
ππ¦+ 3π₯π§2 π
ππ₯
Now π Γ β β = π β3π₯π§2 πβ
ππ§β 2π₯π¦π§
πβ
ππ¦ + π 2π₯π¦π§
πβ
ππ₯β π¦π§2 πβ
ππ§ + π π¦π§2 πβ
ππ¦+ 3π₯π§2 πβ
ππ₯
= π β3π₯π§2 π¦π₯ β 2π₯π¦π§ π₯π§ + π 2π₯π¦π§ π¦π§ β π¦π§2 π₯π¦ + π π¦π§2 π₯π§ + 3π₯π§2 π¦π§
= β5π₯2π¦π§2 π + π₯π¦2π§2π + 4π₯π¦π§3 π
(iii) β Γ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π¦π§2 β3π₯π§2 2π₯π¦π§
= 2π₯π§ + 6π₯π§ π + 2π¦π§ β 2π¦π§ π + (β3π§2 β π§2) π
= 8π₯π§ π + 0 π β 4π§2 π
Now β Γ π Γ g = π π π
8π₯π§ 0 β4π§2
3π₯ 4π§ βπ₯π¦
= 0 + 16π§3 π + β12π₯π§2 + 8π₯2π¦π§ π + (32π₯π§2 β 0) π
= 16π§3 π + β12π₯π§2 + 8π₯2π¦π§ π + 32π₯π§2 π
g β β Γ π = 3π₯ π + 4π§ π β π₯π¦ π β 8π₯π§ π + 0 π β 4π§2 π = 24π₯2π§ + 4π₯π¦π§2
Example 44: Find the directional derivative of π β πβ at the point (1, -2, 1) in the direction of
the normal to the surface ππππ = ππ + ππ where β = πππππππ. [Raipur 2005]
Solution: Given β = 2π₯3π¦2π§4
So ββ = π πβ
ππ₯+ π
πβ
ππ¦+ π
πβ
ππ§ = π 6π₯2π¦2π§4 + π (4π₯3π¦π§4) + π (8π₯3π¦2π§3)
And π = β β ββ = π
ππ₯ 6π₯2π¦2π§4 +
π
ππ¦(4π₯3π¦π§4) +
π
ππ§(8π₯3π¦2π§3)
= 12π₯π¦2π§4 + 4π₯3π§4 + 24 π₯3π¦2π§2
Consider ππππ π = βπ = π ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§
= π 12π¦2π§4 + 12π₯2π§4 + 72π₯3π¦2π§2 + π 24π₯π¦π§4 + 48π₯3π¦π§2
+ π (48π₯π¦2π§3 + 16π₯3π§3 + 48π₯3π¦2π§)
At point (1, -2, 1)
ππππ π = π (48 + 12 + 288) + π β48 β 96 + π 192 + 16 + 192
= 348 π β 144 π + 400 π
Now consider the surface π = π₯π¦2π§ β 3π₯ β π§2 = 0
36
So, βπ = π ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§ = π π¦2π§ β 3 + π 2π₯π¦π§ + π π₯π¦2 β 2π§
The vector normal to the surface (1) at point (1, -2, 1) is given by
π = π β 4π + 2π
And π = 1 + 16 + 4 = 21
So the direction derivative of π = β β ββ at the point (1, -2, 1) in the direction of the normal to the
surface (1) is ππππ π .π
π
= 348 π β 144 π + 400 π .1
21 π β 4π + 2π
=1
21 348 + 576 + 800
= ππππ/ ππ
Example 45: If r is the distance of a point (x, y, z) from the origin, prove that
ππππ π Γ ππππ π
π + ππππ π β ππππ
π
π = π where π is the unit vector in the direction of OZ.
Solution: Here π = π₯ π + π¦ π + π§ π so that π = π₯2 + π¦2 + π§2
Now ππππ 1
π= π
π
ππ₯+ π
π
ππ¦+ π
π
ππ§
1
π= π
π
ππ₯+ π
π
ππ¦+ π
π
ππ§ π₯2 + π¦2 + π§2 β
1
2
= β1
2 π₯2 + π¦2 + π§2 β
3
2(2π₯ π + 2π¦ π + 2π§ π ) = β1
π3 π
ππππ π β ππππ 1
π = ππππ β
1
π3 π§ = βππππ π§
π₯2+π¦2+π§2 3/2
=3π§π₯
π₯2+π¦2+π§2 5/2π +
3π§π¦
π₯2+π¦2+π§2 5/2π +
π₯2+π¦2β2π§2
π₯2+π¦2+π§2 5/2π β¦ (1)
And ππ’ππ π Γ ππππ 1
π = ππ’ππ π Γ β
1
π3 π = ππ’ππ 1
π3 π¦ π β π₯ π
= β3π§π₯
π₯2+π¦2+π§2 5/2π β
3π§π¦
π₯2+π¦2+π§2 5/2π β
π₯2+π¦2β2π§2
π₯2+π¦2+π§2 5/2π β¦ (2)
Adding (1) and (2), ππ’ππ π Γ ππππ 1
π + ππππ π β ππππ
1
π = 0 .
Example 46: Prove that π β π π β ππ
π =
π(π βπ )(π βπ )
ππβ
π βπ
ππ, where π and π are constant vectors.
Solution: From example 45, β1
π= β
1
π3 π = β π₯2 + π¦2 + π§2 β3
2(π₯ π + π¦ π + π§ π )
Let the constant vectors are π = π1 π + π2 π + π3 π and π = π1 π + π2 π + π3 π
So β π β β1
π = β β π₯2 + π¦2 + π§2 β
3
2 π1π₯ + π2π¦ + π3π§
= β π₯2 + π¦2 + π§2 β3
2 β π1π₯ + π2π¦ + π3π§ β π1π₯ + π2π¦ + π3π§ β π₯2 + π¦2 + π§2 β3
2
= β π₯2 + π¦2 + π§2 β3
2 π1 π + π2 π + π3 π +3 π1π₯ + π2π¦ + π3π§ π₯2 + π¦2 + π§2 β5
2(π₯ π + π¦ π + π§ π )
= βπ
π3+
3 π βπ π
π5
37
Now π β β π β β1
π = π β β
π
π3 +3 π βπ π
π5 =3(π βπ )(π βπ )
π5 βπ βπ
π3
Hence Proved.
ASSIGNMENT 4
1. If π = π₯ + π¦ + 1 π + π β (π₯ + π¦)π , show that π β ππ’ππ π = 0.
2. Evaluate (a) πππ£ 3π₯2π + 5π₯π¦2π + π₯π¦π§3π at the point (1, 2, 3).
(b) ππ’ππ [ππ₯π¦π§ π + π + π ].
3. Find the value of βaβ if the vector ππ₯2π¦ + π¦π§ π + π₯π¦2 β π₯π§2 π + 2π₯π¦π§ β 2π₯2π¦2 π has zero
divergence. Find the curl of above vector which has zero divergence .
4. If π£ = π / π₯2 + π¦2 + π§2, show that β β π£ = 2/ π₯2 + π¦2 + π§2 and β Γ π£ = 0 .
5. If π’ = π₯2 + π¦2 + π§2 and π£ = π₯π + π¦π + π§π , show that πππ£ π’ π£ = 5π’.
6. Show that each of following vectors are solenoidal:
(a) π₯ + 3π¦ π + π¦ β 3π§ π + π₯ β 2π§ π (b) 3π¦4π§2π + 4π₯3π§2π + 3π₯2π¦2π (c) βu Γ βπ£
7. If π = π₯π + π¦π + π§π and π = π β 0, show that
(a) β 1/π2 = β2π /π4; β β π /π2 = 1/π2 (b) β β πππ = (π + 3)ππ ; β Γ πππ = 0 .
(c) β β βπ
π = β
2
π3 π
8. Prove that (a) βπ 2 = 2 π β β π + 2π Γ β Γ π , where π is the constant vector.
(b) β Γ π Γ π’ = π β β u β 2π’ β π β β π’ .
9. (a) If β = π₯2 + π¦2 + π§2 βπ , find the πππ£ ππππ β and determine n if πππ£ ππππ β = 0.
(b) Show that ππππ ππ = π(π + 1)ππβ2 , where π2 = π₯2 + π¦2 + π§2.
10. In electromagnetic theory, we have β β π· = π, β β π» = 0, β Γ π· = β1
π
ππ»
ππ‘, β Γ π» =
1
π ππ +
ππ·
ππ‘ .
Prove that β2π» β1
π2
π2π»
ππ‘ 2 = β1
πβ Γ ππ and β2π· β
1
π2
π2π·
ππ‘ 2 = βπ +1
π2
π
ππ‘ ππ .
11. If π’ = π₯2π¦π§, π£ = π₯π¦ β 3π§2 , find (i) β βπ’ β βπ£ (ii) β β βπ’ Γ βπ£ .
12. For a solenoidal vector π , show that ππ’ππ ππ’ππ ππ’ππ ππ’ππ π = β4π .
13. Calculate (i) ππ’ππ ππππ π , given π π₯, π¦, π§ = π₯2 + π¦2 β π§ [BPTU 2006]
(ii) ππ’ππ(ππ’ππ π ), given π = π₯2π¦ π + π¦2π§ π + π§2π¦ π
14. Show that each of the following vectors are solenoidal:
(i) βπ₯2 + π¦π§ π + 4π¦ β π§2π₯ π + (2π₯π§ β 4π§) π
(ii) 3π¦4π§2 π + 4π₯3π§2 π + 3π₯2π¦2 π
(iii) βπ Γ βπ
INTEGRAL VECTOR CALCULUS
38
17.9 INTEGRATION OF VECTORS
If two vector functions π (π‘) and π (π‘) be such that π
ππ‘ π (π‘) = π (π‘), then π (π‘) is called an integral
of π (π‘) with respect to a scalar variable t and we can write π (π‘)ππ‘ = π (π‘).
Indefinite Integral
If π be an arbitrary constant vector and π π‘ =π
ππ‘ π (π‘) =
π
ππ‘ π π‘ + π , then π (π‘)ππ‘ = π π‘ + π .
This is called the indefinite integral of π (π‘).
Definite Integral
If π
ππ‘ π (π‘) = π π‘ for all values of π‘ in the interval π, π , then the definite integral of π π‘ between
π and π is defined and denoted by π π‘ ππ‘π
π= π (π‘) π
π = π π β π π .
Example 47: If π ππ
π ππ= ππ π β ππππ π + π ππ¨π¬ π π , find π , given that
π π
π π= β π β π π and
π = π π + π at π = π.
Solution: Given that π2π
ππ‘2= 6π‘ π β 12π‘2 π + 4 cos π‘ π β¦ (1)
Integrating (1) with respect to t,
π2π
ππ‘2 ππ‘ = 6π‘ π β 12π‘2 π + 4 cos π‘ π ππ‘
Implying ππ
ππ‘ = 3π‘2 π β 4π‘3 π + 4 sin π‘ π + π 1 β¦ (2)
Now, integrating (2) with respect to t,
ππ
ππ‘ππ‘ = 3π‘2 π β 4π‘3 π + 4 sin π‘ π + π 1 ππ‘
Implying π = π‘3 π β π‘4 π β 4 cos π‘ π + π 1π‘ + π 2 β¦ (3)
Also we are given that at π‘ = 0, ππ
ππ‘= β π β 3 π β¦ (4)
π = 2 π + π β¦ (5)
Using (2) and (4), π 1 = β π β 3 π
Using (3) and (5), β4 π + π 2 = 2 π + π => π 2 = 2 π + π + 4 π
Putting the values of the constant vectors π 1 & π 2 in (3), we get
π = π‘3 π β π‘4 π β 4 cos π‘ π + β π β 3 π π‘ + 2 π + π + 4 π
= π‘3 β π‘ + 2 π + 1 β π‘4 π β 4 cos π‘ + 3 π‘ + 4 π
ASSIGNMENT 5
1. For given π π‘ = 5π‘2 β 3π‘ π + 6π‘3 π β 7π‘ π , evaluate π π‘ ππ‘4
2.
2. Given π π‘ = 3π‘2 π + π‘ π β π‘3 π , evaluate π Γπ2π
ππ‘ 2 ππ‘1
0.
39
X O
Y
Z
3. If π π‘ = 2 π β π + 2 π , π€πππ π‘ = 1
3 π β 2π + 4 π , π€πππ π‘ = 2 , show that π β
ππ
ππ‘ ππ‘ = 10
2
1.
4. The acceleration of a particle at any time π‘ β₯ 0 is given by 12 cos 2π‘ π β 8 sin 2π‘ π + 16π‘ π , the
displacement and velocity are initially zero. Find the velocity and displacement at any time.
17.10 LINE INTEGRAL
Let kuzjuyiuxur
)()()()( defines a curve C joining points P1 and P2 where )(ur
is the
position vector of ),,( zyx and the value of u at P1 and P2 is u1 and u2, respectively.
Now if kAjAiAzyxA
321),,( be vector function of defined position and continuous along C,
then the integral of the tangential component of A along C from P1 to P2 written as 2
1
.P
PrdA
is
known as the line integral. Also in terms of Cartesian components, we have
2
1
2
1
)(.)(. 321
P
P
P
PkdzjdyidxkAjAiArdA
)()( 321321
2
1
C
P
PdzAdyAdxAdzAdyAdxA
If A
(vector function of the position) represents the force F
on a particle moving along C, then the
line integral represents the work done by the force F
. If C is a simple closed curve, then the integral
around C is generally written as
1P 2P
C
Fig 17.11
)(. 321 dzAdyAdxArdA
In Fluid Mechanics and Aerodynamics, the above integral is called circulation of A
about C, where
A
represents the velocity of the fluid.
Let gradA , then we have
Q
P
Q
PrdrdA
.)(.
Q
Pdzkdyjdxi
zk
yj
xi )(.
40
Q
Pdz
zdy
ydx
x
PQ
Q
P
Q
Pd ][ β¦ (1)
We see that the integral Q
PrdA
. depends on the value of at the end P and Q and not on the
particular path. In case is single valued and the integral is taken round a closed curve, the terminal
points P and Q coincide and PB .
[Because function is uniform]
The integration along a closed curve is denoted by the sign of circle in the mid of the integral sign
i.e. for a uniform function, we have
C
rd 0.)(
β¦ (2)
The converse of the above result is also true i.e. if there exists a vector A
and its integral round
every closed curve in the region under consideration vanishes, then there exist a point function
such that gradA
.
To prove this consider any closed curve ABCD such that the integral round it is zero, so integral along
ABC must be equal to that along ADC. Similarly, the integral along ABC must be equal to that along
any curve joining A to C, i.e. independent of the path from A to C with A be a fixed point and C a
variable point. Then due to the fact that line integral is independent of the path chosen, the value of
the line integral from A to C must be a scalar point function, say i.e. C
ArdA
.
Now if d is the increment in due to a small displacement ππ of π , then we have rdAd
.
But we already know that rdd
. , so ,.)(. rdrdA
,0.)( rdA
which is true for all rd
and hence A
.
The vector A
is called a potential vector (or gradient vector), and in cartesian component; the
condition that dzAdyAdxArdA 321.
be a perfect differential can be thrown easily into the
form
0,0,0 211332
x
A
y
A
z
A
x
A
y
A
z
A β¦ (3)
Circulation: If A
represents the velocity of a fluid particle, then the line integral CrdA
. is called
the circulation of A
along C.
The vector point function A
, is said to be irrotational in a region, if its circulation along every
closed curve in the region is zero i.e. 0. CrdA
41
Theorem: The necessary and sufficient condition for a vector point function A
to be irrotaional in a
simply connected region is the 0Acurl
at every point of the region.
Work: If A
represents the force acting on a particle moving along an arc PQ then the work done
during the small displacement rd
is equal to rdA
. . Therefore, the total work done by A
during the
displacement from P to Q is given by the line integral Q
PrdA
. .
Example 48: Evaluate the line integral (ππ + ππ)π π + (ππ + ππ)π π , where πͺ is the square
formed by the lines π = Β±π and π = Β±π.
Solution: Curve πΆ is square in the π₯π¦ plane where π§ = 0
β΄ π = π₯π + π¦π in π₯π¦ plane
ππ = ππ₯π + ππ¦π β¦ (1)
Now πΉ. ππ πΆ
= π₯2 + π₯π¦ ππ₯ + π₯2 + π¦2 ππ¦
Path of the integration is shown in figure 17.12, it consists of lines
AB, BC, CD and DA. As curve C is a square, then
On AB, π¦ = β1 β ππ¦ = 0 and π₯ varies from β1 to 1
On BC, π₯ = 1 β ππ₯ = 0 and π¦ varies from β1 to 1
On CD, π¦ = 1 β ππ¦ = 0 and π₯ varies from 1 to β1
On DA, π₯ = β1 β ππ₯ = 0 and π¦ varies from 1 to β1
πΉ . ππ πΆ
= π₯2 β π₯ ππ₯π΄π΅
+ 1 + π¦2 ππ¦π΅πΆ
+ π₯2 + π₯ ππ₯πΆπ·
+ 1 + π¦2 ππ¦π·π΄
= π₯2 β π₯ ππ₯1
β1+ 1 + π¦2 ππ¦
1
β1+ π₯2 + π₯ ππ₯
β1
1+ 1 + π¦2 ππ¦
β1
1
= π₯3
3β
π₯2
2 β1
1
+ 1 + π¦2 ππ¦1
β1+
π₯3
3+
π₯2
2
1
β1
β 1 + π¦2 ππ¦1
β1
= 1
3β
1
2+
1
3+
1
2 + β
1
3+
1
2β
1
3β
1
2 = 0
Example 49: If π = πππ π β πππ , evaluate π . π π πͺ
where πͺ is the arc of the parabola π = πππ
from (π, π) to (π, π).
Solution: Because the integration is performed in the π₯π¦-plane (π§ = 0), we take
π = π₯π + π¦π so that ππ = ππ₯ π + ππ¦ π
β΄ πΉ . ππ = 3π₯π¦ π β π¦2π . ππ₯ π + ππ¦ π = 3π₯π¦ ππ₯ β π¦2ππ¦
On the curve C: π¦ = 2π₯2 from (0, 0) to (1, 2)
42
πΉ . ππ = 3π₯ 2π₯2 ππ₯ β 2π₯2 24π₯ ππ₯ = (6π₯3 β 16π₯5)ππ₯
Also x varies from 0 to 1.
πΉ . ππ πΆ
= (6π₯3 β 16π₯5)ππ₯1
0=
6π₯4
4β
16π₯6
6
0
1
=3
2β
8
3= β
7
6
Note: If the curve is traversed in the opposite direction, that is from (1, 2) to (0, 0), the value of the integral
would be 7
6.
Example 50: A vector field is given by π = π¬π’π§π π + π(π + ππ¨π¬ π)π . Evaluate the line integral
over the circular path given by ππ + ππ = ππ, π = π. [PTU 2003]
Solution: The parametric equations of the circular path are π₯ = π cos π‘ , π¦ = π sin π‘ , π§ = 0 where π‘
varies from 0 to 2π. Since the particle moves in the π₯π¦-plane (π§ = 0), we can take π = π₯ π + π¦ π .
πΉ . ππ πΆ
= sin π¦ π + π₯(1 + cos π¦)π πΆ
. ππ₯ π + ππ¦ π
= sin π¦ ππ₯ + π₯ 1 + cos π¦ ππ¦ πΆ
= sin π¦ ππ₯ + π₯ cos π¦ ππ¦ + π₯ ππ¦ πΆ
= π(π₯ sin π¦)πΆ
+ π₯ ππ¦πΆ
= π a cos π‘ sin a sin π‘ ππ‘ + a cos π‘ . a cos π‘ ππ‘2π
0
2π
0
= a cos π‘ sin(a sin π‘) 02π + π2 cos2 π‘ ππ‘
2π
0
=π2
2 (1 + cos 2π‘)ππ‘
2π
0=
π2
2 π‘ +
sin 2π‘
2
0
2π
=π2
2 2π = ππ2
Example 51: Compute the line integral (πππ π β πππ π)πͺ
about the triangle whose vertices are
(π, π), (0, 1) and (βπ, π). [NIT Uttrakhand 2011]
Solution: Here the closed curve C is a triangle ABC.
On AB: Equation of line AB is
π¦ β 0 =1β0
0β1(π₯ β 1) β π¦ = 1 β π₯
β΄ ππ¦ = βππ₯ and π₯ varies from 1 to 0.
On BC: Equation of line BC is
π¦ β 1 =0β1
β1β0(π₯ β 0) β π¦ = 1 + π₯
β΄ ππ¦ = ππ₯ and π₯ varies from 0 to -1.
On CA: π¦ = 0. Therefore, ππ¦ = 0 and π₯ varies from -1 to 1.
π¦2ππ₯ β π₯2ππ¦ = π¦2ππ₯ β π₯2ππ¦ π΄π΅πΆ
+ π¦2ππ₯ β π₯2ππ¦ π΅πΆ
+ π¦2ππ₯ β π₯2ππ¦ πΆπ΄
= (1 β π₯)2ππ₯ β π₯2(βππ₯) 0
1+ (1 + π₯)2ππ₯ β π₯2ππ₯ + 0 ππ₯
1
β1
β1
0
43
= 2π₯2 β 2π₯ + 1 ππ₯0
1+ 2π₯ + 1 ππ₯
β1
0+ 0
= 2π₯3
3β
2π₯2
2+ π₯
1
0
+ 2π₯2
2+ π₯
0
β1
= β2
3+ 1 β 1 + 1 β 1 = β
2
3
Example 52: If π = πππ + ππ π β πππππ + ππππππ , evaluate π . π π πͺ
where
(i) πͺ is the line joining the point (π, π, π) to (π, π, π)
(ii) πͺ is given by π = π, π = ππ, π = ππ from the point (π, π, π) to (π, π, π).
Solution: (i) Equation of line joining (0,0,0) to (1,1,1) is
π₯β0
1β0=
π¦β0
1β0=
π§β0
1β0= π‘ (say)
β΄ Parametric equations of the line C are π₯ = π‘, π¦ = π‘, π§ = π‘; 0 β€ π‘ β€ 1
β΄ π = π₯π + π¦π + π§π = π‘π + π‘π + π‘π
β π π
π π= π + π + π
Now πΉ = 3π₯2 + 6π¦ π β 14π¦π§π + 20π₯π§2π = 3π‘2 + 6π‘ π β 14π‘2π + 20π‘3π
πΉ . ππ πΆ
= πΉ .ππ
ππ‘ ππ‘
πΆ
= 3π‘2 + 6π‘ π β 14π‘2π + 20π‘3π . π + π + π ππ‘πΆ
= 3π‘2 + 6π‘ β 14π‘2 + 20π‘3 ππ‘1
0
= 3π‘3
3+
6π‘2
2β
14π‘3
3+
20π‘4
4
0
1
= 1 + 3 β14
3+ 5 =
13
3
(ii) Here the curve C is given by π₯ = π‘, π¦ = π‘2 , π§ = π‘3 from the point (0,0,0) to (1,1,1)
β΄ π = π₯π + π¦π + π§π = π‘π + π‘2π + π‘3π
β π π
π π= π + 2π‘ π + 3π‘2 π
Now πΉ = 3π₯2 + 6π¦ π β 14π¦π§π + 20π₯π§2π = 9π‘2π β 14π‘5π + 20π‘7π
πΉ . ππ πΆ
= πΉ .ππ
ππ‘ ππ‘
πΆ
= 9π‘2π β 14π‘5π + 20π‘7π . π + 2π‘ π + 3π‘2 π ππ‘πΆ
= 9π‘2 β 28π‘6 + 60π‘9 ππ‘1
0
= 9π‘3
3β
28π‘7
7+
60π‘10
10
0
1
= 3 β 4 + 6 = 5
Example 53: Find the circulation of π around the curve πͺ where π = ππ + ππ + ππ and πͺ is the
circle ππ + ππ = π, π = π.
44
Solution: Circulation of πΉ along the curve πΆ is πΉ . ππ πΆ
Equation of circle is π₯2 + π¦2 = 1, π§ = 0
Its parameteric equations are π₯ = cos ΞΈ , π¦ = sin ΞΈ , π§ = 0
Now π = π₯π + π¦π + π§π = cos ΞΈ π + sin ΞΈ π + 0π so that
ππ = β sin ΞΈ π + cos ΞΈ π + 0π πΞΈ
Also πΉ = π¦π + π§π + π₯π = sin ΞΈ π + 0π + cos ΞΈπ
β΄ πΉ . ππ πΆ
= sin ΞΈ π + 0π + cos ΞΈπ .πΆ
β sin ΞΈ π + cos ΞΈ π + 0π πΞΈ
= β sin2 ΞΈ2π
0 ππ (since along the circle, π varies from 0 to 2π)
= β 1βcos 2ΞΈ
2
2π
0πΞΈ = β
ΞΈ
2β
sin 2ΞΈ
4
0
2Ο
= β 2Ο
2β 0 β (0 β 0) = βΟ
Example 54: Find the work done in moving a particle once round a circle πͺ in the ππ plane, if
the circle has its centre at the origin and radius 2 units and the force field is given as
π = ππ β π + ππ π + π + π β π π + (ππ β ππ β ππ)π .
Solution: Equation of a circle having centre (0, 0) with radius 2 in π₯π¦ plane is π₯2 + π¦2 = 4 .
Parametric equations of this circle are π₯ = 2 cos π‘ , π¦ = 2 sin π‘ , π§ = 0.
Since integration is to be performed around a circle in π₯π¦ plane,
β΄ π = π₯π + π¦π = 2 cos π‘ π + 2 sin π‘ π β ππ
ππ‘= β2 sin π‘ π + 2 cos π‘ π
Work done, πΉ .ππ
ππ‘ ππ‘
πΆ
= 2π₯ β π¦ + 2π§ π + π₯ + π¦ β π§ π + (3π₯ β 2π¦ β 5π§)π . β2 sin π‘ π + 2 cos π‘ π ππ‘
= 4 cos π‘ β 2 sin π‘ π + 2 cos π‘ + 2 sin π‘ π + (6 cos π‘ β 4 sin π‘)π . β2 sin π‘ π + 2 cos π‘ π ππ‘
In moving round the circle, π‘ varies from 0 to 2π
β΄ Work done = 4 cos π‘ β 2 sin π‘ (β2 sin π‘) + 2 cos π‘ + 2 sin π‘ (2 cos π‘) 2π
0ππ‘
= β8 cos π‘ sin π‘ + 4 sin2 π‘ + 4 cos2 π‘ + 4 sin π‘ cos π‘ 2π
0ππ‘
= 4 β 4 sin π‘ cos π‘ 2π
0ππ‘ = 4π‘ β 4
sin 2 π‘
2
0
2π
= 8π β 2 sin2 2π β (0 β 0) = 8π
ASSIGNMENT 6
45
1. Using the line integral, compute the work done by the force πΉ = 2π¦ + 3 π + π₯π§ π + (π¦π§ β π₯)π ,
when it moves a particle from (0,0,0) to (2,1,1) along the curve π₯ = 2π‘2 , π¦ = π‘, π§ = π‘3.
2. Find the work done in moving a particle in the force field πΉ = 3π₯2π + 2π₯π§ β π¦ π + π§π , along
(a) the straight line from (0,0,0) to (2,1,3).
(b) the curve defined by π₯2 = 4π¦, 3π₯3 = 8π§ from π₯ = 0 to π₯ = 2.
3. If C is a simple closed curve in the π₯π¦ plane not enclosing the origin, show that πΉ . ππ πΆ
= 0,
where πΉ =π¦π βπ₯π
π₯2+π¦2.
4. If π = 5π₯π¦ β 6π₯2 π + 2π¦ β 4π₯ π , evaluate π . ππ πΆ
along the curve C in xy-plane, π¦ = π₯3
from the point 1, 1 to (2, 8).
5. Evaluate (π₯π¦ + π§2)ππ πΆ
where C is the arc of the helix π₯ = cos π‘ , π¦ = sin π‘ , π§ = π‘ which joins
the points 1, 0, 0 πππ (β1, 0, π).
6. If πΉ = 2π¦π β π§π + π₯π , evaluate πΉ πΆ
Γ ππ along the curve π₯ = cos π‘, π¦ = sin π‘, π§ = 2 cos π‘ from
π‘ = 0 to π‘ =π
2.
17.11 SURFACE INTEGRALS AND FLUX
An integral which is to be evaluated over a surface is called a surface integral. Suppose π is a surface
of finite area. Divide the area π into n sub-areas πΏπ1, πΏπ2, β¦β¦ , πΏππ .
In each area πΏππ , choose an arbitrary point ππ π₯π , π¦π , π§π . Let π define
a scalar point function over the area π.
Now from the sum π ππ πΏππππ=1 , where π ππ = π π₯π , π¦π , π§π
Now let us take the limit of the sum as π β β, each sub-area πΏππ
reduces to a point and the limit if it exists is called the surface
integral of π over π and is denoted by πππ
.
Note: If π is piecewise smooth then the function π π₯, π¦, π§ is continuous over π and then the limit exists and is
independent of sub-divisions and choice of the point ππ .
Flux: Suppose π is a piecewise smooth surface so that the vector function πΉ defined over π is
continuous over π. Let π be any point of the surface π and suppose π is a unit vector at π in the
direction of outward drawn normal to the surface π at π. Then the component of πΉ along π is πΉ . π and
the integral of πΉ . π over π is called the surface integral of πΉ over π and is denoted by πΉ . π πππ
. It is
also called flux of πΉ over π.
Different Forms of Surface Integral
(i) Flux of πΉ over π = πΉ . π πππ
β¦ (1)
46
Now let ππ denote a vector (called vector area) whose magnitude is that of differential of
surface area i.e., ππ and whose direction is that of π . Then clearly
ππ = π ππ
From (1), flux of πΉ over π = πΉ . ππ π
β¦. (2)
(ii) Suppose outward drawn normal to the surface π at π makes angles πΌ, π½, πΎ with the
positive direction of axes and if π, π, π denote the direction cosines of this outward drawn
normal, then
π = cos πΌ , π = cos π½ , π = cos πΎ
Therefore, π = cos πΌ π + cos π½ π + cos πΎ π
If πΉ = πΉ1π + πΉ2π + πΉ3π then πΉ . π = πΉ1 cos πΌ + πΉ2 cos π½ + πΉ3 cos πΎ
β΄ From (1), flux of πΉ over π = πΉ1 cos πΌ + πΉ2 cos π½ + πΉ3 cos πΎ πππ
β¦ (3)
Now ππ cos πΌ is the projection of area ππ on the π¦π§ plane, therefore ππ cos πΌ = ππ¦ππ§.
Similarly ππ cos π½ and ππ cos πΎ are the projections of the area ππ on the π§π₯ and π₯π¦ plane
respectively and therefore ππ cos π½ = ππ§ππ₯, ππ cos πΎ = ππ₯ππ¦.
β΄ From (3), πΉ . π πππ
= πΉ 1ππ¦ππ§ + πΉ 2ππ§ππ₯ + πΉ 3ππ₯ππ¦π
β¦ (4)
Note: In order to evaluate surface integral it is convenient to express them as double integrals by
taking the projection of surface π on one of the coordinate planes. This will happen only if any line
perpendicular to co-ordinate plane chosen meets the surface in one point and not more than one
point. Surface S is divided into sub surfaces, if above requirement is not met, so that sub surfaces
may satisfy the above requirement.
(iii) Suppose surface π is such that any line
perpendicular to π₯π¦ plane does not meet π in more
than one point. Let the equation of surface π be
π = π(π₯, π¦).
Let π 1 denotes the orthogonal projection of π on the
π₯π¦ plane. Then projection of ππ on the π₯π¦ plane
= ππ cos πΎ , where πΎ is the acute angle which the
normal to the surface π makes with positive
direction of Z-axis.
β΄ ππ cos πΎ = ππ₯ππ¦ β¦ (5)
But cos πΎ = π .π
π = π . π
Therefore from (5), ππ =ππ₯ ππ¦
π .π β¦ (6)
Thus πΉ . π πππ
= πΉ . π π 1
ππ₯ ππ¦
π .π β¦ (7)
Similarly we have, πΉ . π πππ
= πΉ . π π 2
ππ¦ ππ§
π .π β¦ (8)
πΉ . π πππ
= πΉ . π π 3
ππ§ ππ₯
π .π β¦ (9)
where π 2, π 3 are the projections of π on π§π₯ and π₯π¦ planes, respectively.
47
Example 55: Evaluate π . π π πΊπΊ
where π = π π + π π + ππππ π and S is the surface of the
cylinder ππ + ππ = ππ included in the first octant between π = π and π = π.
Solution: A vector normal to the surface π is given by
π = β π₯2 + π¦2 = 2π₯ π + 2π¦ π β¦ (1)
β΄ π = unit vector normal to surface π at any point π₯, π¦, π§ =2π₯π +2π¦π
4π₯2+4π¦2 β¦ (2)
β΅ π₯2 + π¦2 = 16, therefore π =2π₯π +2π¦π
4(π₯2+π¦2)=
2π₯π +2π¦π
8=
π₯
4π +
π¦
4π β¦ (3)
Now πΉ . π πππ
= πΉ . π ππ₯ ππ§
π .π π
(projection on π₯π¦ plane canβt be taken as the surface π is peerpendicular to π₯π¦ plane)
= π₯π§
4+
π₯π¦
4
ππ₯ ππ§π¦
4π
= π₯π§
π¦+ π₯ ππ₯ ππ§
π , since from 3 , π . π =
π¦
4
= π₯π§
16βπ₯2+ π₯ ππ₯ ππ§
4
π₯=0
5
π§=0=
βπ§
2(β2π₯)
16βπ₯2+ π₯ ππ₯ ππ§
4
π₯=0
5
π§=0
= βπ§
2
16βπ₯2
1 2 +
π₯2
2 π₯=0
4
ππ§5
π§=0= (4π§ + 8)ππ§
5
π§=0=
4π§2
2+ 8π§
π§=0
5
= 90
Example 56: Evaluate π . π π πΊπΊ
where π = ππ π β π π + π π and S is the portion of the plane
ππ + ππ + ππ = ππ in the first octant.
Solution: Vector normal to surface π is given by
β 2π₯ + 3π¦ + 6π§ = 2π + 3π + 6π
β΄ π = unit vector normal to surface π at any point (π₯, π¦, π§) =2π +3π +6π
4+9+36=
2
7π +
3
7π +
6
7π
Now πΉ . π = 6π§ π β 4 π + π¦ π . 2
7π +
3
7π +
6
7π =
12
7π§ β
12
7+
6
7π¦
Taking projection on π₯π¦ plane
πΉ . π πππ
= πΉ . π ππ₯ ππ¦
π .π π = πΉ . π
ππ₯ ππ¦
6 7 π β¦ (1)
where π is the region of projection of π on π₯π¦ plane. π is bounded by x-axis, y-axis and the line
2π₯ + 3π¦ = 12, π§ = 0. In order to evaluate double integral in (1), y varies from 0 to 4 and x varies
from 0 to 12β3π¦
2. Therefore from (1)
πΉ . π πππ
= 2π§ β 2 + π¦ ππ₯ ππ¦,12β3π¦
2π₯=0
4
π¦=0 Find π§ from 2π₯ + 3π¦ + 6π§ = 12
= 2 2 βπ₯
3β
π¦
2 β 2 + π¦ ππ₯ ππ¦
12β3π¦
2π₯=0
4
π¦=0
48
= 2 β2π₯
3 ππ₯ ππ¦
12β3π¦
2π₯=0
4
π¦=0= 2π₯ β
π₯2
3 π₯=0
12β3π¦
2ππ¦
4
0
= 2 Γ12β3π¦
2β
1
3
12β3π¦
2
2
ππ¦4
0
= 12β3π¦ 2
β6+
12β3π¦ 3
108 π¦=0
4
=144
6β
1728
108= 24 β 16 = 8
Example 57: ππΊ
π π πΊ where π =π
ππππ and πΊ is the surface of cylinder ππ + ππ = ππ
included in the first octant between π = π to π = π.
Solution: A vector normal to the surface π is given by
π = β π₯2 + π¦2 = 2π₯π + 2π¦π
β΄ π = unit vector normal to surface π at any point (π₯, π¦, π§) =2π₯π +2π¦π
4π₯2+4π¦2
π =2π₯π +2π¦π
4(π₯2+π¦2)=
2π₯π +2π¦π
8=
π₯
4π +
π¦
4π β΅ π₯2 + π¦2 = 16
Now π π πππ
= 3
8π₯π¦π§
π₯
4π +
π¦
4π
ππ₯ ππ§
π .π π β¦(1)
where π is the region of projection of π on π§π₯ plane. Therefore, from (1)
π π πππ
= 3
32π₯2π¦π§ π +
3
32π₯π¦2π§ π
4
π₯=0
5
π§=0
ππ₯ ππ§
π¦ 4 , where π¦2 = 16 β π₯2
= 3
8π₯2π§ π +
3
8π₯π§ 16 β π₯2 π
4
π₯=0
5
π§=0ππ₯ ππ§
=3
8
π₯3π§
3 π β
π§
2
16βπ₯2 3 2
3 2 π
π₯=0
45
π§=0ππ
=3
8
64
3π§ π +
64
3π§ π
5
π§=0ππ§ = 8
π§2
2π +
π§2
2π
π§=0
5
= 100 π + 100 π
Example 58: Evaluate π . π πΊπΊ
where π = ππ β ππ β ππ π β πππ and πΊ is the triangular
surface with vertices π, π, π , (π, π, π) and (π, π, π).
Solution:The triangular surface π with vertices 2,0,0 , (0,2,0), and (0,0,4) is given by the equation
π₯
2+
π¦
2+
π§
4= 1 β 2π₯ + 2π¦ + π§ = 4 β¦ (1)
Vector normal to surface π is given by β 2π₯ + 2π¦ + π§ = 2π + 2π + π
β΄ π = unit vector normal to surface π at any point (π₯, π¦, π§) =2π +2π +π
4+4+1=
2
3π +
2
3π +
1
3π
Now πΉ . π = π₯π β π§2 β π§π₯ π β π₯π¦π . 2
3π +
2
3π +
1
3π =
2
3π₯ β
2
3 π§2 β π§π₯ β
1
3π₯π¦
Taking projection on π₯π¦ plane
49
πΉ . π πππ
= πΉ . π ππ₯ ππ¦
π .π π = πΉ . π
ππ₯ ππ¦
1 3 π β¦ (2)
where π is the region of projection of π on π₯π¦ plane. π is bounded by x-axis, y-axis and the line
2π₯ + 2π¦ = 4 i.e., π₯ + π¦ = 2, π§ = 0. In order to integrate double integral in (2), y varies from 0 to 2
and x varies from 0 to 2 β π¦. Therefore from (2)
πΉ . π πππ
= 2
3π₯ β
2
3 π§2 β π§π₯ β
1
3π₯π¦ ππ₯ ππ¦
2βπ¦
π₯=0
2
π¦=0 Find π§ from 2π₯ + 2π¦ + π§ = 4
= 2
3π₯ β
2
3 4 β 2π₯ β 2π¦ 2 β 4 β 2π₯ β 2π¦ π₯ β
1
3π₯π¦ ππ₯ ππ¦
2βπ¦
π₯=0
2
π¦=0
=1
3 2π₯ β 2 16 + 4π₯2 + 4π¦2 β 16π₯ + 8π₯π¦ β 16π¦ β 4π₯ + 2π₯2 + 2π₯π¦ β π₯π¦ ππ₯ ππ¦
2βπ¦
π₯=0
2
π¦=0
=1
3 β12π₯2 β 8π¦2 + 42π₯ + 32π¦ β 21π₯π¦ β 32 ππ₯ ππ¦
2βπ¦
π₯=0
2
π¦=0
=1
3 β4π₯3 β 8π₯π¦2 + 21π₯2 + 32π₯π¦ β 21
π₯2π¦
2β 32π₯
π₯=0
2βπ¦
ππ¦2
π¦=0
=1
3 β4 2 β π¦ 3 β 8 2 β π¦ π¦2 + 21 2 β π¦ 2 + 32 2 β π¦ π¦ β 21
2βπ¦ 2π¦
2 β 32 2 β π¦ ππ¦
2
π¦=0
=1
3
3
2π¦3 + 9π¦2 + 18π¦ β 12 ππ¦
2
π¦=0= 38
Example 59: Evalutate π β π πΊ
π π where π = ππππ π β ππ π + ππππ π and S is closed surface of
the region in the first octant bounded by the cylinder ππ + ππ = π and the planes π = π, π = π,
π = π ππ§π π = π.
Solution: The given closed surface S is piecewise smooth and is
comprised of S1 βthe rectangular face OAEB in xy-plane; S2 βthe
rectangular face OADC in xz-plane; S3 β the circular quadrant ABC
in yz-plane; S4 β the circular quadrant AED and S5 β the curved
surface BCDE of the cylinder in the first octant (see Fig. 17.16).
β΄ π β π π
ππ = π β π π1
ππ + π β π π2
ππ + π β π π3
ππ
+ π β π π4
ππ + π β π π5
ππ β¦ (1)
Now π β π π1
ππ = 2π₯2π¦ π β π¦2 π + 4π₯π§2 π β βπ π1
ππ = β4 π₯π§2π1
ππ = 0
(ππ π§ = 0 ππ π₯π¦ β πππππ)
Similarly, π β π π2
ππ = 0 and π β π π3
ππ
π β π π4
ππ = 2π₯2π¦ π β π¦2 π + 4π₯π§2 π β π π4
ππ = 2π₯2π¦π4
ππ
= 8π¦ ππ¦ππ§ 9βπ§2
0
3
0= 4 (9 β π§2)ππ§
3
0= 72
To find π in S5 , we note that β π¦2 + π§2 β 9 = 2π¦ π + 2π§π ,
Implying π =2π¦ π +2π§π
4 π¦2+π§2 =
π¦ π +π§π
3 and π β π =
π§
3 so that ππ = ππ₯ππ¦/(π§/3) (ππ π¦2 + π§2 = 9)
50
π β π π5
ππ = βπ¦3+4π₯π§3
3 ππ₯ππ¦/(π§/3)
3
0
2
0=
βπ¦3
π§+ 4π₯π§2 ππ₯ππ¦
3
0
2
0
Now putting π¦ = 3 sin π , π§ = 3 cos π β΄ ππ¦ = 3 cos π ππ
β27 sin 3 π
3 cos π+ 4π₯ (9 cos2 π) 3 cos π ππππ₯
π
20
2
0
ASSIGNMENT 7
1. If velocity vector is πΉ = π¦ π + 2 π + π₯π§ π m/sec., show that the flux of water through the
parabolic cylinder π¦ = π₯2, 0 β€ π₯ β€ 3, 0 β€ π§ β€ 2 is 69 π3/π ππ.
2. Evaluate πΉ . π πππ
where πΉ = π₯ + π¦2 π β 2π₯ π + 2π¦π§ π and S is the surface of the plane
2π₯ + π¦ + 2π§ = 6 in the first octant.
3. If πΉ = 4π₯π§ π β π¦2 π + π¦π§ π ; evaluate πΉ . π πππ
, where S is the surface of the cube bounded by
π₯ = 0, π₯ = 1, π¦ = 0, π¦ = 1, π§ = 0, π§ = 1.
4. If πΉ = 2π¦ π β 3 π + π₯2 π and S is the surface of the parabolic cylinder π¦2 = 8π₯ in the first
octant bounded by the planes π¦ = 4 and π§ = 6, show that πΉ β π π
ππ = 132.
5. Evaluate πΉ β π π
ππ where πΉ = 6π§ π β 4 π + π¦ π and S is the portion of the plane
2π₯ + 3π¦ + 6π§ = 12 in the first octant.
17.12 VOLUME INTEGRALS
Suppose π is the volume bounded by a surface π. Divide the volume π into sub-volumes πΏπ1, πΏπ2,
β¦, πΏππ . In each πΏππ , choose an arbitrary point ππ whose coordinates are (π₯π , π¦π , π§π). Let π be a single
valued function defined over π. Form the sum π ππ πΏππ , where π ππ = π π₯π , π¦π , π§π .
Now let us take the limit of the sum as π β β, then the limit, if exists, is called the volume integral
of π over π and is denoted as ππ
ππ.
Likewise if πΉ is a vector point function defined in the given region of volume π then vector volume
integral of πΉ over π is πΉ π
ππ.
Note: Above volume integral becomes ππ
ππ₯ ππ¦ ππ§ if we subdivide the volume π into small
cuboids by drawing lines parallel to three co-ordinate axes because in that case ππ = ππ₯ππ¦ππ§.
Example 60: If π = πππ β ππ π β ππππ β πππ , evualuate π Γ π π π½π½
where π½ is the region
bounded by the co-ordinate planes and the plane ππ + ππ + π = π.
Solution: Consider
51
β Γ πΉ =
π π π
π
ππ₯
π
ππ¦
π
ππ§
2π₯2 β 3π§ β2π₯π¦ β4π₯
= 0 π + π β 2π¦ π
Region bounded by 2π₯ + 2π¦ + π§ = 4 and coordinate planes such that
2π₯ β€ 4, 2π₯ + 2π¦ β€ 4, 2π₯ + 2π¦ + π§ β€ 4
i.e. π₯ β€ 2, π¦ β€ 2 β π₯, π§ β€ 4 β 2π₯ β 2π¦
β΄ β Γ πΉ πππ
= π β 2π¦ π ππ§ ππ¦ ππ₯4β2π₯β2π¦
π§=0
2βπ₯
π¦=0
2
π₯=0
= π§ π β 2π¦π§ π π§=0
π§=4β2π₯β2π¦ ππ¦ ππ₯
2βπ₯
π¦=0
2
π₯=0
= (4 β 2π₯ β 2π¦) π β 2π¦(4 β 2π₯ β 2π¦) π ππ¦ ππ₯2βπ₯
π¦=0
2
π₯=0
= 4π¦ β 2π₯π¦ β π¦2 π β 4π¦2 β 2π₯π¦2 β4
3π¦3 π
π¦=0
π¦=2βπ₯2
π₯=0ππ₯
= 8 β 4π₯ β 2π₯ 2 β π₯ β 2 β π₯ 2 π β
4(2 β π₯)2 β 2π₯(2 β π₯)2 β4
3(2 β π₯)3 π
2
π₯=0ππ₯
= 4 β 4π₯ + π₯2 π β β2
3π₯3 + 4π₯2 β 8π₯ +
16
3 π
2
π₯=0ππ₯
= 4π₯ β 2π₯2 +π₯3
3 π₯=0
π₯=2
π β β1
6π₯4 +
4
3π₯3 β 4π₯2 +
16
3π₯
π₯=0
π₯=2
π
=8
3π β
8
3π =
8
3(π β π )
ASSIGNMENT 8
1. Evaluate Ο πππ
where π = 45π₯2π¦ and π is the region bounded by the planes 4π₯ + 2π¦ + π§ =
8, π₯ = 0, π¦ = 0, π§ = 0.
2. If πΉ = 2π₯π§ π β π₯ π + π¦2 π ; evaluate F πππ
where π is the region bounded by the planes
π₯ = 0, π¦ = 0, π₯ = 2, π¦ = 6, π§ = π₯2 , π§ = 4.
17.13 STOKEβS THEOREM (Relation between Line and Surface Integral)
Statement: Let S be a piecewise smooth open surface bounded by a piecewise smooth simple curve
πΆ . If π (π₯, π¦, π§) be a continuous vector function which has continuous first partial derivative in a
region of space which contains π , then π . ππ πΆ
= ππ’ππ π . π πΆ
ππ , where π is the unit normal
vector at any point of π and πΆ is trasversed in positive direction.
Direction of πΆ is positive if an observer walking on the boundary of π in this direction with its head
pointing in the direction of outward normal π to π has the surface on the left.
We may put the statement of Stokeβs theorem in words as under:
52
The line integral of the tangential component of a vector π taken around a simple closed curve πΆ is
equal to the surface integral of normal component of curl of π taken over π having πΆ as its boundary.
Stokeβs Theorem in Cartesian Form:
a) Cartesian Form of Stokeβs Theorem in Plane (or Greenβs Theorem in Plane)
Choose system of coordinate axes such that the plane of the surface is in π₯π¦ plane and normal to the
surface π lies along the z-axis. Normal vector is constant in this case.
Suppose π = π1 π + π2 π + π3 π
β΄ π . ππ πΆ
= π .ππ
ππ ππ
πΆ= π . π‘ ππ
πΆ where π‘ =
ππ
ππ is unit vector tangent to πΆ.
β΄ π . ππ πΆ
= π1 π + π2 π + π3 π . π ππ₯
ππ + π
ππ¦
ππ + π
ππ§
ππ ππ
πΆ
= π1 ππ₯
ππ + π2
ππ¦
ππ + π3
ππ§
ππ ππ
πΆ
But tangent at any point lies in the π₯π¦ plane, so ππ§
ππ = 0
β΄ π . ππ πΆ
= π1 ππ₯
ππ + π2
ππ¦
ππ ππ
πΆ β¦ (1)
Now ππ’ππ π . π π
ππ = ππ’ππ π . π π
ππ (Here normal is along Z-axis)
= ππ2
ππ₯β
ππ1
ππ¦
πππ₯ππ¦ β¦ (2)
Using (1) and (2), Stokeβs theorem is
π1ππ₯ + π2ππ¦ πΆ
= ππ2
ππ₯β
ππ1
ππ¦
πππ₯ππ¦
b) Cartesian Form of Stokeβs Theorem in Space
Suppose π = π1 π + π2 π + π3π and π is an outward drawn normal unit vector of π making angles πΌ,
π½, πΎ with positive direction of axes.
β΄ π = cos πΌ π + cos π½ π + cos πΎ π
Now, β Γ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π1 π2 π3
π. π. ππ’ππ π = ππ3
ππ¦β
ππ2
ππ§ π +
ππ1
ππ§β
ππ3
ππ₯ π +
ππ2
ππ₯β
ππ1
ππ¦ π
β΄ ππ’ππ π . π = ππ3
ππ¦β
ππ2
ππ§ cos πΌ +
ππ1
ππ§β
ππ3
ππ₯ cos π½ +
ππ2
ππ₯β
ππ1
ππ¦ cos πΎ β¦ (1)
Also π . ππ = π1 π + π2 π + π3π . ππ₯ π + ππ¦ π + ππ§ π
53
or π . ππ = π1 ππ₯ + π2 ππ¦ + π3 ππ§
Then Stokeβs theorem is
π1 ππ₯ + π2 ππ¦ + π3 ππ§ πΆ
= ππ3
ππ¦β
ππ2
ππ§ cos πΌ +
ππ1
ππ§β
ππ3
ππ₯ cos π½ +
ππ2
ππ₯β
ππ1
ππ¦ cos πΎ
πππ
Example 61: Verify Stokeβs theorem for π = ππ β π π β ππππ β ππππ when πΊ is the upper half
of the surface of the sphere ππ + ππ + ππ = π and πͺ is its boundary.
Solution: The boundary πΆ of the upper half of the sphere π is circle in the π₯π¦ plane. Therefore,
parametric equations of πΆ are π₯ = cos π‘ , π¦ = sin π‘ , π§ = 0 when 0 β€ t β€ 2Ο
Now, π . ππ πΆ
= π1 ππ₯ + π2 ππ¦ + π3 ππ§ πΆ
= 2π₯ β π¦ ππ₯ β π¦π§2ππ¦ β π¦2π§ ππ§πΆ
= 2 cos π‘ β sin π‘ (β sin π‘) ππ‘πΆ
β΅ π₯ = cos π‘ , β΄ ππ₯ = β sin π‘ ππ‘ and other terms of integrand become zero as π§ = 0
= 2 cos π‘ β sin π‘ + sin2 π‘ ππ‘2π
π‘=0= 2
cos 2 π‘
2 π‘=0
2π
+ sin2 π‘ ππ‘2π
π‘=0
= 1 β 1 + 4 sin2 π‘ ππ‘π 2
π‘=0 (Property of definite integral)
= 0 + 4.1
2.π
2= π β¦ (1)
Now, ππ’ππ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
2π₯ β π¦ βπ¦π§2 βπ¦2π§
= β2π¦π§ + 2π¦π§ π + 0 β 0 π + 0 + 1 π = π
Now, ππ’ππ π . π ππ = π π
. π ππ
= π π
. π ππ₯ ππ¦
π .π [where π is the projection of S on π₯π¦ β plane]
= ππ₯ ππ¦π
Now projection of π on π₯π¦ plane is circle π₯2 + π¦2 = 1.
= ππ₯ 1βπ₯2
π¦=β 1βπ₯2 ππ¦1
π₯=β1= 4 ππ₯
1βπ₯2
π¦=0ππ¦
1
π₯=0 [By definite integral]
= 4 1 β π₯21
π₯=0ππ₯ = 4
π₯ 1βπ₯2
2+
1
2sinβ1 π₯
π₯=0
1
= 4 1
2sinβ1 1 = 4 Γ
π
4 = π β¦ (2)
From (1) and (2), Stokeβs theorem is verified.
Example 62: Verify Stokeβs theorem for the function π = ππ + ππ π β πππ π taken round the
rectangle bounded by π = Β±π, π = π, π = π. [KUK 2006]
54
O B A
D C
π¦ = 0
π¦ = π
π₯ = π π₯ = βπ
X
Y
Solution: Given π = π₯2 + π¦2 π β 2π₯π¦ π
Therefore, π . ππ = π₯2 + π¦2 π β 2π₯π¦ π . ππ₯ π + ππ¦ π = π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦
Fig. 17.17
β΄ π . ππ πΆ
= π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦πΆ
= π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦π·π΄
+ π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦π΄π΅
+ π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦π΅πΆ
+ π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦πΆπ·
β¦ (1)
On DA: π₯ = βπ, β΄ ππ₯ = 0
β΄ π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦π·π΄
= β2 βπ π¦ ππ¦π·π΄
= 2ππ¦ ππ¦0
π¦=π= ππ¦2 π
0 = βππ2
On AB: π¦ = 0, β΄ ππ¦ = 0
β΄ π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦π΄π΅
= π₯2 ππ₯π΄π΅
= π₯2π
βπππ₯ =
2
3π3
On BC: π₯ = π, β΄ ππ₯ = 0
β΄ π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦π΅πΆ
= β2ππ¦ ππ¦π΅πΆ
= β2ππ¦ ππ¦π
π¦=0= βππ2
On CD: π¦ = π, β΄ ππ¦ = 0
β΄ π₯2 + π¦2 ππ₯ β 2π₯π¦ ππ¦πΆπ·
= π₯2 + π2 ππ₯πΆπ·
= π₯2 + π2 ππ₯βπ
π₯=π
= π₯3
3+ π2π₯
π₯=π
βπ
= β2 π3
3+ π2π
Substituting these values in (1), we get
π . ππ πΆ
= βππ2 +2
3π3 β ππ2 β 2
π3
3+ π2π = β4ππ2 β¦ (2)
Now, ππ’ππ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π₯2 + π¦2 β2π₯π¦ 0
= 0 π + 0 π + β2π¦ β 2π¦ π = β4π¦ π
Since Surface lies in xy-plane, therefore π = π .
55
X
Y
B (1,1,0)
O (0,0,0) X
β΄ ππ’ππ π π
. π ππ = β4π¦ π . π π
ππ = β4π¦π
π₯=βπππ¦
π
π¦=0ππ₯ = β4ππ2 β¦ (3)
Hence from (2) and (3), theorem is verified.
Example 63: Evaluate by Stokeβs theorem ππ π π + ππ π π + πππ π πͺ
, where πͺ is the curve
ππ + ππ = π, π = ππ.
Solution: π¦π§ ππ₯ + π₯π§ ππ¦ + π₯π¦ππ§ πΆ
= π¦π§ π + π₯π§ π + π₯π¦ π πΆ
. ππ₯ π + ππ¦ π + ππ§ π
= π πΆ
. ππ , where π = π¦π§ π + π₯π§ π + π₯π¦ π
Now, ππ’ππ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π¦π§ π§π₯ π₯π¦
= π₯ β π₯ π + (π¦ β π¦) π + π§ β π§ π = 0
β΄ π πΆ
. ππ = ππ’ππ π . π π
ππ = 0 β΅ ππ’ππ π = 0
Example 64: Evaluate π πͺ
. π π by Stokeβs theorem, where π = ππ π + ππ π β π + π π and πͺ is
the boundary of triangle with vertices at (π, π, π), (π, π, π), (π, π, π).
Solution: Here,
ππ’ππ π =
π π π
π
ππ₯
π
ππ¦
π
ππ§
π¦2 π₯2 β(π₯ + π§)
= 0 π + π + 2 π₯ β π¦ π
Fig. 17.18
Here triangle is in the xy plane as z co-ordinate of each vertex of the triangle is zero.
β΄ π = π
β΄ ππ’ππ π . π = 0 π + π + 2 π₯ β π¦ π . π = 2 π₯ β π¦
By Stokeβs theorem, π πΆ
. ππ = ππ’ππ π . π π
ππ
= 2 π₯ β π¦ π
ππ¦ ππ₯
Note here the equation of OB is π¦ = π₯, thus for π, x varies from 0 to 1 and π¦ from 0 to π₯.
A (1,0,0)
56
β΄ π πΆ
. ππ = 2(π₯ β π¦)π₯
π¦=0ππ¦
1
π₯=0ππ₯ = 2 π₯π¦ β
π¦2
2 π¦=0
π₯
ππ₯1
π₯=0
= 2 π₯2 βπ₯2
2 ππ₯
1
π₯=0= π₯2ππ₯
1
π₯=0=
1
3
Note: Greenβs Theorem in plane is special case of Stokeβs Theorem: If R is the region in xy plane
bounded by a closed curve πΆ then this is a special case of Stokeβs theorem. In this case π = π and it
is called vector form of Greenβs theorem in plane. Vector form of Greenβs theorem can be written as
β Γ π π
. π ππ = π πΆ
. ππ
Example 65: Evaluate π β π¬π’π§ π π π + ππ¨π¬ ππ π πͺ
, where πͺ is the triangle having vertices
π, π , π
π, π and
π
π, π (i) directly (ii) by using Greenβs theorem in plane [KUK 2011]
Solution:
(i) Here π¦ β sin π₯ ππ₯ + cos π₯ ππ¦ πΆ
= π¦ β sin π₯ π + cos π₯ π . (ππ₯ π + ππ¦ π πΆ
)
= π πΆ
. ππ
where π = π¦ β sin π₯ π + cos π₯ π and ππ = ππ₯ π + ππ¦ π and
C is triangle OAB
Now π πΆ
. ππ = π¦ β sin π₯ ππ₯ + cos π₯ ππ¦ πΆ
= π¦ β sin π₯ ππ₯ + cos π₯ ππ¦ ππ΄
+ π¦ β sin π₯ ππ₯ + cos π₯ ππ¦ π΄π΅
+ π¦ β sin π₯ ππ₯ + cos π₯ ππ¦ π΅π
β¦ (1)
On OA: π¦ = 0, β΄ ππ¦ = 0
β΄ π¦ β sin π₯ ππ₯ + cos π₯ ππ¦ ππ΄
= β sin π₯ ππ₯ ππ΄
= β sin π₯ ππ₯π 2
π₯=0= β1
On AB: π₯ =π
2, β΄ ππ₯ = 0
β΄ π¦ β sin π₯ ππ₯ + cos π₯ ππ¦ π΄π΅
= 0
On BO: π¦ =2
ππ₯, β΄ ππ¦ =
2
πππ₯
β΄ π¦ β sin π₯ ππ₯ + cos π₯ ππ¦ π΅π
= 2
ππ₯ β sin π₯ ππ₯ + cos π₯
2
πππ₯
0
π₯=π 2
= 2
π
π₯2
2+ cos π₯ +
2
πsin π₯
π₯=π 2
0
= 1 β π
4+
2
π
57
Substituting these values in (1), we get
π πΆ
. ππ = β1 + 0 + 1 βπ
4β
2
π= β
π
4+
2
π
(ii) By Greenβs Theorem
π1ππ₯ + π2ππ¦ πΆ
= ππ2
ππ₯β
ππ1
ππ¦
πππ₯ππ¦
= β sin π₯ β 1 ππ¦2π₯ π
π¦=0ππ₯
π 2
π₯=0= β sin π₯ β 1 π¦ π¦=0
2π₯ π ππ₯
π 2
π₯=0
= β2
ππ₯ sin π₯ + 1 ππ₯
π 2
π₯=0= β
2
π π₯ sin π₯ + π₯ ππ₯
π 2
π₯=0
= β2
π π₯ β cos π₯ + sin π₯ +
π₯2
2 π₯=0
π 2
= β2
π β0 + 1 +
π2
8 β (0 + 0 + 0)
= β 2
π+
π
4
Example 66: Verify Greenβs theorem in the plane for ππ + ππ π π + πππ ππͺ
, where πͺ is the
closed curve of the region bounded by π = π and π = ππ.
Solution:
π₯π¦ + π¦2 ππ₯ + π₯2ππ¦πΆ
= π₯π¦ + π¦2 ππ₯ + π₯2ππ¦ ππ΅π΄
+ π₯π¦ + π¦2 ππ₯ + π₯2ππ¦ ππ΄
β¦ (1)
Along curve OBA: π¦ = π₯2 β΄ ππ¦ = 2π₯ ππ₯
β΄ π₯π¦ + π¦2 ππ₯ + π₯2ππ¦ ππ΅π΄
= π₯3 + π₯4 ππ₯ + 2π₯3ππ₯ 1
π₯=0=
19
20
Along curve AO: π¦ = π₯ β΄ ππ¦ = ππ₯
β΄ π₯π¦ + π¦2 ππ₯ + π₯2ππ¦ π΄π
= π₯2 + π₯2 ππ₯ + π₯2ππ₯ π΄π
= 3π₯2 ππ₯0
π₯=1= β1
β΄ from (1), π₯π¦ + π¦2 ππ₯ + π₯2ππ¦πΆ
=19
20β 1 = β
1
20 β¦ (2)
Here π1 = π₯π¦ + π¦2, π2 = π₯2
β ππ1
ππ¦= π₯ + 2π¦,
ππ2
ππ₯= 2π₯
By Greenβs theorem,
π1 ππ₯ + π2 ππ¦ πΆ
= ππ2
ππ₯β
ππ1
ππ¦
πππ¦ ππ₯
= 2π₯ β π₯ β 2π¦ π
ππ¦ ππ₯
= π₯ β 2π¦ ππ¦π¦=π₯
π¦=π₯2 ππ₯1
π₯=0 = π₯π¦ β π¦2
π¦=π₯2π¦=π₯
ππ₯1
π₯=0
= π₯4 β π₯3 ππ₯1
π₯=0=
x5
5β
x4
4
x=0
x=1
=1
5β
1
4= β
1
20 β¦ (3)
Equation (2) and (3) verify the result.
58
ASSIGNMENT 9
1. Verify Greenβs theorem for 3π₯ β 8π¦2 ππ₯ + 4π¦ β 6π₯π¦ ππ¦ πΆ
where C is the boundary of the
region bounded by 0,0 yx and 1 yx . [KUK 2007]
2. Verify Greenβs theorem in plane for 3π₯2 β 8π¦2 ππ₯ + 4π¦ β 6π₯π¦ ππ¦ πΆ
, where C is the
boundary of the region defined by xy and 2xy . [KUK 2008]
3. Apply Greenβs theorem to evaluate 2π₯2 β π¦2 ππ₯ + π₯2 + π¦2 ππ¦ πΆ
, where C is the boundary
of the area enclosed by x-axis and the upper half of the circle 122 yx . [KUK 2010]
4. Evaluate the surface integral SdSnFcurl Λ.
by transforming it into a line integral, S being that
part of the surface of the paraboloid π§ = (1 β π₯2 β π¦2) for which 0z and kxjziyF ΛΛΛ
.
[KUK 2008]
5. Using Stokeβs theorem, evaluate π₯ + π¦ ππ₯ + 2π₯ β π§ ππ¦ + π¦ + π§ ππ§ πΆ
, where C is the
boundary of the triangle with vertices )0,0,2( , )0,3,0( and )6,0,0( . [KUK 2009]
6. Verify Stokeβs theorem for a vector field defined by π = βπ¦3 π + π₯3 π , in the region
π₯2 + π¦2 β€ 1, π§ = 0.
17.14 GAUSSβS DIVERGENCE THEOREM (Relation between Volume and Surface Integral)
Statement: Suppose π is the volume bounded by a closed piecewise smooth surface S. Suppose
π (π₯, π¦, π§) is a vector function which is continuous and has continuous first partial derivatives in
π.Then
β. π π
ππ = π . π ππ
where π is the outward unit normal to the surface π.
In other words: The surface integral of the normal component of a vector π taken over a closed
surface is equal to the integral of the divergence of π over the volume enclosed by the surface.
Divergence Theorem in Cartesian Coordinates
If π = π1π + π2π + π3π then πππ£ π = β. π =ππ1
ππ₯+
ππ2
ππ¦+
ππ3
ππ§ . Suppose πΌ, π½, πΎ are the angles made by
the outward drawn unit normal with the positive direction of axes, then
π = cos πΌ π + cos π½ π + cos πΎ π
Now, π . π = π1 cos πΌ + π2 cos π½ + π3 cos πΎ
Then divergence theorem is
ππ1
ππ₯+
ππ2
ππ¦+
ππ3
ππ§ ππ₯ ππ¦ ππ§
π= π1 cos πΌ + π2 cos π½ + π3 cos πΎ ππ
π
= π1ππ¦ππ§ + π2 ππ§ππ₯ + π3ππ₯ππ¦ π
β΅ cos πΌ ππ = ππ¦ππ§ ππ‘π.
59
Proof: Let π = π1π + π2π + π3π where π1, π2 and π3 and their derivatives in any direction are finite
and continuouos.
Suppose π is a closed surface such that it is possible to
choose rectangular cartesian co-ordinate system such that
any line drawn parallel to coordinate axes does not cut π
in more than two points.
Let π be the orthogonal projection of π on the xy-plane.
Any line parallel to z-axis through a point of π meets the
boundary of S in two points. Let π1 and π2 be the lower
and upper portions of π . Let the equations of these
portions be
π§ = π·1(π₯, π¦) and π§ = π·2(π₯, π¦)
where π·1(π₯, π¦) β₯ π·2(π₯, π¦)
Consider the volume integral
ππ3
ππ§ ππ =
π
ππ3
ππ§ ππ₯ ππ¦ ππ§ =
ππ3
ππ§ππ§
π§=π·1(π₯ ,π¦)
π§=π·2(π₯ ,π¦) ππ₯ ππ¦
ππ3
ππ§ ππ =
π π3(π₯, π¦, π§) π·2(π₯ ,π¦)
π·1(π₯ ,π¦)ππ₯ ππ¦ = π3 π₯, π¦, π·1 β π3 π₯, π¦, π·2 ππ₯ ππ¦ β¦(1)
Let π 1 be the unit outward drawn vector making an acute angle πΎ1 with π for the upper position π1 as
shown in the figure.
Now projection ππ₯ ππ¦ of ππ1 on the π₯π¦ plane is given as ππ₯ππ¦ = ππ1 cos πΎ = ππ1 π . π 1 = π . π 1ππ1
Now π3 π₯, π¦, π·1 π ππ₯ππ¦ = π3 π . π 1ππ1π1
β¦ (2)
Similarly if π 2 be the unit outward drawn normal to the lower surface π2 making an angle πΎ2 with π .
Obviously πΎ2 is an obtuse angle
β΄ ππ₯ππ¦ = ππ2 cos π β πΎ2 = βππ2 cos πΎ2 = βπ . π 2 ππ2
β΄ π3 π₯, π¦, π·2 π ππ₯ ππ¦ = β π3 π . π 2ππ2π2
β¦ (3)
From (1), (2) and (3), we have
ππ3
ππ§ ππ
π= π3 π . π 1ππ1π1
+ π3 π . π 2ππ2π2= π3 π . π ππ
π β¦ (4)
Similarly by projecting π on the other coordinate planes
ππ2
ππ¦ ππ
π= π2 π . π ππ
π β¦ (5)
And ππ1
ππ₯ ππ
π= π1 π . π ππ
π β¦ (6)
60
Adding (4), (5) and (6), we get
ππ1
ππ₯+
ππ2
ππ¦+
ππ3
ππ§ ππ
π= π1 π + π2 π + π3 π . π ππ
π = π . π ππ
π
Note: With the help of this theorem we can express volume integral as surface integral or vice versa.
17.15 GREENβS THEOREM (For Harmonic Functions)
Statement: If π· and π are two scalar point functions having continuous second order derivatives in a
region π bounded by a closed surface π, then
π·β2π β πβ2π· π
ππ = π· βπ β π βπ· . π πππ
Proof: By Gaussβs divergence theorem,
β. π πππ
= π . π π
ππ β¦ (1)
Take π = π· βπ so that β. π = β. π· βπ
= π· β. βπ + βπ·. βπ
= π·β2π β βπ·. βπ β¦ (2)
Now π . π = π· βπ . π . Using this and (2) in (1), we get
π· β2π β βπ·. βπ πππ
= π· βπ . π π
ππ β¦ (3)
Again starting as above by interchanging π· and π, we obtain as in (3)
π β2π· β βπ. βπ· πππ
= π βπ· . π π
ππ β¦ (4)
Subtracting (4) from (3), we get
π·β2π β πβ2π· π
ππ = π· βπ β π βπ· . π πππ
β¦ (5)
Another Form of Greenβs Theorem:
ππ·
ππ and
ππ
ππ denote the direction of derivative of π· and π along the outward drawn normal at any point
of π.
βπ· =ππ·
πππ and βπ =
ππ
πππ
β΄ π· βπ β π βπ· = π·ππ
ππ π β π
ππ·
ππ π
or π· βπ β π βπ· . π = π·ππ
πππ β π
ππ·
πππ . π = π·
ππ
ππβ π
ππ·
ππ
β΄ Equation (5) becomes
π·β2π β πβ2π· π
ππ = π·ππ
ππβ π
ππ·
ππ
πππ
61
Example 67: Evaluate π . π π πΊπΊ
where π = πππ π + ππ π β ππ π and πΊ is the surface of the
cube bounded by the planes π = π, π = π, π = π, π = π, π = π, π = π.
Solution: Here π = 4π₯π¦ π + π¦π§ π β π§π₯ π . By Gaussβs divergence theorem
π . π πππ
= β. π πππ
= π π
ππ₯+ π
π
ππ¦+ π
π
ππ§
π. 4π₯π¦ π + π¦π§ π β π§π₯ π ππ
= 4π¦ + π§ β π₯ πππ
= 4π¦ + π§ β π₯ ππ§ ππ¦ ππ₯2
π§=0
2
π¦=0
2
π₯=0
= 4π¦π§ +π§2
2β π₯π§
π§=0
π§=22
π¦=0
2
π₯=0 ππ¦ ππ₯
= 8π¦ + 2 β 2π₯ 2
π¦=0ππ¦
2
π₯=0ππ₯
= 4π¦2 + 2π¦ β 2π₯π¦ π¦=0π¦=2
ππ₯2
π₯=0= 20 β 4π₯ ππ₯
2
π₯=0
= 20π₯ β 2π₯2 π₯=0π₯=2 = 32
Example 68: Use Gauss theorem to show that ππ β ππ π β ππππ π + ππ . π π πΊπΊ
=ππ
π
where S denotes the surface of the cube bounded by the planes, π = π, π = π, π = π, π = π,
π = π, π = π .
Solution: By Gaussβs divergence theorem
π₯3 β π¦π§ π β 2π₯2π¦ π + 2π . π πππ
= β.π
π₯3 β π¦π§ π β 2π₯2π¦ π + 2π ππ
= 3π₯2 β 2π₯2 ππ₯π
0ππ¦
π
0ππ§
π
0
= π₯2ππ₯π
0 ππ¦
π
0ππ§
π
0=
π₯3
3
0
π
ππ¦π
0ππ§
π
0
= π3
3 ππ¦
π
0ππ§
π
0=
π3
3π¦
0
π
ππ§π
0=
π4
3ππ§
π
0=
π4
3 π§ 0
π =π5
3
Example 69: Evaluate π . π πΊ
π πΊ with the help of Gauss theorem for π = ππ π + ππ + π π β
π π taken over the region πΊ bounded by the surface of the cylinder ππ + ππ = π included
between π = π, π = π, π = π and π = π.
Solution: π = 6π§ π + 2π₯ + π¦ π β π₯ π
β. π = π π
ππ₯+ π
π
ππ¦+ π
π
ππ§ . 6π§ π + 2π₯ + π¦ π β π₯ π = 1
By Gaussβs divergence theorem,
π . π π
ππ = 1 ππ₯ ππ¦ ππ§π
62
= ππ§ 9βπ₯2
π§=0ππ¦
8
π¦=0ππ₯
3
π₯=0= π§ π§=0
π§= 9βπ₯2ππ¦
8
π¦=0ππ₯
3
π₯=0
= 9 β π₯2 ππ¦8
π¦=0ππ₯
3
π₯=0
= 9 β π₯2 π¦ π¦=0
8ππ₯
3
π₯=0= 8 9 β π₯2 ππ₯
3
π₯=0
= 8 π₯ 9βπ₯2
2+
9
2sinβ1 π₯
3 π₯=0
3
= 18 π
Example 70: Evaluate (ππ ππ π + ππ ππ π + ππ ππ π)πΊ
where πΊ is the surface of the sphere
ππ + ππ + ππ = ππ. [KUK 2011, 2009]
Solution: By Gaussβs divergence theorem
(π₯ππ¦ππ§ + π¦ππ§ππ₯ + π§ππ₯ππ¦)π
= ππ₯
ππ₯+
ππ¦
ππ¦+
ππ§
ππ§ ππ₯ ππ¦ ππ§ =
π 3 ππ₯ ππ¦ ππ§
π
= 3 ππ₯ ππ¦ ππ§π
= 3 Γ volume of the sphere x2 + y2 + z2 = a2
= 3 Γ4
3Οa3 = 4Οa3
Example 71: Show that π πΊ
π πΊ = π for any closed surface πΊ.
Solution: Let πΆ be any closed vector.
β΄ πΆ π π
ππ = πΆ.π
π ππ = πππ£ πΆπ
ππ
β΄ πΆ π π
ππ = 0 β΅ πΆ is constant, therfore πππ£ πΆ = 0
β π π
ππ = 0
Example 72: Prove that π
πππ π½
π½=
π .π
πππ πΊ
πΊ, where π = π π + π π + π π and π = π.
Solution: π .π
π2 πππ
= π
π2 . π πππ
= β.π
π2 πππ β¦ (1)
Now, β. π
π2 =1
π2 β. π + π . β
1
π2 β¦ (2)
Also, π = π₯ π + π¦ π + π§ π β π2 = π₯2 + π¦2 + π§2
β΄ 2πππ
ππ₯= 2π₯; 2π
ππ
ππ¦= 2π¦; 2π
ππ
ππ§= 2π§
β ππ
ππ₯=
π₯
π;
ππ
ππ¦=
π¦
π;
ππ
ππ§=
π§
π β¦ (3)
Now, β 1
π2 = π
π
ππ₯+ π
π
ππ¦+ π
π
ππ§
1
π2 =
β2
π3 π
ππ
ππ₯+ π
ππ
ππ¦+ π
ππ
ππ§ using (3)
=β2
π4 π₯ π + π¦ π + π§ π =
β2
π4π
63
Also, β. π = 1 + 1 + 1 = 3
Substituting these values in (2), we have
β. π
π2 =1
π2 . 3 + π . β2
π4 π =3
π2 β2
π4 π2 =1
π2 β΅ π . π = π2
β΄ From 1 , π .π
π2ππ
π=
1
π2ππ
π
ASSIGNMENT 10
1. Find SdSnF Λ.
where kzyjyxzizxF Λ)2(Λ)(Λ)2( 2
and S is the surface of the sphere
having centre )2,1,3( and radius 3 units. [KUK 2006]
2. Use Divergence theorem to evaluate SSdF
. where kzjyixF ΛΛΛ 333
and S is the outer
surface of the sphere 1222 zyx . [KUK 2007]
3. Verify Divergence theorem for kxyzjzxyiyzxF Λ)(Λ)(Λ)( 222
taken over rectangular
parallelopiped ,0 ax ,0 by .0 cz [KUK 2010]
ANSWERS
ASSIGNMENT 1
1. β4(π + 2π ) 3. π₯ β π/ 2 = π¦ β π/ 2 = π§ βππ
4tan πΌ / 2 π‘ππ πΌ
4. π‘ π + 2π β 2π‘ β 3 π / (5π‘2 β 12π‘ + 13) ; 1
3(2 π + 2π + π )
5. (a) π’ π2 sec πΌ (b) π3 tan πΌ; (β cos πΌ sin π‘ π + cos πΌ cos π‘ π + sin πΌ π )
6. (a) π π + π + 2π π + π + π π ; 1
6 βπ β π + 2 π (b) π + π π + π π + 2π π ;
1
5 2π β π
17. (a) ππ/ π2 sin2 π‘ + π2 cos2 π‘ 3/2 (b) 1/4 2
ASSIGNMENT 2
1. π£ = 37 and π = 325 at π‘ = 0 3. 8
7 14 ;
1
7 14 4. π = Β±
1
6 7.
70
29;
436
29
8. 21.29 knots/hr. in the direction 740 47β South of East 9. 17 meter per hour in the direction
tanβ1 0.25 North of East
ASSIGNMENT 3
1. (a) 2(π₯ π + π¦ π + π§ π )/ (π₯2 + π¦2 + π§2) 3. 15
17 4.
37
3 5.
1
3(2 π + 2 π β π ) 6. 11
7. cosβ1 β1
30 8. cosβ1 1/ 22 9. cosβ1
8
3 21 10. π = 4 πππ π = 1
64
ASSIGNMENT 4
2. (a) 80 (b) ππ₯π¦π§ π₯ π§ β π¦ π + π¦ π₯ β π§ π + π§ π¦ β π₯ π
3. π = β2; 4π₯ π§ β π₯π¦ π + π¦ 1 β 2π§ + 4π₯π¦ π + (2π₯2 + π¦2 β π§2 β π§)π
9. (a) 2π 2πβ1
π₯2+π¦2+π§2 π+1; π =
1
2
11. (i) 2 π¦3 + 3π₯2π¦ β 6π₯π¦2 π§ π + 2 3π₯π¦2 + π₯3 β 6π₯2π¦ π§ π + 2 π₯π¦2 + π₯3 β 3π₯2π¦ π¦ π (ii) Zero
13. (i) 0 (ii) 2 π₯ + π§ π + 2π¦ π
ASSIGNMENT 5
1. 226
3 π + 360 π β 42 π 2. β2 π + 3 π β 3 π
4. π£ = 6 sin 2π‘ π + 4(cos 2π‘ β 1) π + 8π‘2 π and π = 3 1 β cos 2π‘ π + 2 sin 2π‘ π +8π‘3
3 π
ASSIGNMENT 6
1. 88
35 2. 16, 16 4. 35 5.
π3 2
3
6. 2 βπ
4 π β π β
1
2 π
ASSIGNMENT 7
2. 81 3. 3/2 5. 8
ASSIGNMENT 8
1. 128 2. 128π β 24π + 384π
ASSIGNMENT 9
3. 4/3 5. 21
ASSIGNMENT 10
1. 108Ο 2. 56Οa2/9