uicm002 & engineering mathematics ii unit ii vector
TRANSCRIPT
UICM002 & Engineering Mathematics - II
Vector Calculus 1
SRI RAMAKRISHNA INSTITUTE OF TECHNOLOGY
(AN AUTONOMOUS INSTITUTION)
COIMBATORE- 641010
UICM002 & Engineering Mathematics – II
Unit II – Vector Calculus
Course Material
History of Vectors
Vector calculus was developed from quaternion analysis by J. Willard Gibbs and
Oliver Heaviside at the end of the 19th century. The terminology and the notation were
established by Gibbs and Edwin Bidwell Wilson in 1901.
Vector calculus is a branch of mathematics concerned with differentiation and
integration of vector fields, primarily in 3-dimensional Euclidean space.
Vector calculus plays an important role in differential geometry and in the study
of partial differential equations. It is used extensively in physics and engineering,
especially in the description of electromagnetic fields, gravitational fields and fluid
flow.
Introduction to vector
A vector is an object that has both a magnitude and a direction. Geometrically, we
can picture a vector as a directed line segment, whose length is the magnitude of the
vector and with an arrow indicating the direction. The direction of the vector is from its
tail to its head.
Two vectors are the same if they have the same magnitude and direction. This
means that if we take a vector and translate it to a new position (without rotating it),
then the vector we obtain at the end of this process is the same vector we had in the
beginning.
UICM002 & Engineering Mathematics - II
Vector Calculus 2
Examples of everyday activities that involve vectors include:
Breathing (your diaphragm muscles exert a force that has a magnitude and
direction)
Walking (you walk at a velocity of around 6 km/h in the direction of the bathroom)
Going to school (the bus has a length of about 20 m and is headed towards your
school)
Lunch (the displacement from your class room to the canteen is about 40 m in a
northerly direction)
Gradient, divergence and curl
Definition: Laplacian operator
Definition: Gradient of a scalar function
Let ( ) be a scalar point function and is continuously differentiable then
the vector
(
)
is called gradient of the function and is denoted as .
Example:
If , find at ( )
Answer:
UICM002 & Engineering Mathematics - II
Vector Calculus 3
( ) ( ) ( )
( ) ( ( )) ( ) ( )
( )
Example:
If , find
Answer:
Example:
If Find | | at ( )
Answer:
( ) ( ) ( )
( ) [ ( ) ( ) ( )] [ ( ) ] [ ( ) ]
( )
| | √ ( ) ( ) √
Example:
Find the gradient of the function ( ) .
Answer:
UICM002 & Engineering Mathematics - II
Vector Calculus 4
( )
[ ( ) ( ) ( )]
[ ]
Example:
If is the position vector of the point ( ), find
1. 2. 3.
Answer:
(
) ( )
| | √
(
)
√
√ [
(√ )
√ ]
UICM002 & Engineering Mathematics - II
Vector Calculus 5
√
√
(
√ ) (
√ ) (
√ )
√
[ ]
(
)
(
)
(
) *
+
[
]
[ ]
Definition: Directional Derivative
The directional derivative of the scalar point function at a point is defined by
the dot product of and the unit normal vector through that point.
Definition: Unit tangent vector
|
|
UICM002 & Engineering Mathematics - II
Vector Calculus 6
Example:
Find the unit tangent vector to the following surfaces at the specified points
.
Answer:
( ) ( ) ( )
( )
(
)
|
| √
|
|
Definition: Unit normal to the surface
| |
Example:
Find the unit vector normal to the surface ( ).
Answer:
UICM002 & Engineering Mathematics - II
Vector Calculus 7
( )
| | √ √ √
| |
√
√
Example:
Find the unit vector normal to the surface ( ).
Answer:
( ) ( ) ( )
( ) ( ( ) ) ( ) ( )
( )
| | √ √
| |
√
Example:
Find the directional derivative of at that point ( ) in the
direction of .
Answer:
| |
UICM002 & Engineering Mathematics - II
Vector Calculus 8
√
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) (
)
Definition: Angle between the surfaces
The angle between the surfaces is given by
| || |
Example:
Find the angle between the normal to the surface at the points ( )
and ( ).
Answer:
UICM002 & Engineering Mathematics - II
Vector Calculus 9
( ) ( ) ( )
( )
( )
| | √ ( ) ( ) √
( )
( ) ( ) ( ) ( )
| | √ ( ) ( ) √ The angle between two surfaces is
| || |
( ) ( )
√ √
√ √
(
√ √ )
Example:
Find the angle between the normal to the surface at the points
( ) and ( ).
Answer:
( ) ( ) ( )
( )
( ) [( ) ( ) ] [ ( )( ) ( ) ] [ ( )( ) ]
UICM002 & Engineering Mathematics - II
Vector Calculus 10
| | √( ) ( ) √
( )
( ) [ ( ) ] [ ( ) ( ) ] [ ( ) ( )]
| | √ ( ) √
The angle between two surfaces is
| || |
( ) ( )
√ √
√ √
(
√ √ )
Example:
Find the angle between the surfaces
at ( )
Answer:
( )
| | √ ( ) ( ) √
UICM002 & Engineering Mathematics - II
Vector Calculus 11
( )
| | √ ( )
The angle between two surfaces is
| || |
( ) ( )
√
√
√
(
√ )
Example:
Find the angle between the surfaces and at
( )
Answer:
( )
| | √
UICM002 & Engineering Mathematics - II
Vector Calculus 12
( )
| | √
The angle between two surfaces is
| || |
( ) ( )
(
)
Definition: Divergence of a scalar point function
If ( ) is a continuously differentiable vector point function in a region of
space then the divergence is defined by
( )
(
) ( )
Definition: Curl of a vector function
If ( ) is a continuously differentiable vector point function defined in each
point ( ) then the of is defined by
UICM002 & Engineering Mathematics - II
Vector Calculus 13
||
||
(
) (
) (
)
Example:
If , then find .
Answer:
( ) ( ) ( )
( )
||
||
||
||
(
) (
) (
)
Example:
If ( ) , then find at ( ).
Answer:
( )
UICM002 & Engineering Mathematics - II
Vector Calculus 14
( ) ( ) ( )
[ ]( )
( ( )) ( ) ( ( ) )
[ ]( )
( )
||
||
||
||
(
( )
) (
( )
)
(
)
( ) ( ) ( )
[ ]( )
[ ( )] [( ) ] [ ( )]
Definition: Laplace Equation
is called the Laplace equation.
Definition: Solenoidal vector
UICM002 & Engineering Mathematics - II
Vector Calculus 15
Definition: Irrotational vector
Definition: Scalar Potential
If is irrotational vector, then there exists a scalar function such that
Such a scalar function is called scalar potential of .
Example:
Find the value of such that the vector is both solenoidal and irrotational.
Answer:
( )
(
) ( )
( )
( )
( )
(
) (
) (
)
( )
[
]
[ ]
( )
is solenoidal if
||
||
∑ (
( )
( ))
UICM002 & Engineering Mathematics - II
Vector Calculus 16
∑ (
)
∑ ( (
) (
) )
is irrotational for all values of .
Example:
Prove that is Solenoidal.
Answer:
Hence is Solenoidal.
Example:
Prove that is Solenoidal.
Answer:
Hence is Solenoidal.
UICM002 & Engineering Mathematics - II
Vector Calculus 17
Example:
If ( ) ( ) ( ) is Solenoidal find the value of .
Answer:
( ) ( ) ( )
Example:
Prove that is Irrotational.
Answer:
||
||
||
||
(
) (
) (
)
( ) ( ) ( )
Hence is Irrotational.
UICM002 & Engineering Mathematics - II
Vector Calculus 18
Example:
Find the constants so that ( ) ( )
( ) is Irrotational.
Answer:
( ) ( ) ( )
||
||
||
||
[
( )
( )]
[
( )
( )]
[
( )
( )]
( ) ( ) ( )
Example:
Prove that ( ) ( ) ( ) is Solenoidal as well as
Irrotational. Also find the scalar potential of .
Answer:
( ) ( ) ( )
UICM002 & Engineering Mathematics - II
Vector Calculus 19
Hence is Solenoidal.
( ) ( ) ( )
||
||
||
||
(
( )
( )) (
( )
( ))
(
( )
( ))
( ) ( ) ( )
Hence is Irrotational.
To find scalar potential:
To find such that
( ) ( ) ( )
Integrating w.r.t Integrating w.r.t Integrating w.r.t
UICM002 & Engineering Mathematics - II
Vector Calculus 20
∫( )
(
) ( )
( )
∫( )
(
) ( )
( )
∫( )
(
) ( )
( )
Combining, we get ( )
where is a constant.
Therefore is a scalar potential.
Example:
Prove that ( ) ( ) ( ) is Irrotational. Also
find the scalar potential of .
Answer:
||
|| ||
||
(
( )
( )) (
( )
( ))
(
( )
( ))
( ) ( ) ( )
Hence is Irrotational.
To find scalar potential:
To find such that
( ) ( ) ( )
UICM002 & Engineering Mathematics - II
Vector Calculus 21
Integrating w.r.t Integrating w.r.t Integrating w.r.t
∫( )
(
) ( )
( )
∫( )
( )
∫( )
(
) ( )
( )
( )
where is a constant. Therefore is a scalar potential.
Example:
Prove that ( ) ( ) is Irrotational. Also find the scalar
potential of .
Answer:
||
|| ||
||
(
( )
( )) (
( )
( ))
(
( )
( ))
( ) ( ) [ ( )]
Hence is Irrotational.
To find scalar potential:
To find such that
( ) ( )
UICM002 & Engineering Mathematics - II
Vector Calculus 22
( )
Integrating w.r.t Integrating w.r.t
∫( )
(
) ( )
( )
∫( )
*
+ ( )
( )
( )
where is a constant. Therefore is a scalar potential.
Example:
Prove that ( ) ( ) is Irrotational. Also find the scalar
potential.
Answer:
||
|| ||
||
(
( )
( )) (
( )
( ))
(
( )
( ))
( ) ( ) ( )
Hence is Irrotational.
To find scalar potential:
To find such that
UICM002 & Engineering Mathematics - II
Vector Calculus 23
( ) ( )
Integrating w.r.t Integrating w.r.t
∫( )
( )
∫( )
( )
( )
where is a constant. Therefore is a scalar potential.
Vector integration
Line integral
Let be a vector field in space and let be a curve described in the sense to .
∑
∫
If the line integral along the curve then it is denoted by
∫
∮
Example:
Find the work done by the moving particle in the force field
( ) from along the curve
Solution:
( )
( )
UICM002 & Engineering Mathematics - II
Vector Calculus 24
∮
∫[ ( ) ( ) ( ( ) ) ( )]
∫[ [ ] ]
∫[ ]
*
+
Example:
Find the work done when a force ( ) ( ) moves a
particle from the origin to the point ( ) along .
Solution:
( ) ( )
( ) ( )
Since varies from ( ) to ( ).
∮
∫[( ) ( ) ]
∫[( ( ) ) ( ) ]
∫[ ]
UICM002 & Engineering Mathematics - II
Vector Calculus 25
*
+
[
( )]
Greens theorem in plane
If ( ) ( ) are continuous functions with continuous, partial
derivatives in a region of the plane bounded by a simple closed curve then
∫
∬(
)
where is the curve described in the positive direction.
Example:
Evaluate using Green’s theorem in the plane for ∫( ) where is
the closed curve of the region bounded by and .
Answer:
∫
∬(
)
∫
∬(
)
∬( )
UICM002 & Engineering Mathematics - II
Vector Calculus 26
Limits: to
or √
( ) to ( )
∫∫ ( )
√
∫*
+
√
∫*((√ )
√ ) (
)+
∫*(
) (
)+
∫[(
)]
(
( )
(
)
)
(
(
)
)
Example:
Verify Green’s theorem in the plane for ∫ ( )
where is the curve
in the plane given by ( ).
Answer:
∫
∬(
)
𝒙𝟐 𝒚 𝒙 𝒚
(𝟎 𝟎)
𝒙
𝒚
(𝟏 𝟏)
UICM002 & Engineering Mathematics - II
Vector Calculus 27
Limits: to
to
∬(
)
∫∫
∫ [ ]
( ) *
+
(
)
∬(
)
∫
∫
∫
∫
∫
Region Equation ∫
Along ∫
*
+
Along ∫
*
+
(
)
Along ∫
*
+
Along
∫
∫
∬(
)
Hence Green’s theorem is verified.
𝑪(𝟎 𝒂) 𝒚 𝒂
𝒚 𝟎
𝒙 𝟎
𝑶(𝟎 𝟎) 𝑨(𝒂 𝟎)
𝒙 𝒂
𝑩(𝒂 𝒂)
𝒙
𝒚
UICM002 & Engineering Mathematics - II
Vector Calculus 28
Example:
Verify Green’s theorem in the plane for ∫ ( ) ( )
where is the boundary of the region bounded by
Answer:
∫
∬(
)
( )
∬(
)
∫ ∫
∫ [ ]
∫ ( )
∫[ ]
(
)
(
)
∬(
)
Limits: to
( ) ( )
to
∫
∫
∫
∫
𝒚 𝟎
𝒙 𝟎
𝑶(𝟎 𝟎) 𝑨(𝟏 𝟎)
𝒙 𝒚 𝟏
𝑩(𝟎 𝟏)
𝒙
𝒚
UICM002 & Engineering Mathematics - II
Vector Calculus 29
Region Equation ∫( ) ( )
Along ;
∫( )
∫
*
+
Along ; [ ]
∫( ( ) ) ( ( ) ( ))( )
∫( ( )) ( )
∫( )
∫( )
*
+
(
)
Along ∫
*
+
(
)
∫
∫
∬(
)
Hence Green’s theorem is verified.
UICM002 & Engineering Mathematics - II
Vector Calculus 30
Example:
Verify Green’s theorem in the plane for ∫ ( ) ( )
where
is the square bounded by
Answer:
∫
∬(
)
( )
Limits: to
to
∬(
)
∫ ∫
∫ *
+
( ( )
) ∫
(
) ( )
(
) [ ( )]
( )
∬(
)
∫
∫
∫
∫
∫
𝑶(𝟎 𝟎)
𝑫( 𝒂 𝒂) 𝒚 𝒂
𝒚 𝒂
𝒙
𝒂
𝑨( 𝒂 𝒂) 𝑩(𝒂 𝒂)
𝒙 𝒂
𝑪(𝒂 𝒂)
𝒙
𝒚
UICM002 & Engineering Mathematics - II
Vector Calculus 31
Region Equation ∫ ( ) ( )
Along
∫ ( )
( ) *
+
( )( ( )
)
( )(
)
Along
∫( )
*
+
(
) (
( )
( ))
(
)
Along
∫ ( )
( ) *
+
( )(( )
)
( )(
)
Along ∫ ( )
*
+
UICM002 & Engineering Mathematics - II
Vector Calculus 32
(( )
( )) (
)
∫
∬(
)
Hence Green’s theorem is verified.
Example:
Using Green’s theorem evaluate ∫ ( ) ( )
where is
the boundary of the region defined by
Answer:
∫
∬(
)
( )
√
∬(
)
∫ ∫
√
∫ ( ) √
∫ (√ )
UICM002 & Engineering Mathematics - II
Vector Calculus 33
∫(
)
∫( )
(
)
(
( )
)
∫( ) ( )
Stoke’s theorem
If is a open surface bounded by a simple closed curve and if a vector function
is continuous and has continuous partial derivatives in and so on , then
∬
∫
where is the unit vector normal to the surface. That is, the surface integral of the
normal component of is equal to the line integral of the tangential component of
taken around .
Example:
Verify Stoke’s theorem for a vector field defined by ( ) in the
rectangular region in the plane bounded by the lines
Answer:
∫
∬
( )
UICM002 & Engineering Mathematics - II
Vector Calculus 34
( )
||
|| ||
||
(
( )
( )) (
( )
( ))
(
( )
( ))
( ) ( ) ( )
Here the surface denotes the rectangle and the unit outward normal to the
vector is .
That is
( )
Limits: to
to
∬
∬
∫∫
∫ [ ]
( ) *
+
( )
∬
𝑪(𝟎 𝒃) 𝒚 𝒃
𝒚 𝟎
𝒙 𝟎
𝑶(𝟎 𝟎) 𝑨(𝒂 𝟎)
𝒙 𝒂
𝑩(𝒂 𝒃)
𝒙
𝒚
UICM002 & Engineering Mathematics - II
Vector Calculus 35
∫
∫( )
∫
∫
∫
∫
Region Equation ∫( )
Along ∫
*
+
(
)
Along ∫
*
+
(
)
Along
∫( )
*
+
(
)
Along
∫
∫
∬
Example:
Verify Stoke’s theorem when ( ) ( ) and is the
boundary of the region enclosed by the parabolas and
Answer:
∫
∬
( ) ( )
( ) ( )
UICM002 & Engineering Mathematics - II
Vector Calculus 36
||
|| ||
||
(
( )
( )) (
( )
( ))
(
( )
( ))
( ) ( ) ( )
Here the surface denotes the rectangle and the unit outward normal to the
vector is .
That is ( )
√
∬
∫ ∫
√
∫*
+
√
∫*(√ ) ( ) +
∫[ ]
*
+
(
)
∫
∫
∫
UICM002 & Engineering Mathematics - II
Vector Calculus 37
Region Equation ∫( ) ( )
Along
∫( ( ) ) ( ( ) )
∫( )
∫( )
*
+
(
)
Along
∫[ ( ) ] [( ) ]
∫( )
∫( )
*
+
(
)
∫
∬
Hence Stoke’s theorem is verified.
UICM002 & Engineering Mathematics - II
Vector Calculus 38
Example:
Verify Stoke’s theorem for the function integrated around the
square in the plane whose sides are along the line .
Answer:
∫
∬
||
|| ||
||
(
( )
( )) (
( )
( )) (
( )
( ))
( ) ( ) ( )
Here the surface denotes the rectangle and the unit outward normal to the
vector is .
That is ( )
∬
∫∫
∫ ( )
( ) *
+
(
)
𝑪(𝟎 𝒂) 𝒚 𝒂
𝒚 𝟎
𝒙 𝟎
𝑶(𝟎 𝟎) 𝑨(𝒂 𝟎)
𝒙 𝒂
𝑩(𝒂 𝒂)
𝒙
𝒚
UICM002 & Engineering Mathematics - II
Vector Calculus 39
∫
∫
∫
∫
∫
Region Equation ∫
Along ∫
*
+
(
)
Along ∫
*
+
(
)
Along ∫
*
+
Along
∫
∬
Hence Stoke’s theorem is verified.
Example:
Evaluate by Stoke’s theorem ∮ ( )
where the curve is
.
Answer:
Here
By Stoke’s theorem,
UICM002 & Engineering Mathematics - II
Vector Calculus 40
∮
∬
∬
∬
where is the surface
is the surface
||
||
(
( )
( )) (
( )
( )) (
( )
( ))
( ) ( ) ( )
∬
∬
∮
Example:
Evaluate ∮
where , is the circle
Answer:
By Stoke’s theorem,
∮
∬
||
||
(
( )
( )) (
( )
( )) (
( )
( ))
( ) ( ) ( )
( ) ( )
UICM002 & Engineering Mathematics - II
Vector Calculus 41
( ( ) ( ))
| |
| |
∬
∬
( )
∮
Example:
Verify Stoke’s theorem for the vector field ( )
over the upper half surface of bounded by its projection on the -
plane.
Answer:
By Stoke’s theorem,
∮
∬
𝑦
𝑥
𝑧
𝑥 𝑦 𝑧
0 𝐴 𝐵
𝐶
𝑧
𝑦
𝑥
𝑥 𝑦
( ) ( )
( )
( )
UICM002 & Engineering Mathematics - II
Vector Calculus 42
R.H.S
||
||
(
( )
( )) (
( )
( ))
(
( )
( ))
( ) ( ) ( ( ))
| |
| |
∬
∬
( )
L.H.S
(( ) ) ( )
( )
Here
( )
Put
∮
∫( )
∫ ( )
( )
UICM002 & Engineering Mathematics - II
Vector Calculus 43
∫ ( )
∫
∫
∫
[
]
∫
( ) (
)
( )
∮
Stoke’s theorem is verified.
Gauss Divergence theorem
Statement:
The surface integral of the normal component of a vector function over a closed
surface enclosing the volume is equal to the volume of integral of the divergence of
taken throughout the volume .
∬
∭
Example:
Verify Gauss divergence theorem for over the cube
bounded by
Answer:
∬
∭
UICM002 & Engineering Mathematics - II
Vector Calculus 44
∭
∫∫∫( )
∫∫[ ]
∫∫[ ]
∫*
+
∫(
)
(
)
∭
∬
∬ ∬ ∬ ∬ ∬ ∬
Surface Equation
( )
( )
( )
UICM002 & Engineering Mathematics - II
Vector Calculus 45
( )
( )
( )
∬
∬
∬
∬
∫∫
∫∫
∫∫
∫ ( )
∫( )
∫ ( )
( ) (
)
( )( ) ( )(
)
(
) ( ) (
)
∬
∭
Hence Gauss divergence theorem is verified.
Example:
Verify Gauss divergence theorem for ( ) ( ) ( )
over the rectangular parallelepiped .
Answer:
∬
∭
( ) ( ) ( )
UICM002 & Engineering Mathematics - II
Vector Calculus 46
( )
∭
∫∫∫ ( )
∫∫*
+
∫∫*
+
∫*
+
∫(
)
*
+
(
)
( )
∬
∬ ∬ ∬ ∬ ∬ ∬
Surface Equation
( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
UICM002 & Engineering Mathematics - II
Vector Calculus 47
∬
∫∫( )
∫(
)
∫(
)
(
)
∬
∫∫
∫(
)
∫(
)
(
)
∬
∫∫( )
∫(
)
∫(
)
(
)
∬
∫∫
∫(
)
∫(
)
(
)
∬
∫∫( )
∫(
)
∫(
)
∬
∫∫
∫(
)
∫(
)
UICM002 & Engineering Mathematics - II
Vector Calculus 48
(
)
(
)
∬
∬
( ) ∭
Hence Gauss divergence theorem is verified.
Example:
Verify Gauss divergence theorem for ( ) ( ) ( )
over the rectangular parallelepiped .
Answer:
In previous example replace , hence the theorem is verified.
Example:
Verify Gauss divergence theorem for over the cuboid
formed by the planes
Answer:
∬
∭
( )
UICM002 & Engineering Mathematics - II
Vector Calculus 49
∭
∫∫∫ ( )
∫∫*
+
∫∫*
+
∫*
+
∫(
)
*
+
(
)
( )
∬
∬ ∬ ∬ ∬ ∬ ∬
Surface Equation
( )
( )
( )
( )
( )
( )
∬
∬
∬
∬
UICM002 & Engineering Mathematics - II
Vector Calculus 50
∫∫
∫∫
∫∫
∫
∫
∫
∫
∫
∫
( ) ( )
( ) ( )
( ) ( )
∬
( ) ∭
Hence Gauss divergence theorem is verified.
Example:
Verify Gauss divergence theorem for taken over the cube
bounded by the planes
Answer:
In previous example replace , we prove the theorem.
Example:
Verify Gauss divergence theorem for over the cube
formed by the planes
Answer:
∬
∭
UICM002 & Engineering Mathematics - II
Vector Calculus 51
∭
∫ ∫ ∫( )
∫ ∫*
+
∫ ∫[ ( )]
∫ ∫
∫ (
)
∫( )
∬
∬ ∬ ∬ ∬ ∬ ∬
Surface Equation
( )
( )
( )
( )
( )
( )
∬
∬
∬
∬
∬
∬
∬
UICM002 & Engineering Mathematics - II
Vector Calculus 52
∫ ∫
∫ ∫
∫ ∫
∫ ∫
∫ ∫
∫ ∫
∫ ( )
( ( )) *
+
( ) * ( )
+
∬
∭
Hence Gauss divergence theorem is verified.
Example:
Verify Gauss divergence theorem for the vector function ( )
over the cube bounded by and .
Answer:
∬
∭
( )
∭
∫∫∫
UICM002 & Engineering Mathematics - II
Vector Calculus 53
∫∫*
+
*
+∫∫
∫
∫
( )
( )
( ) ( )
∭
∬
∬ ∬ ∬ ∬ ∬ ∬
Surface Equation
( )
( ) ( )
( )
( )
( )
( )
∬
∫∫( )
∫(
)
∫(
)
∬
∫∫
∫(
)
∫(
)
UICM002 & Engineering Mathematics - II
Vector Calculus 54
(
)
(
)
∬
∫∫
∫(
)
( )∫
( )
∬
∫∫
∫( )
( )∫
( )
∬
∫∫
∫( )
( )∫
( )
∬
(
)
∬
∭
Gauss divergence theorem is verified.
UICM002 & Engineering Mathematics - II
Vector Calculus 55
Example:
Evaluate ∬
where and is the surface bounding
the region and
Answer:
By Gauss divergence theorem,
∬
∭
( )
( )
( )
∬
∭( )
To find Limits:
to
( ) √
√ to √
Put ( )
to
𝑦
𝑥
𝑧
𝑧
𝑧
𝑦
𝑥
𝑥 𝑦
( ) ( )
( )
( )
UICM002 & Engineering Mathematics - II
Vector Calculus 56
∫ ∫ ∫( )
√
√
∫ ∫ *
+
√
√
∫ ∫ * ( ) ( ) ( )
+
√
√
∫ ∫ ( )
√
√
∫ ∫ ( )
√
√
∫ ∫
√
√
∫ ∫
√
√
∫ ∫
√
∫[ ] √
∫√
∫√
[
√
(
)]
(
)
UICM002 & Engineering Mathematics - II
Vector Calculus 57
Example:
By transferring into triple integral, evaluate ∬ ( )
where is the closed surface consisting of the cylinder and the circular
discs and
Answer:
By Gauss divergence theorem,
∬
∭
Here
( )
( )
( )
∬
∭( )
To find Limits:
to
( ) √
√ to √
𝑦
𝑥
𝑧
𝑧 𝑏
𝑧
𝑦
𝑥
𝑥 𝑦 𝑎
( 𝑎 ) (𝑎 )
( 𝑎)
( 𝑎)
UICM002 & Engineering Mathematics - II
Vector Calculus 58
Put ( )
to
∭( )
∫ ∫ ∫( )
√
√
∫ ∫ [ ]
√
√
∫ ∫
√
√
We change to polar co-ordinates ( ) so that
To find Limits:
to
to
∭( )
∫ ∫
∫ ∫
∫ *
+
∫
∫
UICM002 & Engineering Mathematics - II
Vector Calculus 59
∫
Example:
Use divergence theorem to evaluate ∬
where ,
and is the surface of the sphere
Answer:
By Gauss divergence theorem,
∬
∭
( )
( )
( )
∬
∭( )
∭( )
𝑦
𝑥
𝑧
𝑥 𝑦 𝑧 𝑎
UICM002 & Engineering Mathematics - II
Vector Calculus 60
We change to spherical polar co-ordinates ( ) so that
To find Limits:
to ; to to
∬
∫ ∫ ∫
∫ ∫ ∫
∫ ∫ *
+
∫ ∫
∫ [ ]
∫ ( )
∫ ( ( ) )
∫
[ ]
∬
UICM002 & Engineering Mathematics - II
Vector Calculus 61
Example:
Evaluate ∬
where , and is upper part of
the sphere above plane.
Answer:
The surface of the region is comprised of two surfaces
the region AB, plane
the surface ACB of the sphere above XOY plane
By Gauss divergence theorem,
∬
∬
∭
∬
∭
∬
( )
( )
( ) ∭
∬
∬
𝑦
𝑥
𝑧
𝑥 𝑦 𝑧 𝑎
0 𝐴 𝐵
𝐶
𝑦
𝑥
𝑥 𝑦 𝑎
( 𝑎 ) (𝑎 )
( 𝑎)
( 𝑎)
UICM002 & Engineering Mathematics - II
Vector Calculus 62
For the surface
| |
( )
| | | | | |
∬
∬
We change to polar co-ordinates ( ) so that
To find Limits:
to ; to
∫ ∫
∫ ∫
∫ *
+
∫
UICM002 & Engineering Mathematics - II
Vector Calculus 63
∫
∫ ( )
∫( )
(
)
(
)
(
)
∬
∬
Example:
Evaluate ∬
where , and is the surface of the
sphere in the first octant.
Answer:
UICM002 & Engineering Mathematics - II
Vector Calculus 64
The surface of the region is comprised of
four surfaces
the region OAB, plane
the region OCA, plane
the region OBC, plane
the surface ABC of the sphere in the first
octant
By Gauss divergence theorem,
∬
∬
∬
∬
∭
∬
∭
∬
∬
∬
( )
( )
( )
∭
∬
∬
∬
∬
For the surface
| |
( )
| | | | | |
For the surface
| |
( )
| | | | | |
For the surface
| |
( )
| | | | | |
𝑦
𝑥
𝑧
0
𝐴
𝐵
𝐶
UICM002 & Engineering Mathematics - II
Vector Calculus 65
We change to polar co-ordinates ( ) so that
;
To find Limits:
∫ ∫
∫ ∫
∫ *
+
∫
∫
[
]
[ ]
[ ( ) ]
∬
∬
∬
∬
∬
∬
UICM002 & Engineering Mathematics - II
Vector Calculus 66
Two Marks
1. Find ( ) where and | |
Answer:
( ) (
)
(
)
(
) [
]
( ) [ ]
[
]
[ ]
(
)
Answer:
(
) (
) (
)
(
)
(
)
(
)
(
) (
)
(
) (
) [
]
(
)(
) [ ]
(
)
UICM002 & Engineering Mathematics - II
Vector Calculus 67
Answer:
(
)
(
) (
) (
)
(
) (
) (
)
( ) (
) (
)
∑
(
) * (
)
+
∑( ( ) (
)
( ) )
∑( (
)
)
(
) (
) (
)
( )
[ ]
4. Find the unit normal to the surface ( ).
Answer:
UICM002 & Engineering Mathematics - II
Vector Calculus 68
( )
( ) ( )
| | √ √
| |
√
5. Find the unit normal vector to the surface ( ).
Answer:
( ) ( )
| | √ ( ) ( ) √
| |
√
6. Prove that is irrotational.
Answer:
||
|| ||
||
[
( )
( )] [
( )
( )] [
( )
( )]
[ ] [ ] [ ]
UICM002 & Engineering Mathematics - II
Vector Calculus 69
7. Define Solenoidal vector function. If ( ) ( ) ( ) is
Solenoidal, find the value of .
Answer:
( ) ( ) ( )
8. Find the value such that ( ) ( ) ( )
is Solenoidal.
Answer:
( ) ( ) ( )
9. Find the constants so that ( ) ( )
( ) may be Irrotational.
Answer:
UICM002 & Engineering Mathematics - II
Vector Calculus 70
||
|| ||
||
[
( )
( )] [
( )
( )]
[
( )
( )]
( ) ( ) ( )
10. Find the directional derivative of ( ) at that point ( )
in the direction of the vector .
Answer:
| |
√
√
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) (
√ )
√
√
UICM002 & Engineering Mathematics - II
Vector Calculus 71
11. Find the directional derivative of ( ) at that point ( ) in the
direction of the vector .
Answer:
| |
√
√
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) (
√ )
√
√
12. In what direction from ( ) is the directional derivative of
maximum? Find also the magnitude of this maximum.
Answer:
( ) ( ) ( )
( ) ( ( ) ) ( ( ) ) ( ( ) )
Hence the directional derivative of is maximum in the direction of
Magnitude of this maximum directional derivative | |
√ ( )
√
UICM002 & Engineering Mathematics - II
Vector Calculus 72
13. Is the position vector is Irrotational? Justify.
Answer:
||
|| ||
||
(
) (
) (
)
Yes the position vector is is Irrotational. Because it satisfies
.
14. Find if is Irrotational.
Answer:
||
|| ||
||
(
) (
) (
)
( ) ( ) ( )
15. Prove that and .
Answer:
(
) ( )
UICM002 & Engineering Mathematics - II
Vector Calculus 73
( )
||
|| [
( )
( )] [
( )
( )] [
( )
( )]
16. Prove that ( ) .
Answer:
( ) ( ) (
)
|
|
|
| ∑ [
(
)
(
)]
∑ *
+
( )
17. State Stoke’s theorem.
Answer:
If is a open surface bounded by a simple closed curve and if a vector function
is continuous and has continuous partial derivatives in and so on , then
∬
∫
where is the unit vector normal to the surface.
18. State Green’s theorem.
If ( ) ( ) are continuous functions with continuous, partial
derivatives in a region of the plane bounded by a simple closed curve then
UICM002 & Engineering Mathematics - II
Vector Calculus 74
∫
∬(
)
where is the curve described in the positive direction.
19. State Gauss divergence theorem.
The surface integral of the normal component of a vector function over a closed
surface enclosing the volume is equal to the volume of integral of the divergence of
taken throughout the volume .
∬
∭
∭
of the cube enclosed by .
Answer:
( )
∭
∫∫∫( )
∫∫*
+
∫∫[
]
∫ *
+
∫ [
]
*
+
[
]
UICM002 & Engineering Mathematics - II
Vector Calculus 75
21. Prove by Green’s theorem that the area bounded by a simple closed C curve
∫( )
Answer:
By Green’s theorem
∫
∬(
)
Given and
∫
∬( )
∬
[ ]
∫
Applications of vector calculus in various engineering fields:
Vector calculus is applied in electrical engineering especially with the use of
electromagnetics. It is also applied in fluid dynamics, as well as statics.
Vector calculus is applied in electronics and communication engineering, the
theory of radio waves and waveguides is explained in terms of equations in the form
of vector calculus. Examples are Maxwell's equations.
In medicine, determine the concentration of a medicine in a person's body over
time, taking into account how much substance and how frequently it is taken and
how fast it metabolises.
In fluid mechanics, the velocity at each point in the fluid is a vector. If the fluid is
compressible, the divergence of the velocity vector is nonzero in general. In a vortex
the curl is nonzero.
UICM002 & Engineering Mathematics - II
Vector Calculus 76
Prepared by
M. Vijaya Kumar, AP / S & H / SRIT; [email protected]; [email protected]