Unit – 5
Vector Integral Calculus and Applications
Let- F be vector function defined throughout some region of space and let C be any curve in that region. ṝis the position vector of a point p (x,y,z) on C then the integral ƪ F .dṝ is called the line integral of F taken over
Now, since ṝ =xi+yi+zk
And if F͞ =F1i + F2 j+ F3 K
Q1. Evaluate 
where F= cos y.i-x siny j and C is the curve y=
in the xy plae from (1,0) to (0,1)
Solution : The curve y=
i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.
= 
= 
=
=-1
Q2. Evaluate 
where 
= (2xy +z2) I +x2j +3xz2 k along the curve x=t, y=t2, z= t3 from (0,0,0) to (1,1,1).
Solution : F x dr = 
Put x=t, y=t2, z= t3
Dx=dt ,dy=2tdt, dz=3t2dt.
F x dr = 
=(3t4-6t8) dti – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k
=
t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k
=
=
+
Example 4: Prove that ͞͞͞F = [y2 cos x +z3] i+(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for͞͞͞F (ii) the work done in moving an object in this field from (0, 1, -1) to (
/ 2,-1, 2)
Sol. : (a) The fleld is conservative if cur͞͞͞͞͞͞F = 0.
Now, curl͞͞͞F = 
̷̷
X 
/ 
y 
/ 
z
Y2COS X +Z3 2y sin x-4 3xz2 + 2
; Cur 
= (0-0) – (3z2 – 3z2) j + (2y cos x- 2y cos x) k = 0
; F is conservative.
(b) Since F is conservative there exists a scalar potential ȸ such that
F = ȸ
(y2 cos x=z3) i + (2y sin x-4) j + (3xz2 + 2) k = 
i + 
j + 
k
= y2 cos x + z3, 
= 2y sin x – 4, 
= 3xz2 + 2
Now, 
= 
dx + 
dy + 
dz
= (y2 cos x + z3) dx +(2y sin x – 4)dy + (3xz2 + 2)dz
= (y2 cos x dx + 2y sin x dy) +(z3dx +3xz2dz) +(- 4 dy) + (2 dz)
=d(y2 sin x + z3x – 4y -2z)
ȸ = y2 sin x +z3x – 4y -2z
(c) now, work done =
.d ͞r
= 
dx + (2y sin x – 4) dy + ( 3xz2 + 2) dz
= 
(y2 sin x + z3x – 4y + 2z) (as shown above)
= [ y2 sin x + z3x – 4y + 2z ]( 
/2, -1, 2)
= [ 1 +8 
+ 4 + 4 ] – { - 4 – 2} =4
+ 15
Sums Based on Line Integral
1. Evaluate 
where 
=yz i+zx j+xy k and C is the position of the curve.
= (a cost)i+(b sint)j+ct k , from y=0 to t=π/4.
Soln. 
= (a cost)i+(b sint)j+ct k
The parametric eqn. of the curve are x= a cost, y=b sint, z=ct (i)
=
Putting values of x,y,z from (i),
dx=-a sint
dy=b cost
dz=c dt
=
=
=
= 
2. Find the circulation of 
around the curve C where 
=yi+zj+xk and C is circle 
.
Soln. Parametric eqn of circle are:
x=a cos
y=a sin
z=0
=xi+yj+zk = a cos
i + b cos
+ 0 k
d
=(-a sin
i + a cos
j)d
Circulation =
=
+zj+xk). d
=
-a sin
i + a cos
j)d
=
= 
The surface integral of a vector point function 
over a surface S is defined as the integral of the normal component of 
taken over the surface S.
Consider surface S as shown in fig.
Let 
act at p enclosed by an element of area 
is a unit vector normal to the surface at P. normal component of 
is given by 
The surface integral can be expressed as
or 
If we write 
, then the above integral is also be written as
Q1. Find the work done in moving a particle once round the complete circle x2+y2=a2 , z=0 in the force filed given by 
= sin yi +(x + x cos y )j.
Solution: work done= 
= 
(x+x cos y )dy
Using parametric equation x=a cos θ, y=a sinθ
Work done =
= 
.
If P and Q are two functions of x, y and their partial derivatives 
,
are continuous single valued functions over the closed region bounded by a curve C then
} dx dy.
EXAMPLE – 1 Verify green’s theorem for 
and C is the triangle having verticles A (0,2 ) , B (2,0 ) , C (4,2 ).
SOLUTION : By green theorem.
Here ,
(a) Along AB , since the equation of AB is
Putting 
Along BC , since the equation of BC , 
.
Along CA , since the equation of CA, is y = 2 , dy = 0.
(b) 
.
From (1) and (2) , the theorem is verified .
Example 2 :Evaluate by Green ‘s theorem 
= - xy (xi –yj) and c is r= a (1+ cos 
)
Sol : By Green’s Theorem , 
) dx dy
Now, 
.dṝ = 
2yi + xy2 j ) . (d xi + dy j) = 
2y dx + xy2dy )
By comparison p= - x2y, Q = xy2
2, 
= - x2
2 + x2) dx dy
To evaluate the integral , we put x = r cos 
, y = r sin 
for the cardioid r = a ( 1 + cos 
), we take the integral from 
2 .rdr d
= 2 
3dr d
= 
]a ( 1 + cos 
) dθ = 
(1+ cos θ)4dθ
=8a4
= 
a4
Example Based on Green’s Theorem
1. Verify Green’s Theorem in the plane for 
where C is the closed curve of region bounded by y=x and y=
.
Soln. By Green’s Theorem,
A(1,1)
y=x
y=
B
(a) Along 
, y=
and dy=2x dx and x varies from 0 to 1.
Along 
y=x and dy=dx and x varies from 1 to 0.
=
=
= 1
= L.H.S.
RHS=
The divergence theorem states that the surface integral of the normal component of a vector point function “F” over a closed surface “S” is equal to the volume integral of the divergence of F⃗ taken over the volume “V” enclosed by the surface S. Thus, the divergence theorem is symbolically denoted as:
∬v∫▽F⃗ .dV=∬sF⃗ .n⃗ .dS
Sums on surface and volume
EXAMPLE – 1 Use divergence theorem to show that 
where Sisanyclosed surface enclosing a volume V.
SOLUTION: By divergence theorem
Here ,
= 6V
EXAMPLE – 2Show that 
SOLUTION: By divergence theorem, 
..…(1)
Comparing this with the given problem let
Hence, by (1)
………….(2)
Now ,
Hence,from (2),Weget,
Example Based on Gauss Divergence Theorem
Soln. We have Gauss Divergence Theorem
By data, F=
=(n+3)
2 Prove that 
=
Soln. By Gauss Divergence Theorem,
=
=
=
.[
= 
The integral of the normal component of the curl of a vector F͞ over a surface S is equal to the line integral of the tangential component of F͞ around the curve bounding S i.e
F͞ )ds =
Q1.Use stoke’s theorem to evaluate 
SOLUTION : We have by stoke’s theorem 
Now ,
Applications to problems in fluid mechanics are mainly in
- STREAMLINES
- EQUATION OF CONTINUITY
- BERNOULLI’S EQUATION
- EQUATION OF MOTION
applications of vectors and fluid mecahnies
5.8 streamlines
Definition:-an imaginary curve drawn in the fluid such that at any instant of time, the tangent at any point of it is along the velocity vector at the point is called stream line.
Stream line indicates the dim of motion at each as shown in fig. from the definition it is clear that these can be no flaw area a stream line
To find the equation of stream line , consider two points 
on the stream line .
Represent unit tangent vector F at P .
Let 
be velocity vector at P.
From the definition ,
(k is some constant)
Equating 
component , we get
represents the diffential equation of stream line
question 1 :- velocity distribution for a fluid flow is given by
find the equation of
Stream line passing through the point (1,1,2)
Ans:- the diffential equation of stream line are given by
Putting for u,v,w we get
These are two independent equations, consider
Integrating 
-------1
As stream the passes through (1,1,2)
Put x=1 , y=1 in eq. 1
Consider now, 
Integrating gives , 
Putting 
we get 
Equation 2 &4 together represent curve of intersection of surface
Which represents the stream line
Question 2:- if the velocity of an incompressible fluid at (x,y,z) is given by
Where 
Then determine the stream line at motion
Answer:- equation of stream lines are given by
Similarly
Integrating , 
Eq 1 & 2 together represents stream line
Equation of continuity is the expression of the law of constellation of energy
Let S be the closed surface drawn in the fluid & taken fixed in a space. suppose it encloses a volume v at the fluid.
Let 
be the fluid density.
Let 
be the surface element & 
be the unit outward drawn normal to 
.
is the fluid velocity at 
integrating 
Equation 1 & 2 together represents stream line
Using Gauss divergence theorem
Total rate of mass flow= 
At time t , the mass of fluid with the element is 
Rate of increase of mass with V =
For conservation we have
Valid for all volume V,
Which is called equation of continuity in cartesion form,
Equation of continuity
we know that Euler’s equation of motion as
Now tell us consider following conditions:-
( v is for potential)
Using vector identify results , we can write
But 
( motion is imtational)
Euler’s equation of motion now takes the form
Taking dot product with 
we get
But dv=
Integrating above equation
Eq. is called Bernoulli’s equation
Que :-1 given the velocity fields
What is the acceleration of a particle at (3,0,2) at time t=1
Ans :- we have u=
, V = 
, W =25
The component is given by
Substituting x=3, y=0 & z=2 &t=1
Question 2:- a liquid is in equilibrium under the action of field 
per unit mass given by
Find the pressure at any point on the field
Answer :- let v be the force potential
From the Bernoullis eq
Which gives pressure at any point
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