Unit – 5

Vector Integral Calculus and Applications

5.1 Line

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 cosi + b cos + 0 k

 d=(-a sin i + a cos j)d

 Circulation = =+zj+xk). d

  =-a sin i + a cos j)d

  = =

5.2 Surface and Volume integrals

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

5.3 Work-done

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 =

          = .

5.4 Green’s Lemma

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, = - x22 + 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=

5.5 Gauss’s Divergence theorem

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

Show that

Soln.  We have Gauss Divergence Theorem

 By data, F=  

  =(n+3)

2 Prove that  =

Soln. By Gauss Divergence Theorem,

 =

= =

 .[

  =

5.6 Stoke’s 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 ,

5.7 Applications to problems in Fluid Mechanics

Applications to problems in fluid mechanics are mainly in

-         STREAMLINES

-         EQUATION OF CONTINUITY

-         BERNOULLI’S EQUATION

-         EQUATION OF MOTION

5.8 Streamlines

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

5.9 Continuity equations

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

Rate of normal flux of fluid mass per unit time across

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

5.10 Bernoulli’s equation

we know that Euler’s equation of motion as

Now tell us consider following conditions:-

fluid motion is steady so that

motion is irrational giving

external force

is conservation such that

( 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

5.11 Equations of motion

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

Reference Books