Module 5

Vector Calculus

5.1 Vector identities (without proof)

Vector identities:

Identity-1:  grad uv = u grad v + v grad u

Identity-2: 

Identity-3

Identity-4

Identity-5                  curl (u

Identity-6:

Identity-7:

5.2 Vector differentiation: Gradient, Curl and Divergence and their Physical interpretation

Vector function- A vector function can be defined as below-

If a vector ‘r’ is a function of a scalar variable ‘t’, then-

We define the derivative of a vector function as-

We can denote it as-

Similarly is the second order derivative of

Note- gives the velocity and gives acceleration.

Rules for differentiation-

1.

2.

3.

4.

5.

Example-1: A particle moves along the curve , here ‘t’  is the time. Find its velocity and acceleration at t = 2.

Sol. Here we have-

Then, velocity

Velocity at t = 2,

                                  =

Acceleration =

Acceleration at t = 2,

Example-2: If and then find-

1.

2.

Sol. 1. We know that-

2.

Gradient of a scalar field-

Suppose f(x, y, z) be the scalar function and it is continuously differentiable then the vector-

Is called gradient of f and we can write is as grad f.

So that-

Here is a vector which has three components

Properties of gradient-

Property-1:

Proof:

First we will take left hand side

                                               L.H.S =

                                                         =

                                                         =

                                                          =

Now taking R.H.S,

                                       R.H.S. =

                                                   =

                                                      =

Here-                     L.H.S. = R.H.S.

         Hence proved.

Property-2: Gradient of a constant (

Proof:

Suppose

                                              Then

We know that the gradient-

                                                    = 0

Property-3: Gradient of the sum and difference of two functions-

If f and g are two scalar point functions, then

Proof:   

L.H.S                                

        Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

Proof:

So that-

        Hence proved.

Property-5: Gradient of the quotient of two functions-

If f and g are two scalar point functions, then-

Proof:

(f/g) = [(i ∂/∂x + j ∂/∂y + k ∂/∂z)] (f/g)

=i ∂/∂x (f/g) + j ∂/∂y (f/g) + k ∂/∂z (f/g)

= i. g ∂f/ (f/g) + j ∂/∂y (f/g)+ k ∂/∂z (f/g)

= i. g ∂f/ – f ∂g/∂x / g2 + j . g ∂f/ – f ∂g/∂y / g2 + k . g ∂f/ – f ∂g/∂z / g2

=g [ i ∂f/∂z  + j ∂f/∂y + k ∂f/∂z ] – f [ [ i ∂g/∂x  + j ∂g/∂y + k ∂g/∂z] / g2

So that-

(f/g) = g - f / g 2

Example-1: If , then show that

1.

2.

Sol.

Suppose   and

Now taking L.H.S,

Which is

                                                                                               Hence proved.

2.

So that

Example: If then find grad f at the point (1,-2,-1).

Sol.

Now grad f at (1 , -2, -1) will be-       

Example: If then prove that grad u , grad v and grad w are coplanar.

Sol.

Here-                     

Now-

Apply

Which becomes zero.

So that we can say that grad u, grad v and grad w are coplanar vectors.

Divergence and Curl of a vector field-

Divergence (Definition)-

Suppose is a given continuous differentiable vector function then the divergence of this function can be defined as-

Curl (Definition)-

Curl of a vector function can be defined as-

Note- Irrotational vector-

If then the vector is said to be irrotational.

Example-1: Show that-

1.

2.

Sol. We know that-

2. We know that-

i[ ∂/∂y(z) - ∂/∂z(y)] + j[∂/∂z(x0 - ∂/∂x(z)] + k[∂/∂x(y) - ∂/∂y(x)]

= 0

Example-2: If then find the divergence and curl of .

Sol.  we know that-

Now-

i{∂/∂y(2yz 4) + ∂/∂z( 2x2yz} -j{∂/∂x(2yz4) - ∂/∂z(xz3)} + k{∂/∂x(-2x2yz) - ∂/∂y(xz3)

=i(2z4 +  2x2y) – j(0-3z2x) + k(-4xyz-0)

=2(x2y + z4)i + 3z2xj) – 4xyz.k

Example-3: Prove that

Note- here is a constant vector and

Sol. here    and

So that

Now-

So that-

Example: Find the curl of F(x,y,z) = 3i+2zj-xk

Curl F =

=

= i -

= (0-2)i-(-1-0)j+(0-0)k

= -2i+j

Example: What is the curl of the vector field F= ( x +y +z ,x-y-z,)?

Solution:

Curl F =

=

=

= (2y+1)i-(2x-1)j+(1-1)k

= (2y+1)i+(1-2x)j+0k

= (2y+1, 1-2x,0)

Example 7:

Find the curl of F = ()i +4zj +

Solution:

Curl F=

=

=(0-4)i-(2x-0)j+(0+1)k

=(-4)i – (2x)j+1k

=(-4,-2x,1)

5.3 Directional derivatives

Let ϕ be a scalar point function and let ϕ(P) and ϕ(Q) be the values of ϕ at two neighbouring points P and Q in the field. Then,

,are the directional derivative of ϕ in the direction of the coordinate axes at P.

The directional derivative of ϕ in the direction l, m, n=l + m+

The directional derivative of ϕ in the direction of =

Example: Find the directional derivative of 1/r in the direction where

Sol. Here

Now,

And

We know that-

So that-

Now,

            Directional derivative =

Example: Find the directional derivative of

At the points (3, 1, 2) in the direction of the vector .

Sol. Here it is given that-

Now at the point (3, 1, 2)-

Let be the unit vector in the given direction, then

at (3, 1, 2)

Now,

Example: Find the directional derivatives of at the point P(1, 1, 1) in the direction of the line

Sol. Here

Direction ratio of the line are 2, -2, 1

Now directions cosines of the line are-

Which are

Directional derivative in the direction of the line-

5.4 Vector Integration: Line integral, Surface integral, Volume integral

The Line Integral

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 3: Prove that   ͞͞͞F = [y2cos 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

                            Y2 cos 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

(y2cos x=z3) i + (2y sin x-4) j + (3xz2 + 2) k =   i +   j + k

= y2cos x + z3, = 2y sin x – 4, = 3xz2 + 2

Now, = dx + dy + dz

= (y2cos x + z3) dx +(2y sin x – 4)dy + (3xz2 + 2)dz

= (y2cos 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

  = =

Surface integrals-

An integral which we evaluate over a surface is called a surface integral.

                   Surface integral =

Volume integrals-

The volume integral is denoted by

And defined as-

If  , then

Note-

If in a conservative field 

Then this is the condition for independence of path.

Example: Evaluate , where S is the surface of the sphere in the first octant.

Sol. Here-

(y 2/2) 0 √a 2 – z 2  dz + (z 2/2) 0 √a 2 – z 2 dx + (x 2/2) 0 √a 2 – z 2 dy

= ½ (a2 – z2) dz + ½   (a 2  - x2 ) dx + ½   (a 2  - y 2 ) dy

= ½ ( a2 z 2 / 2 – z 4 / 4 ) 0 a + ½ ( a2 x2 /2 – x 4/4) a 0 + ½ ( a2y2 /2 – y4/4) 0 a

Which becomes-

Example: Evaluate , where and V is the closed reason bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.

Sol.

Here-   4x + 2y + z = 8

Put y = 0 and z = 0 in this, we get

4x = 8 or x = 2

Limit of x varies from 0 to 2 and y varies from 0 to 4 – 2x

And z varies from 0 to 8 – 4x – 2y

So that-

45 dy

45  2 y(8-4x-2y) dx dy

= 45   2 [ 4y 2- 2xy2 – 2/3 y3] 0 4-2x dx

= 45   2 [ 4(4-2x) 2 - 2x(4-2x)2 – 2/3 (4-2x)3] 0 4-2x dx

= 45/3   2 [ (4-2x) 3  dx

=15   2 [ (64-8x 3 -96x + 48 x2   dx

= 15[ 64 x3 / 3 – 8x6/6 – 96x4/4 + 48x5/5] 0 2

= 15[ 512/3 – 256/3 -384 + 1536/5]  = 128

So that-

Example: Evaluate if V is the region in the first octant bounded by and the plane x = 2 and .

Sol.

x varies from 0 to 2

The volume will be-

4xyz – 2yz + 4xz2] 0 √9-y2  dy dx

= (4xy – 2y) √9-y2   + 4x(9-y2)  dy dx

= 9(4x – 2) + 72x]dx

= [ 18 x2 – 18x + 36 x2 ] 0 2

= 180.

Example: Find the surface area for the following plane using Integration.

Solution:

z = dA

Using polar coordinates  

=

= r drd=  

=  

=  

Example: Find the volume enclosed by the hyperboloid

Solution:

Considering the plane z=2 and we have or =3 and also we no that the hyperboloid interest .In the x y plane this equation is a circle of radius .we are trying to find the volume of a drum bounded on the bottom by z= and with a top given by z=2.Therefore the volume is given in the triple integral (cylindrical coordinates).

                                                                          =6

              =6,         u=

              =6-

5.5 Gauss’s Divergence theorem, Green’s theorem, Stoke’s theorem (without proof) and their applications.

Green’s theorem in a plane

If C be a regular closed curve in the xy-plane and S is the region bounded by C then,

Where P and Q are the continuously differentiable functions inside and on C.

Green’s theorem in vector form-

Example-1: Apply Green’s theorem to evaluate where C is the boundary of the area enclosed by the x-axis and the upper half of circle

Sol. We know that by Green’s theorem-

And it it given that-

Now comparing the given integral-

P =   and   Q =

Now-

and

So that by Green’s theorem, we have the following integral-

Example-2: Evaluate by using Green’s theorem, where C is a triangle formed by

Sol. First we will draw the figure-

Here the vertices of triangle OED are (0,0), (

Now by using Green’s theorem-

Here P = y – sinx, and Q  =cosx

So that-

and

Now-

= -2/π [ x2 / 2 + x(-cosx) – 1 (-sinx)] 0 π/2

= -2/π[ ½ . π 2/4 +1]

                              =

Which is the required answer.

Example-3: Verify green’s theorem in xy-plane for where C is the boundary of the region enclosed by

Sol.

On comparing with green’s theorem,

We get-

                  P = and Q =

and

By using Green’s theorem-

  ………….. (1)

And left hand side=

= ( 2xy – x2) dx + (x2 + y2) dy] + ( 2xy – x2) dx + (x2 + y2) dy]

………….. (2)

Now,

Along

( 2xy – x2) dx + (x2 + y2) dy]  = x 3 – x 2 + 2 x 3 + 2 x 5 ) dx

x 3 – x 2 + 2 x 5  dx = [ x 4 – x 3 / 3 + x 6 /3 ] 0 1 = 1

Along

(2xy – x 2) dx + ( x2 + y2 ) dy] = ( 2x . x ½ - x 2 ) dx + ( x2 + x) dx / 2 x ½ ]

= ( 2x 3/2 - x 2 ) dx + (x2 + y2) dy] =  ( 2x 3/2  - x 2  + ½  x 3/2 + ½ x ½ ) dx

= ( 5/2 x 3/2  - x 2  +  ½ x ½ ) dx

= 5/2 [ x 5/2 / 5/2 ] 0 1 – [ x 3 / 3] 0 1   + ½ [ x 3/2 / 3/2]  0 1

= -1 + 1/3 – 1/3 = -1

Put these values in (2), we get-

               L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

Stoke’s theorem (without proofs) and their verification-

If is any continuously differentiable vector point function and S is a surface bounded by a curve C, then-

Example-1: Verify stoke’s theorem when and surface S is the part of sphere , above the xy-plane.

Sol.

We know that by stoke’s theorem,

Here C is the unit circle-

So that-

Now again on the unit circle C, z = 0

dz = 0

Suppose, 

And           

Now

  ……………… (1)

Now-

                 Curl

Using spherical polar coordinates-

. dS = sin cos ɸ + sin sin ɸ + cos ) sin d dɸ

= - sin sin ɸ + sin cos ɸ + ɸcos ) ] 0 2πsin d

= 2 π sin cos ) ]  d

= - π sin ]  d

= π/2 [cos 2] 0 π/2

= - ----------------------(2)

From equation (1) and (2), stoke’s theorem is verified.

Example-2: If and C is the boundary of the triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0), then evaluate by using Stoke’s theorem.

Sol. here we see that z-coordinates of each vertex of the triangle is zero, so that the triangle lies in the xy-plane and

Now,

              Curl

                       Curl

The equation of the line OB is y = x

Now by stoke’s theorem,

Example-3: Verify Stoke’s theorem for the given function-

Where C is the unit circle in the xy-plane.

Sol. Suppose-

Here

We know that unit circle in xy-plane-

Or

So that,

- 2 sin2 ɸ + sin ɸ cos ɸ + 3 cos 2 ɸ} dɸ

=   3 cos 2 ɸ - 2 sin  2 ɸ} dɸ

=   3 cos 2 ɸ - 2(1- cos  2 ɸ} dɸ

= 5 cos 2 ɸ - 2) dɸ

= 5 (1+cos 2 ɸ/2 - 2)] dɸ

= ½ [ ɸ + 5/2 sin 2ɸ] 0 2π

= ½ [ 2π + 0] = π

Now

 Curl

Now,

Hence the Stoke’s theorem is verified.

Gauss divergence theorem

If V is the volume bounded by a closed surface S and is a vector point function with continuous derivative-

Then it can be written as-

where unit vector to the surface S.

Example-1: Prove the following by using Gauss divergence theorem-

1.

2.

Where S is any closed surface having volume V and

Sol. Here we have by Gauss divergence theorem-

Where V is the volume enclose by the surface S.

We know that-

                                             = 3V

2.

Because 

Example – 2 Show that 

Sol

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,

 =

= =

.[

=

Reference Books-

1.E. Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, 2005.

2.Peter V. O’Neil, Advanced Engineering Mathematics, Thomson (Cengage) Learning, 2007.

3.Maurice D. Weir, Joel Hass, Frank R. Giordano, Thomas, Calculus, Eleventh Edition,Pearson.

4.D. Poole, Linear Algebra: A Modern Introduction, 2nd Edition, Brooks/Cole, 2005.

5.Veerarajan T., Engineering Mathematics for the first year, Tata McGraw-Hill, New Delhi, 2008.

6.Ray Wylie C and Louis C Barret, Advanced Engineering Mathematics, Tata Mc-Graw-Hill, Sixth Edition.

7.P. Sivaramakrishna Das and C. Vijayakumari, Engineering Mathematics, 1st Edition, PearsonIndia Education Services Pvt. Ltd

8. Advanced Engineering Mathematics. Chandrika Prasad, Reena Garg, 2018.

9. Engineering Mathematics – I. Reena Garg, 2018.