Unit - 2

Vector Calculus

2.1 Scalar and Vector point functions

Vector function-

Suppose be a function of a scalar variable t, then-

Here vector varies corresponding to the variation of a scalar variable t that its length and direction be known as value of t is given.

Any vector  can be expressed as-

Here , ,    are the scalar functions of t.

Differentiation of a vector-

Note-

1. Velocity =  

2. Acceleration =        

Example: A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

Sol. Suppose

Now,

At t = 0	4
At t = π/2	- 4
At t = 0	|v|=
At t = π/2	|v|=

Again acceleration-

Now-

At t = 0
At t = π/2
At t = 0	|a|= (- 4 )2 = 4
At t = π/2	|a|=(- 4 )2 = 4

Scalar point function-

If for each point P of a region R, there corresponds a scalar denoted by f(P), in that case f is called scalar point function of the region R.

Note-

Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).

Vector point function-

If for each point P of a region R, then there corresponds a vector then is called a vector point function for the region R.

Vector field-

Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).

Del operator-

The del operated is defined as-

∇ =

Example: show that    where

Sol. Here it is given-

Therefore-

Note-

Hence proved.

2.2 Directional derivative

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,

δ/δx, δ/δy, δ/δz 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 δ/δx+ m δ/δy+ nδ/δz

The directional derivative of ϕ in the direction of            

2.3 Gradient

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-

Grad f = =

Here    is a vector which has three components f/x, f/y, f/z

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 (𝛁∅ = 0)

Proof:

Suppose ∅(x, y, z) = c

Then ∅/x = ∅/y = ∅/z = 0

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

∇(f ± g) = ∇f ± ∇g

Proof:   

∇(f ± g) =

L.H.S                             

=    

=    

=    

∇(f ± g) = ∇f ± ∇g

Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

∇(fg) = f∇g + g∇f

Proof:

∇(fg) =

So that-

∇(fg) = f∇g + g∇f

Hence proved.

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

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

∇(f/g) =

Proof:

∇(f/g) =

So that-

∇(f/g) =

Example-1: If         , then show that

1. ∇(

2. Grad r =

Sol.

Suppose
          

Now taking L.H.S,

∇(

Which is

Hence proved.

2. Grad r = r =

So that

Grad r =

Example: If f = 3x2y – y3z2 then find grad f at the point (1,-2,-1).

Sol.

Grad f = (3x2y – y3z2)

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

= - 12

Example: If u = x+ y+ z, v = x2 + y2 + z2 and w = yz + zx + xy then prove that grad u , grad v and grad w are coplanar.

Sol.

Here-        grad u =        

Grad v =         

Grad w =

Now-

Grad u (grad v  grad w) =

Apply R2  R2 + R3

Which becomes zero.

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

2.4 Divergence and curl

Divergence (Definition)-

Suppose (x, y, z) 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-

Curl

Note- Irrotational vector-

If curl then the vector is said to be irrotational.

Vector identities:

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

Proof: 

Grad (uv) = ∇ (uv)

So that

Grad uv = u grad v + v grad u

Identity-2: grad( .

Proof:

Grad(

Interchanging , we get-

We get by using above equations-

Grad(

Identity-3   div (u = u div  

Proof:  div = ∇.(u

So that-

Div(u

Identity-4  div ( 

Proof:

Div (

=

=

=

=

= ( curl ) . - (curl ).

So that,

Div (

Identity-5 

Proof:

=

=

=

= (grad u)  + u curl

So that

Identity-6: div curl = 0

Proof:

Div curl

So that-

Div curl

Identity-7: div grad f = ∇ . (∇f) =

Proof:

Div grad f = ∇ . (∇f)

So that-

Divgrad f = ∇2 f

Example-1: Show that-

1. Div

2. Curl

Sol. We know that-

Div

                                                  =  1 + 1 + 1 = 3

2. We know that

Curl

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

Sol.  we know that-

Div =

Now-

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

= 2 (x2y + z4) I + 3z2xj – 4 xyz.k

Example-3: Prove that

Note- here is a constant vector and                  

Sol. Here

So that

∇ 

Now-

So that-

2.5 Line integrals

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 ƪF .dṝ is called the line integral of F taken over

Now, since ṝ =xi+yi+zk        d    

And if F͞ =F1i + F2 j+ F3  K

Q1. Evaluate where F= cos y.i-x siny j and C is the curve y= 1 – x2 in the xyplae from (1,0) to (0,1)

Solution: The curve y=1 – x2 i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

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) dt i  – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

= t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

= [(3/5 -2/3)I + (1/1 + 1/3 – 3/8)j + (4/5 + 1/4 – 1/3)k]                     

= - 1/15 i + 23/24 j + 43/60 k

Example 3: 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 (π/ 2,-1, 2)

Sol.: (a) The field is conservative if cur͞͞͞͞͞͞F = 0.

Now, curl͞͞͞F = |

Y2COS X +Z3        2y sin x-4     3xz2 + 2

Cur F = (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 = Φ/x  i +  Φ/y  j +  Φ/z k

 Φ/x = y2 cos x + z3,  Φ/y  = 2y sin x – 4,  Φ/z = 3xz2 + 2

Now, Φ = Φ/x  dx +  Φ/y  dy +  Φ/z 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 =          

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

=   (y2 sin x + z3x – 4y + 2z)    (as shown above)

= [ y2 sin x + z3x – 4y + 2z ](π/2, -1, 2)

= [ 1 +8 π/2+ 4 + 4 ] – { - 4 – 2} =4π + 15 

Sums Based on Line Integral

1. Evaluate where =yzi+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 x2 + y2 = a2.

Soln. Parametric eqn of circle are:

x=a cos θ

y=a sin θ

z=0

=xi+yj+zk = a cos θ i + b cos θj + 0 k

d=(-a sinθi + a cosθ j)dθ

Circulation = +zj+xk). d

= -a sinθi + a cosθj)dθ

=           

2.6 Surface integrals, Volume integrals

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 x2 + y2 + z2 = a2 in the first octant.

Sol. Here-

Which becomes-

Example: Evaluate  , where ∅ = 45 x2y 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-

= 128

So that-

Example: Evaluate if V is the region in the first octant bounded by y2 + z2 = 9 and the plane x = 2 and .

Sol.

x varies from 0 to 2

The volume will be-

                                                    = 180

2.7 Green's theorem, Stokes theorem and Gauss divergence theorem (without proofs)

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 x2 + y2 = a2.

Sol. We know that by Green’s theorem-

And it it given that-

Now comparing the given integral-

P = 2x2 – y2  and   Q = x2 + y2

Now-

Q/x = 2x     and P/x = -2y

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 y = 0, x = π/2, y = 2x/π

Sol. First we will draw the figure-

Here the vertices of triangle OED are (0,0), (π/2 , 0) and (π/2, π/2)

Now by using Green’s theorem-

Here P = y – sinx, and Q  =cosx

So that-

Q/x = - sin x   and P/x = 1

Now-

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 y = x2 and y2 = x

Sol.

On comparing with green’s theorem,

We get-

P = 2xy – x2 and Q = x2 + y2

Q/x = 2x                  and           P/x = 2x

By using Green’s theorem-

And left hand side =                           

Now,

Along C1 : y = x2 that means dy = 2x dx where x varies from 0 to 1

Along C2 : x2 = x that means 2ydy = dx   or   dy = dx/2x1/2 where x varies from 0 to

1

Put these values in (2), we get-

L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

Gauss’s 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 n 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 r2 = x2 + v2 + z2  

Sol. Here we have by Gauss divergence theorem-

Where V is the volume enclose by the surface S.

We know that-

Div r = 3

= 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), We get,

Example Based on Gauss Divergence Theorem

1. Show that

Soln.  We have Gauss Divergence Theorem

By data, F = rnr ∇.F = ∇.(rn. r)

                                        = nrn – 2 ( n2 + y2 + z2) + 3rn

                                      = nrn – 2r2 + 3rn

                                      ∇. rn. r = nrn + 3rn

=(n+3)rn

2. Prove that              

Soln. By Gauss Divergence Theorem,

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 x2 + y2 + z2 = 1 , above the xy-plane.

Sol.

We know that by stoke’s theorem,

Here C is the unit circle-x2 + y2 = 1, z = 0

So that-

Now again on the unit circle C, z = 0

Dz = 0

Suppose, x = cos ∅ so that dx = - sin ∅. d∅             

And     y = sin ∅ so that dy = cos ∅. d∅      

Now

  ……………… (1)

Now-

Using spherical polar coordinates-

= - π  ………………… (2)

From equation (1) and (2), stokes 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,

The equation of the line OB is y = x

Now by stokes theorem,

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

= (x+2y) dx + (y + 3x) dy

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

Sol. Suppose-

                                                      = F1dx + F2 dy + F3 dz

Here F1 = x+2y, F2 = y+ 3x, F3 = 0

We know that unit circle in xy-plane-

Or

x = cos ∅,                dx = - sin ∅ d∅

y = sin ∅,               dy = cos ∅ d∅

So that,

                                                      = ½ [ 2π + 0] = π

Now

Now,

Hence the Stokes theorem is verified.

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

Unit - 2

Vector Calculus

2.1 Scalar and Vector point functions

Vector function-

Suppose be a function of a scalar variable t, then-

Here vector varies corresponding to the variation of a scalar variable t that its length and direction be known as value of t is given.

Any vector  can be expressed as-

Here , ,    are the scalar functions of t.

Differentiation of a vector-

Note-

1. Velocity =  

2. Acceleration =        

Example: A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

Sol. Suppose

Now,

At t = 0	4
At t = π/2	- 4
At t = 0	|v|=
At t = π/2	|v|=

Again acceleration-

Now-

At t = 0
At t = π/2
At t = 0	|a|= (- 4 )2 = 4
At t = π/2	|a|=(- 4 )2 = 4

Scalar point function-

If for each point P of a region R, there corresponds a scalar denoted by f(P), in that case f is called scalar point function of the region R.

Note-

Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).

Vector point function-

If for each point P of a region R, then there corresponds a vector then is called a vector point function for the region R.

Vector field-

Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).

Del operator-

The del operated is defined as-

∇ =

Example: show that    where

Sol. Here it is given-

Therefore-

Note-

Hence proved.

2.2 Directional derivative

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,

δ/δx, δ/δy, δ/δz 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 δ/δx+ m δ/δy+ nδ/δz

The directional derivative of ϕ in the direction of            

2.3 Gradient

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-

Grad f = =

Here    is a vector which has three components f/x, f/y, f/z

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 (𝛁∅ = 0)

Proof:

Suppose ∅(x, y, z) = c

Then ∅/x = ∅/y = ∅/z = 0

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

∇(f ± g) = ∇f ± ∇g

Proof:   

∇(f ± g) =

L.H.S                             

=    

=    

=    

∇(f ± g) = ∇f ± ∇g

Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

∇(fg) = f∇g + g∇f

Proof:

∇(fg) =

So that-

∇(fg) = f∇g + g∇f

Hence proved.

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

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

∇(f/g) =

Proof:

∇(f/g) =

So that-

∇(f/g) =

Example-1: If         , then show that

1. ∇(

2. Grad r =

Sol.

Suppose
          

Now taking L.H.S,

∇(

Which is

Hence proved.

2. Grad r = r =

So that

Grad r =

Example: If f = 3x2y – y3z2 then find grad f at the point (1,-2,-1).

Sol.

Grad f = (3x2y – y3z2)

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

= - 12

Example: If u = x+ y+ z, v = x2 + y2 + z2 and w = yz + zx + xy then prove that grad u , grad v and grad w are coplanar.

Sol.

Here-        grad u =        

Grad v =         

Grad w =

Now-

Grad u (grad v  grad w) =

Apply R2  R2 + R3

Which becomes zero.

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

2.4 Divergence and curl

Divergence (Definition)-

Suppose (x, y, z) 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-

Curl

Note- Irrotational vector-

If curl then the vector is said to be irrotational.

Vector identities:

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

Proof: 

Grad (uv) = ∇ (uv)

So that

Grad uv = u grad v + v grad u

Identity-2: grad( .

Proof:

Grad(

Interchanging , we get-

We get by using above equations-

Grad(

Identity-3   div (u = u div  

Proof:  div = ∇.(u

So that-

Div(u

Identity-4  div ( 

Proof:

Div (

=

=

=

=

= ( curl ) . - (curl ).

So that,

Div (

Identity-5 

Proof:

=

=

=

= (grad u)  + u curl

So that

Identity-6: div curl = 0

Proof:

Div curl

So that-

Div curl

Identity-7: div grad f = ∇ . (∇f) =

Proof:

Div grad f = ∇ . (∇f)

So that-

Divgrad f = ∇2 f

Example-1: Show that-

1. Div

2. Curl

Sol. We know that-

Div

                                                  =  1 + 1 + 1 = 3

2. We know that

Curl

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

Sol.  we know that-

Div =

Now-

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

= 2 (x2y + z4) I + 3z2xj – 4 xyz.k

Example-3: Prove that

Note- here is a constant vector and                  

Sol. Here

So that

∇ 

Now-

So that-

2.5 Line integrals

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 ƪF .dṝ is called the line integral of F taken over

Now, since ṝ =xi+yi+zk        d    

And if F͞ =F1i + F2 j+ F3  K

Q1. Evaluate where F= cos y.i-x siny j and C is the curve y= 1 – x2 in the xyplae from (1,0) to (0,1)

Solution: The curve y=1 – x2 i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

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) dt i  – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

= t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

= [(3/5 -2/3)I + (1/1 + 1/3 – 3/8)j + (4/5 + 1/4 – 1/3)k]                     

= - 1/15 i + 23/24 j + 43/60 k

Example 3: 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 (π/ 2,-1, 2)

Sol.: (a) The field is conservative if cur͞͞͞͞͞͞F = 0.

Now, curl͞͞͞F = |

Y2COS X +Z3        2y sin x-4     3xz2 + 2

Cur F = (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 = Φ/x  i +  Φ/y  j +  Φ/z k

 Φ/x = y2 cos x + z3,  Φ/y  = 2y sin x – 4,  Φ/z = 3xz2 + 2

Now, Φ = Φ/x  dx +  Φ/y  dy +  Φ/z 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 =          

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

=   (y2 sin x + z3x – 4y + 2z)    (as shown above)

= [ y2 sin x + z3x – 4y + 2z ](π/2, -1, 2)

= [ 1 +8 π/2+ 4 + 4 ] – { - 4 – 2} =4π + 15 

Sums Based on Line Integral

1. Evaluate where =yzi+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 x2 + y2 = a2.

Soln. Parametric eqn of circle are:

x=a cos θ

y=a sin θ

z=0

=xi+yj+zk = a cos θ i + b cos θj + 0 k

d=(-a sinθi + a cosθ j)dθ

Circulation = +zj+xk). d

= -a sinθi + a cosθj)dθ

=           

2.6 Surface integrals, Volume integrals

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 x2 + y2 + z2 = a2 in the first octant.

Sol. Here-

Which becomes-

Example: Evaluate  , where ∅ = 45 x2y 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-

= 128

So that-

Example: Evaluate if V is the region in the first octant bounded by y2 + z2 = 9 and the plane x = 2 and .

Sol.

x varies from 0 to 2

The volume will be-

                                                    = 180

2.7 Green's theorem, Stokes theorem and Gauss divergence theorem (without proofs)

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 x2 + y2 = a2.

Sol. We know that by Green’s theorem-

And it it given that-

Now comparing the given integral-

P = 2x2 – y2  and   Q = x2 + y2

Now-

Q/x = 2x     and P/x = -2y

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 y = 0, x = π/2, y = 2x/π

Sol. First we will draw the figure-

Here the vertices of triangle OED are (0,0), (π/2 , 0) and (π/2, π/2)

Now by using Green’s theorem-

Here P = y – sinx, and Q  =cosx

So that-

Q/x = - sin x   and P/x = 1

Now-

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 y = x2 and y2 = x

Sol.

On comparing with green’s theorem,

We get-

P = 2xy – x2 and Q = x2 + y2

Q/x = 2x                  and           P/x = 2x

By using Green’s theorem-

And left hand side =                           

Now,

Along C1 : y = x2 that means dy = 2x dx where x varies from 0 to 1

Along C2 : x2 = x that means 2ydy = dx   or   dy = dx/2x1/2 where x varies from 0 to

1

Put these values in (2), we get-

L.H.S. = 1 – 1 = 0

So that the Green’s theorem is verified.

Gauss’s 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 n 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 r2 = x2 + v2 + z2  

Sol. Here we have by Gauss divergence theorem-

Where V is the volume enclose by the surface S.

We know that-

Div r = 3

= 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), We get,

Example Based on Gauss Divergence Theorem

1. Show that

Soln.  We have Gauss Divergence Theorem

By data, F = rnr ∇.F = ∇.(rn. r)

                                        = nrn – 2 ( n2 + y2 + z2) + 3rn

                                      = nrn – 2r2 + 3rn

                                      ∇. rn. r = nrn + 3rn

=(n+3)rn

2. Prove that              

Soln. By Gauss Divergence Theorem,

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 x2 + y2 + z2 = 1 , above the xy-plane.

Sol.

We know that by stoke’s theorem,

Here C is the unit circle-x2 + y2 = 1, z = 0

So that-

Now again on the unit circle C, z = 0

Dz = 0

Suppose, x = cos ∅ so that dx = - sin ∅. d∅             

And     y = sin ∅ so that dy = cos ∅. d∅      

Now

  ……………… (1)

Now-

Using spherical polar coordinates-

= - π  ………………… (2)

From equation (1) and (2), stokes 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,

The equation of the line OB is y = x

Now by stokes theorem,

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

= (x+2y) dx + (y + 3x) dy

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

Sol. Suppose-

                                                      = F1dx + F2 dy + F3 dz

Here F1 = x+2y, F2 = y+ 3x, F3 = 0

We know that unit circle in xy-plane-

Or

x = cos ∅,                dx = - sin ∅ d∅

y = sin ∅,               dy = cos ∅ d∅

So that,

                                                      = ½ [ 2π + 0] = π

Now

Now,

Hence the Stokes theorem is verified.

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

Unit - 2

Vector Calculus

Unit - 2

Vector Calculus

2.1 Scalar and Vector point functions

Vector function-

Suppose be a function of a scalar variable t, then-

Here vector varies corresponding to the variation of a scalar variable t that its length and direction be known as value of t is given.

Any vector  can be expressed as-

Here , ,    are the scalar functions of t.

Differentiation of a vector-

Note-

1. Velocity =  

2. Acceleration =        

Example: A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

Sol. Suppose

Now,

At t = 0	4
At t = π/2	- 4
At t = 0	|v|=
At t = π/2	|v|=

Again acceleration-

Now-

At t = 0
At t = π/2
At t = 0	|a|= (- 4 )2 = 4
At t = π/2	|a|=(- 4 )2 = 4

Scalar point function-

If for each point P of a region R, there corresponds a scalar denoted by f(P), in that case f is called scalar point function of the region R.

Note-

Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).

Vector point function-

If for each point P of a region R, then there corresponds a vector then is called a vector point function for the region R.

Vector field-

Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).

Del operator-

The del operated is defined as-

∇ =

Example: show that    where

Sol. Here it is given-

Therefore-

Note-

Hence proved.

2.2 Directional derivative

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,

δ/δx, δ/δy, δ/δz 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 δ/δx+ m δ/δy+ nδ/δz

The directional derivative of ϕ in the direction of            

2.3 Gradient

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-

Grad f = =

Here    is a vector which has three components f/x, f/y, f/z

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 (𝛁∅ = 0)

Proof:

Suppose ∅(x, y, z) = c

Then ∅/x = ∅/y = ∅/z = 0

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

∇(f ± g) = ∇f ± ∇g

Proof:   

∇(f ± g) =

L.H.S                             

=    

=    

=    

∇(f ± g) = ∇f ± ∇g

Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

∇(fg) = f∇g + g∇f

Proof:

∇(fg) =

So that-

∇(fg) = f∇g + g∇f

Hence proved.

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

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

∇(f/g) =

Proof:

∇(f/g) =

So that-

∇(f/g) =

Example-1: If         , then show that

1. ∇(

2. Grad r =

Sol.

Suppose
          

Now taking L.H.S,

∇(

Which is

Hence proved.

2. Grad r = r =

So that

Grad r =

Example: If f = 3x2y – y3z2 then find grad f at the point (1,-2,-1).

Sol.

Grad f = (3x2y – y3z2)

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

= - 12

Example: If u = x+ y+ z, v = x2 + y2 + z2 and w = yz + zx + xy then prove that grad u , grad v and grad w are coplanar.

Sol.

Here-        grad u =        

Grad v =         

Grad w =

Now-

Grad u (grad v  grad w) =

Apply R2  R2 + R3

Which becomes zero.

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

2.4 Divergence and curl

Divergence (Definition)-

Suppose (x, y, z) 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-

Curl

Note- Irrotational vector-

If curl then the vector is said to be irrotational.

Vector identities:

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

Proof: 

Grad (uv) = ∇ (uv)

So that

Grad uv = u grad v + v grad u

Identity-2: grad( .

Proof:

Grad(

Interchanging , we get-

We get by using above equations-

Grad(

Identity-3   div (u = u div  

Proof:  div = ∇.(u

So that-

Div(u

Identity-4  div ( 

Proof:

Div (

=

=

=

=

= ( curl ) . - (curl ).

So that,

Div (

Identity-5 

Proof:

=

=

=

= (grad u)  + u curl

So that

Identity-6: div curl = 0

Proof:

Div curl

So that-

Div curl

Identity-7: div grad f = ∇ . (∇f) =

Proof:

Div grad f = ∇ . (∇f)

So that-

Divgrad f = ∇2 f

Example-1: Show that-

1. Div

2. Curl

Sol. We know that-

Div

                                                  =  1 + 1 + 1 = 3

2. We know that

Curl

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

Sol.  we know that-

Div =

Now-

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

= 2 (x2y + z4) I + 3z2xj – 4 xyz.k

Example-3: Prove that

Note- here is a constant vector and                  

Sol. Here

So that

∇ 

Now-

So that-

2.5 Line integrals

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 ƪF .dṝ is called the line integral of F taken over

Now, since ṝ =xi+yi+zk        d    

And if F͞ =F1i + F2 j+ F3  K

Q1. Evaluate where F= cos y.i-x siny j and C is the curve y= 1 – x2 in the xyplae from (1,0) to (0,1)

Solution: The curve y=1 – x2 i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

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) dt i  – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

= t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

= [(3/5 -2/3)I + (1/1 + 1/3 – 3/8)j + (4/5 + 1/4 – 1/3)k]                     

= - 1/15 i + 23/24 j + 43/60 k

Example 3: 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 (π/ 2,-1, 2)

Sol.: (a) The field is conservative if cur͞͞͞͞͞͞F = 0.

Now, curl͞͞͞F = |

Y2COS X +Z3        2y sin x-4     3xz2 + 2

Cur F = (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 = Φ/x  i +  Φ/y  j +  Φ/z k

 Φ/x = y2 cos x + z3,  Φ/y  = 2y sin x – 4,  Φ/z = 3xz2 + 2

Now, Φ = Φ/x  dx +  Φ/y  dy +  Φ/z 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 =          

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

=   (y2 sin x + z3x – 4y + 2z)    (as shown above)

= [ y2 sin x + z3x – 4y + 2z ](π/2, -1, 2)

= [ 1 +8 π/2+ 4 + 4 ] – { - 4 – 2} =4π + 15 

Sums Based on Line Integral

1. Evaluate where =yzi+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 x2 + y2 = a2.

Soln. Parametric eqn of circle are:

x=a cos θ

y=a sin θ

z=0

=xi+yj+zk = a cos θ i + b cos θj + 0 k

d=(-a sinθi + a cosθ j)dθ

Circulation = +zj+xk). d

= -a sinθi + a cosθj)dθ

=           

2.6 Surface integrals, Volume integrals

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 x2 + y2 + z2 = a2 in the first octant.

Sol. Here-

Which becomes-

Example: Evaluate  , where ∅ = 45 x2y 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-

= 128

So that-

Example: Evaluate if V is the region in the first octant bounded by y2 + z2 = 9 and the plane x = 2 and .

Sol.

x varies from 0 to 2

The volume will be-

                                                    = 180

Unit - 2

Vector Calculus

2.1 Scalar and Vector point functions

Vector function-

Suppose be a function of a scalar variable t, then-

Here vector varies corresponding to the variation of a scalar variable t that its length and direction be known as value of t is given.

Any vector  can be expressed as-

Here , ,    are the scalar functions of t.

Differentiation of a vector-

Note-

1. Velocity =  

2. Acceleration =        

Example: A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

Sol. Suppose

Now,

At t = 0	4
At t = π/2	- 4
At t = 0	|v|=
At t = π/2	|v|=

Again acceleration-

Now-

At t = 0
At t = π/2
At t = 0	|a|= (- 4 )2 = 4
At t = π/2	|a|=(- 4 )2 = 4

Scalar point function-

If for each point P of a region R, there corresponds a scalar denoted by f(P), in that case f is called scalar point function of the region R.

Note-

Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).

Vector point function-

If for each point P of a region R, then there corresponds a vector then is called a vector point function for the region R.

Vector field-

Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).

Del operator-

The del operated is defined as-

∇ =

Example: show that    where

Sol. Here it is given-

Therefore-

Note-

Hence proved.

2.2 Directional derivative

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,

δ/δx, δ/δy, δ/δz 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 δ/δx+ m δ/δy+ nδ/δz

The directional derivative of ϕ in the direction of            

2.3 Gradient

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-

Grad f = =

Here    is a vector which has three components f/x, f/y, f/z

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 (𝛁∅ = 0)

Proof:

Suppose ∅(x, y, z) = c

Then ∅/x = ∅/y = ∅/z = 0

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

∇(f ± g) = ∇f ± ∇g

Proof:   

∇(f ± g) =

L.H.S                             

=    

=    

=    

∇(f ± g) = ∇f ± ∇g

Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

∇(fg) = f∇g + g∇f

Proof:

∇(fg) =

So that-

∇(fg) = f∇g + g∇f

Hence proved.

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

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

∇(f/g) =

Proof:

∇(f/g) =

So that-

∇(f/g) =

Example-1: If         , then show that

1. ∇(

2. Grad r =

Sol.

Suppose
          

Now taking L.H.S,

∇(

Which is

Hence proved.

2. Grad r = r =

So that

Grad r =

Example: If f = 3x2y – y3z2 then find grad f at the point (1,-2,-1).

Sol.

Grad f = (3x2y – y3z2)

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

= - 12

Example: If u = x+ y+ z, v = x2 + y2 + z2 and w = yz + zx + xy then prove that grad u , grad v and grad w are coplanar.

Sol.

Here-        grad u =        

Grad v =         

Grad w =

Now-

Grad u (grad v  grad w) =

Apply R2  R2 + R3

Which becomes zero.

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

2.4 Divergence and curl

Divergence (Definition)-

Suppose (x, y, z) 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-

Curl

Note- Irrotational vector-

If curl then the vector is said to be irrotational.

Vector identities:

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

Proof: 

Grad (uv) = ∇ (uv)

So that

Grad uv = u grad v + v grad u

Identity-2: grad( .

Proof:

Grad(

Interchanging , we get-

We get by using above equations-

Grad(

Identity-3   div (u = u div  

Proof:  div = ∇.(u

So that-

Div(u

Identity-4  div ( 

Proof:

Div (

=

=

=

=

= ( curl ) . - (curl ).

So that,

Div (

Identity-5 

Proof:

=

=

=

= (grad u)  + u curl

So that

Identity-6: div curl = 0

Proof:

Div curl

So that-

Div curl

Identity-7: div grad f = ∇ . (∇f) =

Proof:

Div grad f = ∇ . (∇f)

So that-

Divgrad f = ∇2 f

Example-1: Show that-

1. Div

2. Curl

Sol. We know that-

Div

                                                  =  1 + 1 + 1 = 3

2. We know that

Curl

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

Sol.  we know that-

Div =

Now-

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

= 2 (x2y + z4) I + 3z2xj – 4 xyz.k

Example-3: Prove that

Note- here is a constant vector and                  

Sol. Here

So that

∇ 

Now-

So that-

2.5 Line integrals

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 ƪF .dṝ is called the line integral of F taken over

Now, since ṝ =xi+yi+zk        d    

And if F͞ =F1i + F2 j+ F3  K

Q1. Evaluate where F= cos y.i-x siny j and C is the curve y= 1 – x2 in the xyplae from (1,0) to (0,1)

Solution: The curve y=1 – x2 i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

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) dt i  – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

= t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

= [(3/5 -2/3)I + (1/1 + 1/3 – 3/8)j + (4/5 + 1/4 – 1/3)k]                     

= - 1/15 i + 23/24 j + 43/60 k

Example 3: 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 (π/ 2,-1, 2)

Sol.: (a) The field is conservative if cur͞͞͞͞͞͞F = 0.

Now, curl͞͞͞F = |

Y2COS X +Z3        2y sin x-4     3xz2 + 2

Cur F = (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 = Φ/x  i +  Φ/y  j +  Φ/z k

 Φ/x = y2 cos x + z3,  Φ/y  = 2y sin x – 4,  Φ/z = 3xz2 + 2

Now, Φ = Φ/x  dx +  Φ/y  dy +  Φ/z 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 =          

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

=   (y2 sin x + z3x – 4y + 2z)    (as shown above)

= [ y2 sin x + z3x – 4y + 2z ](π/2, -1, 2)

= [ 1 +8 π/2+ 4 + 4 ] – { - 4 – 2} =4π + 15 

Sums Based on Line Integral

1. Evaluate where =yzi+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 x2 + y2 = a2.

Soln. Parametric eqn of circle are:

x=a cos θ

y=a sin θ

z=0

=xi+yj+zk = a cos θ i + b cos θj + 0 k

d=(-a sinθi + a cosθ j)dθ

Circulation = +zj+xk). d

= -a sinθi + a cosθj)dθ

=           

2.6 Surface integrals, Volume integrals

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 x2 + y2 + z2 = a2 in the first octant.

Sol. Here-

Which becomes-

Example: Evaluate  , where ∅ = 45 x2y 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-

= 128

So that-

Example: Evaluate if V is the region in the first octant bounded by y2 + z2 = 9 and the plane x = 2 and .

Sol.

x varies from 0 to 2

The volume will be-

                                                    = 180

2.7 Green's theorem, Stokes theorem and Gauss divergence theorem (without proofs)

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 x2 + y2 = a2.

Sol. We know that by Green’s theorem-

And it it given that-

Now comparing the given integral-

P = 2x2 – y2  and   Q = x2 + y2