Unit - 2

Vector Calculus

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.

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

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.

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-

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

=

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

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:

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

Unit - 2

Vector Calculus

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.

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

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.

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-

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

=

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

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:

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

Unit - 2

Vector Calculus

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.

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

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.

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-

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

=

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

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:

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

Unit - 2

Vector Calculus

Unit - 2

Vector Calculus

Vector function-

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

Any vector can be expressed as-

Here , , are the scalar functions of t.

Differentiation of a vector-

Note-

1. Velocity =

2. Acceleration =

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-

Note-

Vector point function-

Vector field-

Del operator-

The del operated is defined as-

∇ =

Example: show that where

Sol. Here it is given-

Therefore-

Note-

Hence proved.

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

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

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.

Divergence (Definition)-

∇. = . =

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-

Now, since ṝ =xi+yi+zk d

And if F͞ =F1i + F2 j+ F3 K

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

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

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

=

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-

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-

Sol.

x varies from 0 to 2

The volume will be-

= 180

Unit - 2

Vector Calculus

Vector function-

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

Any vector can be expressed as-

Here , , are the scalar functions of t.

Differentiation of a vector-

Note-

1. Velocity =

2. Acceleration =

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-

Note-

Vector point function-

Vector field-

Del operator-

The del operated is defined as-

∇ =

Example: show that where

Sol. Here it is given-

Therefore-

Note-

Hence proved.

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

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

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.

Divergence (Definition)-

∇. = . =

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-

Now, since ṝ =xi+yi+zk d

And if F͞ =F1i + F2 j+ F3 K

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

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

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

=

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-

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-

Sol.

x varies from 0 to 2

The volume will be-

= 180

Unit - 2

Vector Calculus

Vector function-

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

Any vector can be expressed as-

Here , , are the scalar functions of t.

Differentiation of a vector-

Note-

1. Velocity =

2. Acceleration =

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-

Note-

Vector point function-

Vector field-

Del operator-

The del operated is defined as-

∇ =

Example: show that where

Sol. Here it is given-

Therefore-

Note-

Hence proved.

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

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

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.

Divergence (Definition)-

∇. = . =

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-

Now, since ṝ =xi+yi+zk d

And if F͞ =F1i + F2 j+ F3 K

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

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

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

=

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-

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-

Sol.

x varies from 0 to 2

The volume will be-

= 180

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-

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-

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.

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

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

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.

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:

- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.