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
Unit - 2
Vector Calculus
Unit - 2
Vector Calculus
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.
