Vector Calculus
Afizudeen S
Assistant Professor
Vellore Institute of Technology
September 25, 2025
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
1 / 28
Vector Calculus
Vector Differentiation module (A Vector equation in curve C)
As a particle moves along a curve C in space, its position P changes with
time t. Accordingly, the co-ordinates (x, y , z) of P are functions of t.
Then the position vector ⃗r of P becomes
⃗r = x⃗i + y⃗j + z ⃗k
or,
(1)
⃗r = x(t)⃗i + y (t)⃗j + z(t)⃗k
This is called vector equation along the curve C
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
2 / 28
Vector Calculus
Velocity and Acceleration
Since, the 1 determines the position of a particle moving along the curve
C at any time t, the rate of change of position is given by,
dx ⃗ dy ⃗ dz ⃗
d⃗r
=
(t)i +
(t)j +
(t)k
dt
dt
dt
dt
(2)
This gives the velocity of the particle at time t.
Unit tangent vector to the curve C is,
d⃗r
dt
d⃗r
dt
(3)
and acceleration of the particle at time t is
d d⃗r
d 2⃗r
= 2
dt dt
dt
Afizudeen S (VIT Chennai)
Vector Calculus
(4)
September 25, 2025
3 / 28
Example problems
1
2
Find the unit tangent vector to the curve ⃗r (t) = 4 sin(t)⃗i + 4 cos(t)⃗j + 3t ⃗k
A particle moving along to the curve x = e−t , y = 2 cos(3t),
z = 2 sin(3t). Determine the velocity and acceleration and their
magnitude at any time t
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
4 / 28
Vector Calculus
Vector point function
Let S ⊆ R, if to each t in S there corresponds a unique vector ⃗u (t) then we
say that ⃗u is a vector function defined on domain S.
Scalar point function
If to each P(x, y , z) of a region R there corresponds a scalar quantity
ϕ(x, y, z) then ϕ is called scalar point function
Vector differential operator ∇
The vector differential operator ∇ is defined by,
⃗i ∂ϕ + ⃗j ∂ϕ + ⃗k ∂ϕ
∂x
∂y
∂z
Afizudeen S (VIT Chennai)
Vector Calculus
(5)
September 25, 2025
5 / 28
Vector Calculus
Gradient of a scalar point function
Let ϕ(x, y , z) be a scalar point function defined in some region of space.
⃗ ∂ϕ ⃗ ∂ϕ
Then the vector function ⃗i ∂ϕ
∂x + j ∂y + k ∂z is called the gradient of ϕ and is
denoted by gradϕ or ∇ϕ
∂ϕ ⃗ ∂ϕ ⃗ ∂ϕ
gradϕ = ∇ϕ = ⃗i
+j
+k
∂x
∂y
∂z
Afizudeen S (VIT Chennai)
Vector Calculus
(6)
September 25, 2025
6 / 28
Directional derivative at a point
The rate of change of f (x, y ) in the direction of the unit vector ⃗(u) = ⟨a, b⟩
is called the directional derivative and it is denoted by D⃗u f (x, y). The
directional derivative of ϕ in the direction of unit vector ⃗u is goven by
∇ϕ.⃗u
(7)
|∇ϕ||⃗u |cosθ
(8)
or
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
7 / 28
(a) Gradient vector field
Afizudeen S (VIT Chennai)
(b) Directional derivative
Vector Calculus
September 25, 2025
8 / 28
Angle between two surface
If φ1 and φ2 are two surfaces then the angle between them can be
calculated as
∇φ1 .∇φ2
cosθ =
|∇φ1 ||∇φ2 |
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
(9)
9 / 28
Example Problems
Prove that ∇r n = nr n−2⃗r where ⃗r = x⃗i + y⃗j + z ⃗k
1
2 Find ∇. r ⃗
r ⃗r = x⃗i + y⃗j + z ⃗k
1
ans: 2r
3
Find unit normal vector to the surface x 3 + y 3 + 3xyz = 3 at the point
(1, 2, -1)
⃗
⃗
⃗
√j+2k
ans: −i+3
4
4
Find the Directional derivative of the function f (x, y , z) = 2xy + z 2 at
the point (1, -1, 3) in the direction of the vector ⃗i + 2⃗j + 2⃗k
ans: 14/3
5
Find the Directional derivative of xyz 2 + xz at (1, -1, 1) in the direction
of the normal to the surface 3xy 2 + y = z at (0, -1, 1)
√
ans: 4/ 11
6
Find the Directional derivative of φ = xy 2 + yz 3 at the point (2, -1, 1) in
the direction of PQ where Q is the point (3, 1, 3)
ans: -11/3
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
10 / 28
Example Problems
1
Find the angle between the surface at (2, -1, 2) the surface are
x 2 + y 2 + z 2 = 9 and z = x 2 + y 2 − 3
ans: cos−1 3√821
2
Find the angle between the surfaces xlog(z) = y 2 − 1 and x 2 y = 2 − z
at the point (1, 1, 1)
ans: cos−1 √130
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
11 / 28
DIVERGENCE and CURL
Divergence of vector function
⃗ (x, y , z) is a continuously differentiable vector point function in a given
If F
⃗ is defined by
region of space, then the divergence of F
∂
∂
∂
⃗
⃗
⃗
⃗
⃗
+j
+k
.(F1⃗i + F2⃗j + F3⃗k )
(10)
∇.F = div F = i
∂x
∂y
∂z
or
⃗ =
div F
∂F1
∂F2
∂F3
+
+
∂x
∂y
∂z
(11)
⃗ is a scalar point function
Note: ∇ × F
Solenoidal vector
⃗ is said to be solenoidal if div F
⃗ =0
A vector F
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
12 / 28
DIVERGENCE and CURL
Curl of a vector function
⃗ (x, y , z) is a differentiable vector point function defines at each point
If F
⃗ is defined by,
(x, y, z) in some region of space, then the curl of F
⃗ =∇×F
⃗ =
Curl F
⃗i
⃗j
⃗k
∂
∂x
∂
∂y
∂
∂z
F1
F2
F3
(12)
⃗ is a vector point function
Note: ∇ × F
Irrotational vector
⃗ = 0.
A vector is said to be irrotational if Curl F
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
13 / 28
Scalar potential
⃗ is an irrotational vector, then there exists a scalar function φ such that
If F
⃗
⃗
F = ∇φ. Such a scalar function is called scalar potential of F
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
14 / 28
Example Problems
Show that ∇⃗r = 3 where ⃗r = x⃗i + y⃗j + z ⃗k
⃗ = (x + 3y)⃗i + (y − 3z)⃗j + (x − 2z)⃗k , prove that F
⃗ is solenoidal
2 If F
1
3
⃗ and curl F
⃗ where F
⃗ = grad(x 3 + y 3 + z 3 − 3xyz)
Find div F
Prove that div (r n⃗r ) = (n + 3)r n . Hence show that r⃗r3 is solenoidal
⃗
⃗ = 3xz 2⃗i − yz⃗j + (x + 2z)⃗k , find curl curl F
5 If F
4
Show that curl{⃗r f (r ) = 0} where ⃗r = x⃗i + y⃗j + z ⃗k
2
2
2
7 Show that the vector (x − yz)⃗i + (y − zx)⃗j + (z − xy )⃗
k is irrotational
and find its scalar potential ϕ such that F = ∇ϕ
6
8
Find the constants a, b, and c so that
⃗ = (x + 2y + az)⃗i + (bx − 3y − z)⃗j + (4x + cy + 2z)⃗k is irrotational.
F
⃗ = ∇ϕ
Also find ϕ such that F
9
Find ∇2 (log(r ))
2
′′
′
10 Prove that ∇ f (r ) = f (r ) + (2/r )f (r )
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
15 / 28
Line Integral in vector fields
Line Integral in vector fields
An integral which is evaluated along a curve then it is called line integral.
Let C be the curve in same region of space described by a vector valued
⃗ = F1⃗i + F2⃗j + F3⃗k be
function. ⃗r = x⃗i + y⃗j + z ⃗k of a point (x, y, z) and let F
a continuous vector valued function defined alomga curve C. Then the
⃗ over C is denoted by
line integral F
Z
⃗ .d⃗r
F
(13)
C
Work done by a force
⃗ (x, y, z) is force acting on a particle which moves along a given curve C,
If F
R
⃗ .d⃗r gives the total work done by the force F
⃗ in the displacement
then C F
along C. Hence, work done by force,
Z
⃗
⃗ .d⃗r
F =
F
(14)
C
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
16 / 28
Line Integral in vector fields
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
17 / 28
Example problems
R
⃗ = (5xy − 6x 2 )⃗i + (2y − 4x)⃗j, evaluate F
⃗ .d⃗r along the curve C in
If F
c
3
xy-plane y = x from the point (1, 1) to (2, 8)
⃗ = (x 2 − y 2 + x)⃗i − (2xy + y)⃗j
2 Find the work done, when a force F
moves a particle from the origin to the point (1, 1) along y 2 = x
1
Find the work done in moving a particle in the force field
⃗ = 3x 2⃗i + (2xz − y )⃗j + z ⃗k along the straight line from (0, 0, 0) to (2, 1,
F
3)
⃗ = (y − 2x)⃗i + (3x + 2y)⃗j, calculate the circulation of F
⃗ about the
4 If F
2
2
circle C in the plane x + y = 4 oriented in the anti-clockwise
direction.
3
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
18 / 28
Surface integral (or Flux Integral) in vector field
Surface Integral
An integral which is evaluated over a surface is called a surface integral.
⃗ be a vector valued function which is defined at
Consider a surface S. Let F
each point on the surface and let P be any point on the surface and ⃗n be
⃗
the unit outward normal to the surface at P. The normal component of F
⃗
at P is F .⃗n.
RR
⃗ is denoted by
⃗ .⃗nds and is
The integral of the normal component of F
F
s
called the surface integral.
⃗ = F1⃗i + F2⃗j + F3⃗k, then we have
If F
ZZ
ZZ
ZZ
ZZ
⃗ .⃗nds =
F
F1 dy dz +
F2 dz dx +
F3 dx dy
s
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
19 / 28
Evaluation of surface integral
Let R1 be the projection of S on the xy -plane, ⃗k is the unit vector normal to
the xy -plane then ds = dx⃗ ⃗dy
n.k
ZZ
⃗ .⃗nds =
F
ZZ
s
If R2 be the projection of s on yz-plane
ZZ
ZZ
⃗
⃗
F .nds =
s
If R3 be the projection of s on xz-plane
ZZ
ZZ
⃗
⃗
F .nds =
s
Afizudeen S (VIT Chennai)
⃗ .⃗n dx dy
F
⃗n.⃗k
R1
(15)
⃗ .⃗n dx dy
F
⃗n.⃗i
R2
(16)
⃗ .⃗n dx dy
F
⃗n.⃗j
R3
(17)
Vector Calculus
September 25, 2025
20 / 28
Surface integral
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
21 / 28
Example Problems
RR
⃗ .⃗nds where F
⃗ = (x + y 2 )⃗i − 2x⃗j + 2yz ⃗k and S is the
Evaluate s F
surface of the plane 2x + y + 2z = 6 in the first octant.
RR
⃗ .⃗nds where F
⃗ = z⃗i + x⃗j + 3y 2 z ⃗k and S is the surface of
2 Evaluate
F
1
s
the cylinder x 2 + y 2 = 16 included in the first octant between z = 0
and z = 5
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
22 / 28
Volume Integral in vector field
Volume Integral
An integral which is evaluated over a volume bounded by a surface is
called a volume integral.
⃗ = F1⃗i + F2⃗j + F3⃗k is a vector field in V , then the volume integral is
If F
defined by
ZZZ
⃗ dv
F
(18)
V
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
23 / 28
Example Problems
RRR
⃗ dv , where V is the volume of the
⃗ = 2xz⃗i − x⃗j + y 2⃗k, evaluate
F
If F
V
region bounded by the surfaces x = 0, x = 1, y = 0, y = 6, z = x 2 , and
z=4
RRR
⃗ = (2x 2 − 3z)⃗i − 2xy⃗j − 4x ⃗k , evaluate
⃗ dv , where V is the
2 If F
curl F
V
closed region bounded by the planes x = 0, y = 0, z = 0 and
2x + 2y + z = 4
1
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
24 / 28
Green’s Theorem
Green’s theorem relates a line integral to the double integral taken
over the region bounded by the closed curve
Statement
If R is a closed region in xy -plane bounded by a simple closed curve C and
if M(x, y) and N(x, y ) are continuous functions having continuous
derivatives in R, then
I
ZZ ∂N
∂M
−
dx dy
(19)
Mdx + Ndy =
∂x
∂y
C
R
where C is traversed in positive (anticlockwise) direction
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
25 / 28
Example Problems
R
Verify Green’s theorem in a plane for c (3x 2 − 8y 2 )dx + (4y − 6xy )dy
where C is the boundary of the region by x = 0, y = 0, x + y = 1.
R
2 Verify Green’s theorem in xy -plane for
(xy + x 2 )dx + x 2 dy where C
c
the boundary of the region bounded by y = x, y = x 2
R
3 Verify Green’s theorem in a plane for
(3x 2 − 8y 2 )dx + (4y − 6xy )dy
c
where C is the boundary of the region defined by y = x 2 , x = y 2 .
R
4 Verify Green’s theorem in a plane for
(x − 2y )dx + xdy taken around
c
the circle x 2 + y 2 = 1
R −x
[e (sinydx + cosydy )] where C being
5 Evaluate by Green’s theorem
c
the rectangle with vertices (0, 0), (π, 0), (π, π2 ) and (0, π2 )
1
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
26 / 28
Stokes Theorem
Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s
Theorem. Whereas Green’s Theorem relates a double integral over a plane
region to a line integral around its plane boundary curve, Stokes’ Theorem relates a surface integral over a surface S to a line integral around the
boundary curve of S
Statement
⃗ is any
If S is an open surface bounded by a close curve C and if F
continuously differentiable vector field, then
I
ZZ
⃗
⃗ .⃗nds
⃗
F dr =
curl F
C
(20)
S
where ⃗n is outward drawn unit normal at any point of S
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
27 / 28
Example Problems
⃗ = (2xy − x 2 )⃗i − (x 2 − y 2 )⃗j where
Verify Stokes theorem in a plane for F
C is the boundary region bounded by the parabolas y = x 2 and x 2 = y
⃗ = (2x − y )⃗i − yz 2⃗j − y 2 z ⃗k where S is the
2 Verify Stokes theorem for F
2
upper half of the sphere x + y 2 + z 2 = 1 and C is the circular
boundary on z = 0 plane
⃗ = xy⃗i − 2yz⃗j − zx ⃗k , where S is
3 Verify Stokes theorem in a plane for F
the open surface of the rectangular parallelopiped formed by the
planes x = 0, x = 1, y = 0, y = 2, z = 0, and z = 3 above the xy -plane
1
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
28 / 28
Gauss Divergence Theorem
Gauss Divergence theorem enables us to convert a surface integral of a
vector function on a closed surface into volume integral.
Statement
If V is the volume bounded by a closed surface S and if a vector function
⃗ is continuous and has continuous partial derivatives in V and on S, then
F
ZZ
⃗ .⃗nds =
F
ZZZ
S
⃗ dv
∇.F
(21)
V
where ⃗n is the unit outward normal to the surface S and dV = dxdydz
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
29 / 28
Example Problems
⃗ = (x 3 − yz)⃗i − 2x 2 y⃗j + z ⃗k
Verify Gauss Divergence theorem for F
taken over the surface of the cube bounded by the planes
x = y = z = 2 and the coordinates planes.
⃗ = 4x⃗i − 2y 2⃗j + z 2⃗k taken over
2 Verify Gauss Divergence theorem for F
the surface bounded by the cylinder x 2 + y 2 = 4 and x 2 + y 2 = 4 and
z = 0, z = 3.
1
Afizudeen S (VIT Chennai)
Vector Calculus
September 25, 2025
30 / 28