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Vectors Coordinate Systems VC - Differential Elements VC - Differential Operators Important Theorems Summary Problems

Vector Calculus

S. R. [email protected]

School of Electronics EngineeringVellore Institute of Technology

July 16, 2013

Vector Calculus EE533, School of Electronics Engineering, VIT

mailto:[email protected]


Outline

1 Vectors

2 Coordinate Systems

3 VC - Differential Elements

4 VC - Differential Operators

5 Important Theorems

6 Summary

7 Problems



Norm (Absolute/Modulus/Magnitude) ∗ ∗ ∗

Definition

Given a vector space V over a subfield F of the complex numbers, a norma on V is a function‖‖ : V → R with the following properties:For all a ∈ F and all~u,~v ∈ V,

1 ‖a~v‖ = |a| ‖~v‖ (positive scalability).

2 ‖~u +~v‖ ≤ ‖~u‖+ ‖~v‖ (triangular inequality)

3 If ‖~v‖ = 0 then~v is the zero vector~0 (separates points)

aSometimes the vertical line, Unicode Ux007c (|), is used (e.g., |v|), but this latter notation isgenerally discouraged, because it is also used to denote the absolute value of scalars and thedeterminant of matrices.



Norm - A Few Examples ∗ ∗ ∗

Euclidean Norm

• On an n-dimensional Euclidean space Rn, the intuitive notion of length of the vectorx = (x1, x2, . . . , xn) is captured by the formula

‖x‖2 :=√

x21 + x2

2 + . . . + x2n. (1)

• On an n-dimensional complex space Cn, the most common norm is

‖z‖2 :=√|z1|2 + |z2|2 + . . . + |zn|2. (2)

Taxicab Norm / Manhattan Norm

• The name relates to the distance a taxi has to drive in a rectangular street grid to get from theorigin to the point x. It is defined as

‖x‖1 :=n

∑i=1|xi| . (3)



Norm - A Few Examples ∗ ∗ ∗

Maximum Norm

• Maximum norm is defined as

‖x‖∞ := max (|x1| , |x2| , . . . , |xn|) . (4)

p-norm

• p-norm is defined as

‖x‖p :=

(n

∑i=1|xi|p

)1/p

. (5)



Norm - The Concept of Unit Circle ∗ ∗ ∗

x 1x 2 x ∞



Addition & Subtraction

b

b

aaa+

b

Addition

b

b

a-ba-ba

Subtraction



Dot or Scalar Product

Definition

The dot product of two vectors,~a = [a1, a2, . . . , an] and~b = [b1, b2, . . . , bn] in a vector space of dimen-sion n is defined as

~a ·~b =n

∑i=1

aibi = a1b1 + a2b2 + . . . + anbn = ‖~a‖∥∥∥~b∥∥∥ cos θ. (6)

Properties

• ~a ·~b =~b ·~a (commutative)

• ~a ·(~b +~c

)=~a ·~b +~a ·~c (distributive over vector addition)

• ~a ·(

r~b +~c)= r

(~a ·~b

)+~a ·~c (bilinear)



Dot or Scalar Product - Physical Interpretation

Projection of~a in the direction of~b, ab is given by

ab =~a ·~b∥∥∥~b∥∥∥ (7)

Corollary

If~a ·~b =~a ·~c and~a 6= ~0, then we can write: ~a ·(~b−~c

)= 0 by the distributive law; the result above

says this just means that~a is perpendicular to(~b−~c

), which still allows

(~b−~c

)6=~0, and therefore

~b 6=~c.



Cross or Vector Product

Definition

The cross product~a×~b is defined as a vector~c that is perpendicular to both~a and~b, with a directiongiven by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectorsspan.

~a×~b =(‖~a‖

∥∥∥~b∥∥∥ sin θ)~n (8)

Properties

• ~a×~b = −~b×~a (anti-commutative)

• ~a×(~b +~c

)=~a×~b +~a×~c (distributive over vector addition)

• ~a×(

r~b +~c)= r

(~a×~b

)+~a×~c (bilinear)



Cross or Vector Product - Physical Interpretation



Cross or Vector Product - Why the Name CrossProduct?

~a×~b =

∣∣∣∣∣∣x y zax ay azbx by bz

∣∣∣∣∣∣



Scalar Triple Product

Definition

The scalar triple product of three vectors is defined as the dot product of one of the vectors with thecross product of the other two,

~a ·(~b×~c

)=~b · (~c×~a) =~c ·

(~a×~b

). (9)

Properties

• ~a ·(~b×~c

)= −~a ·

(~c×~b

)• ~a ·

(~b×~c

)=

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣



Scalar Triple Product - Physical Interpretation

base

a

b

c

θ

hα

Corollary

If the scalar triple product is equal to zero, then the three vectors~a,~b, and~c are coplanar, since the

parallelepiped defined by them would be flat and have no volume.



Vector Triple Product

Definition

The vector triple product is defined as the cross product of one vector with the cross product of theother two,

~a×(~b×~c

)=~b (~a ·~c)−~c

(~a ·~b

). (10)



Vectors - Independency & Orthogonality



Outline

1 Vectors





6 Summary

7 Problems



Remember Complex Numbers?

Cartesian Polar

Euler’s formula is our jewel and one of the most remarkable, almost astounding, formulas in all

of mathematics - Richard Feynman



Typical 2D Coordinate Systems

Cartesian Polar

x = ρ cos φ

y = ρ sin φ

ρ =√

x2 + y2

φ = tan−1( y

x

)



2D Coordinate Transformations

[Aρ

Aφ

]=

[cos φ sin φ− sin φ cos φ

] [AxAy

][

AxAy

]=

[cos φ − sin φsin φ cos φ

] [Aρ

Aφ

]



Typical 3D Coordinate Systems (RHS)

X

Y

Z

O

xy

z

(x,y,z)

Cartesian

O

ρφ

z

(ρ,φ,z)

X

Y

Z

Cylendrical

x = ρ cos φ

y = ρ sin φ

z = z

ρ =√

x2 + y2

φ = tan−1( y

x

)z = z



Typical 3D Coordinate Systems (RHS)

Spherical

x = r sin θ cos φ

y = r sin θ sin φ

z = r cos θ

r =√

x2 + y2 + z2

θ = cos−1

(z√

x2 + y2 + z2

)

φ = tan−1( y

x

)



Cross Product of Standard Basis Vectors

O

ρφ

z

(ρ,φ,z)

X

Y

Z

x× y = z

y× z = x

z× x = y

x× x = 0

ρ× φ = z

φ× z = ρ

z× ρ = φ

ρ× ρ = 0

and so on ...

r× θ = φ

θ × φ = r

φ× r = θ

r× r = 0



Dot Product of Standard Basis Vectors

O

ρφ

z

(ρ,φ,z)

X

Y

Z

x · x = y · y = z · z = 1

x · y = y · z = x · z = 0

ρ · ρ = φ · φ = z · z = 1

ρ · φ = φ · z = z · ρ = 0

r · r = θ · θ = φ · φ = 1

r · θ = θ · φ = φ · r = 0

and so on ...



3D Coordinate TransformationsCartesian⇐⇒ Cylindrical

O

ρφ

z

(ρ,φ,z)

X

Y

Z

Aρ

Aφ

Az

=

cos φ sin φ 0− sin φ cos φ 0

0 0 1

AxAyAz

Ax

AyAz

=

cos φ − sin φ 0sin φ cos φ 0

0 0 1

Aρ

Aφ

Az



3D Coordinate TransformationsCartesian⇐⇒ Spherical

ArAθ

Aφ

=

sin θ cos φ sin θ sin φ cos θcos θ cos φ cos θ sin φ − sin θ− sin φ cos φ 0

AxAyAz

Ax

AyAz

=

sin θ cos φ cos θ cos φ − sin φsin θ sin φ cos θ sin φ cos φ

cos θ − sin θ 0

ArAθ

Aφ



3D Coordinate TransformationsCylindrical⇐⇒ Spherical

O

ρφ

z

(ρ,φ,z)

X

Y

Z

ArAθ

Aφ

=

sin θ 0 cos θcos θ 0 − sin θ

0 1 0

Aρ

Aφ

Az

Aρ

Aφ

Az

=

sin θ cos θ 00 0 1

cos θ − sin θ 0

ArAθ

Aφ



Would you like to see a few more coordinate systems?



Parabolic Coordinate System



Curvilinear Coordinate System

e1

e2

b1

b2

b1

b2



Outline

1 Vectors





6 Summary

7 Problems



Infinitesimal Differential Elements - Cartesian - ~dl

~dl = dxx + dyy + dzz



Infinitesimal Differential Elements - Cartesian - ~ds

~ds = ±dxdyz (or) ± dydzx (or) ± dzdxy



Infinitesimal Differential Elements - Cartesian - dv

dv = dxdydz



Infinitesimal Differential Elements - Cylindrical - ~dl

~dl = dρρ + ρdφφ + dzz



Infinitesimal Differential Elements - Cylindrical - ~ds

~ds = ±ρdφdρz



Infinitesimal Differential Elements - Cylindrical - ~ds

~ds = ±ρdφdzρ



Infinitesimal Differential Elements - Cylindrical - dv

dv = ρdρdφdz



Infinitesimal Differential Elements - Spherical - ~dl

~dl = drr + rdθθ + r sin θdφφ



Infinitesimal Differential Elements - Spherical - ~ds

~ds = ±r2 sin θdθdφr



Infinitesimal Differential Elements - Spherical - dv

dv = r2 sin θdrdθdφ



Outline

1 Vectors





6 Summary

7 Problems



Divergence

Definition

The divergence of a vector field ~F at a point P is defined as the limit of the net flow of ~F across thesmooth boundary of a three dimensional region V divided by the volume of V as V shrinks to P.Formally,

div(~F (P)

)= ∇ ·~F = lim

V→{P}

‹S(V)

~F · n|V| ds = lim

V→{P}

‹S(V)

~F · ~ds|V| . (11)

Properties

• ∇ ·(

k1~A + k2~B)= k1∇ ·~A + k2∇ ·~B (linearity)

• ∇ ·(

w~A)= w∇ ·~A +~A · ∇w

• ∇ ·(~A×~B

)= ~B ·

(∇×~A

)−~A ·

(∇×~B

)



Divergence - Physical Interpretation

V

Sn

nn

n

∇ ·~F =∂Fx

∂x+

∂Fy

∂y+

∂Fz

∂z



Curl

Definition

If n is any unit vector, the curl of ~F is defined to be the limiting value of a closed line integral ina plane orthogonal to n as the path used in the integral becomes infinitesimally close to the point,divided by the area enclosed.

curl(~F (P)

)= ∇×~F = lim

A→0

˛C

~F · ~dl|A| n. (12)

Properties

• ∇×(

k1~A + k2~B)= k1∇×~A + k2∇×~B (linearity)

• ∇×(

w~A)= w∇×~A−~A×∇w

• ∇×(~A×~B

)=[~A(∇ ·~B

)−~B

(∇ ·~A

)]−[(

~A · ∇)~B−

(~B · ∇

)~A]



Curl - Physical Interpretation

∇×~F =

(∂Fz

∂y−

∂Fy

∂z

)x +

(∂Fx

∂z− ∂Fz

∂x

)y +

(∂Fy

∂x− ∂Fx

∂y

)z



Gradient

Definition

In vector calculus, the gradient of a scalar field is a vector field that points in the direction of thegreatest rate of increase of the scalar field, and whose magnitude is that rate of increase,

grad (w) = ∇w =∂w∂x

x +∂w∂y

y +∂w∂z

z. (13)

Properties

• ∇ (k1v + k2w) = k1∇v + k2∇w (Linearity)

• ∇ (vw) = v∇w + w∇v (Product Rule)



Gradient - Physical Interpretation

∇w =∂w∂x

x +∂w∂y

y +∂w∂z

z



Solenoidal and Lamellar Fields

Definition

In vector calculus a solenoidal vector field (also known as an incompressible vector field) is a vectorfield~v with divergence zero at all points in the field:

∇ ·~v = 0. (14)

Definition

A vector field is said to be lamellar or irrotational if its curl is zero. That is, if

∇×~v =~0. (15)



Curvilinear Coordinate Systems - Divergence, Curl,and Gradient

∇ ·~v =1

h1h2h3

[∂

∂q1(h2h3v1) +

∂

∂q2(h3h1v2) +

∂

∂q3(h1h2v3)

]

∇×~v =1

h1h2h3

∣∣∣∣∣∣h1 q1 h1 q2 h1 q3

∂∂q1

∂∂q2

∂∂q3

h1v1 h2v2 h3v3

∣∣∣∣∣∣∇w = ∑

i

(qi

1hi

∂w∂qi

)

where

• when (q1, q2, q3) = (x, y, z) =⇒ (h1, h2, h3) = (1, 1, 1),• when (q1, q2, q3) = (ρ, φ, z) =⇒ (h1, h2, h3) = (1, ρ, 1), and

• when (q1, q2, q3) = (r, θ, φ) =⇒ (h1, h2, h3) = (1, r, r sin θ).



Second Order Derivatives - DCG Chart

∇2w = 4w = ∇ · (∇w)

∇×∇×~A = ∇(∇ ·~A

)−∇2~A



Scalar Laplacian - Curvilinear Coordinate System

∇2w =1

h1h2h3

[∂

∂q1

(h2h3

h1

∂w∂q1

)+

∂

∂q2

(h3h1

h2

∂w∂q2

)+

∂

∂q3

(h1h2

h3

∂w∂q3

)]



Outline

1 Vectors





6 Summary

7 Problems



Open and Closed Surfaces

‚&˝ ˜

&¸



Divergence Theorem

Definition

Suppose V is a subset of Rn (in the case of n = 3, V represents a volume in 3D space) which is compactand has a piecewise smooth boundary S. If~F is a continuously differentiable vector field defined ona neighborhood of V, then we have

˚V

(∇ ·~F

)dv =

‹S

(~F · n

)ds =

‹S~F · ~ds. (16)



Divergence Theorem - Physical Interpretation

[F (y + ∆y)− F (y)]∆x∆z =(∇ ·~F

)vol1× vol1

[F (y + 2∆y)− F (y + ∆y)]∆x∆z =(∇ ·~F

)vol2× vol2

Sum : [F (y + 2∆y)− F (y)]∆x∆z = ∑i

(∇ ·~F

)voli× voli



Stokes’ Theorem

Definition

The surface integral of the curl of a vector field over a surface S in Euclidean three-space is relatedto the the line integral of the vector field over its boundary as

¨S

(∇×~F

)· ~ds =

˛C~F · ~dl. (17)



Stokes’ Theorem - Physical Interpretation

˛1=(∇×~F

)1· ~ds1

˛2=(∇×~F

)2· ~ds2

Sum : ∑i

˛i= ∑

i

(∇×~F

)i· ~dsi



Outline

1 Vectors





6 Summary

7 Problems



Important Vectorial Identities

• A · B = B ·A = ‖A‖ ‖B‖ cos θ

• AB = A·B‖B‖

B‖B‖

• A× B = −B×A = (‖A‖ ‖B‖ sin θ)~n =

∣∣∣∣∣∣x y z

Ax Ay AzBx By Bz

∣∣∣∣∣∣• A · (B×C) = B · (C×A) = C · (A× B) =

∣∣∣∣∣∣Ax Ay AzBx By BzCx Cy Cz

∣∣∣∣∣∣• A× (B×C) = B (A ·C)−C (A · B)• (A× B) · (C×D) = (A ·C) (B ·D)− (B ·C) (A ·D) ***

• (A× B)× (C×D) = (A · B×D)C− (A · B×C)D ***



Coordinate Transformations (Point)

x = ρ cos φ

y = ρ sin φ

ρ =√

x2 + y2

φ = tan−1( y

x

)x = r sin θ cos φ

y = r sin θ sin φ

z = r cos θ

r =√

x2 + y2 + z2

θ = cos−1

(z√

x2 + y2 + z2

)



Coordinate Transformations (Vector)

Aρ

Aφ

Az

=

cos φ sin φ 0− sin φ cos φ 0

0 0 1

AxAyAz

Ax

AyAz

=

cos φ − sin φ 0sin φ cos φ 0

0 0 1

Aρ

Aφ

Az

Ar

Aθ

Aφ

=

sin θ cos φ sin θ sin φ cos θcos θ cos φ cos θ sin φ − sin θ− sin φ cos φ 0

AxAyAz

Ax

AyAz

=

sin θ cos φ cos θ cos φ − sin φsin θ sin φ cos θ sin φ cos φ

cos θ − sin θ 0

ArAθ

Aφ

Ar

Aθ

Aφ

=

sin θ 0 cos θcos θ 0 − sin θ

0 1 0

Aρ

Aφ

Az

Aρ

Aφ

Az

=

sin θ cos θ 00 0 1

cos θ − sin θ 0

ArAθ

Aφ



Differential Elements

Cartesian Coordinate System:

~dl = dxx + dyy + dzz

~ds = ±dxdyz (or) ± dydzx (or) ± dzdxy

dv = dxdydz

Cylindrical Coordinate System:

~dl = dρρ + ρdφφ + dzz

~ds = ±ρdφdρz (or) ± ρdφdzρ

dv = ρdρdφdz

Spherical Coordinate System:

~dl = drr + rdθθ + r sin θdφφ

~ds = ±r2 sin θdθdφr

dv = r2 sin θdrdθdφ



Divergence, Curl, and Gradient

∇ ·~v =1

h1h2h3

[∂

∂q1(h2h3v1) +

∂

∂q2(h3h1v2) +

∂

∂q3(h1h2v3)

]

∇×~v =1

h1h2h3

∣∣∣∣∣∣h1 q1 h2 q2 h3 q3

∂∂q1

∂∂q2

∂∂q3

h1v1 h2v2 h3v3

∣∣∣∣∣∣∇w = ∑

i

(qi

1hi

∂w∂qi

)

∇2w =1

h1h2h3

[∂

∂q1

(h2h3

h1

∂w∂q1

)+

∂

∂q2

(h3h1

h2

∂w∂q2

)+

∂

∂q3

(h1h2

h3

∂w∂q3

)]

where,

(q1, q2, q3) (v1, v2, v3) (h1, h2, h3)

Catersian (x, y, z)(vx, vy, vz

)(1, 1, 1)

Cylindrical (ρ, φ, z)(vρ , vφ , vz

)(1, ρ, 1)

Spherical (r, θ, φ)(vr, vθ , vφ

)(1, r, r sin θ)



Important Differential Identities

• ∇ (vw) = v∇w + w∇v• ∇ (A · B) =

(A · ∇)B + (B · ∇)A + A× (∇× B) + B× (∇×A)***

• ∇ · (wA) = w∇ ·A + A · ∇w• ∇ · (A× B) = B · (∇×A)−A · (∇× B)• ∇× (wA) = w∇×A−A×∇w ***• ∇× (A× B) =

[A (∇ · B)− B (∇ ·A)]− [(A · ∇)B− (B · ∇)A] ***• ∇×∇×A = ∇ (∇ ·A)−∇2A• ∇ |r| = r

|r| ***

• ∇ 1|r| = −

r|r|3

***

• ∇.(

r|r|3

)= −∇2

(1|r|

)= 4πδ (r) ***



Important Integral Identities

• ˝V

(∇ ·~F

)dv =

‚S~F · ~ds (Divergence Theorem)

• ˜S

(∇×~F

)· ~ds =

¸C~F · ~dl (Stokes’ Theorem)



Outline

1 Vectors





6 Summary

7 Problems



Divergence

1 Evaluate the ∇ ·~G at

• PCart(2,−3, 4), if ~G = xx + y2y + z3 z;• PCyl(2, 110◦ ,−1), if ~G = 2ρz2 sin2 φρ + ρz2 sin 2φφ + 2ρ2z sin2 φz; and• PSpherical(1.5, 30◦ , 50◦), if ~G = 2r sin θ cos φr + r cos θ cos φθ − r sin φφ.

Ans: 43; 9.06; 1.28 [H1, D3.7, P73]

2 Given the electric flux density, ~D = 0.3r2 r nC/m2 in free space:

• find~E(=

~Dε , where ε≈ 8.8542× 10−12F/m

)at point P (r = 2, θ = 25◦ , φ = 90◦);

• find the total charge(

ρv = ∇ · ~D)

within the sphere ‖~r‖ = 3;• find the total electric flux leaving the sphere ‖~r‖ = 4.

Ans: 135.5r V/m; 305 nC; 965 nC [H1, D3.3, P61]



Curl

1 Calculate the value of the vector current density(~Je = ∇× ~H

):

• in Cartesian coordinates at PCart(2, 3, 4), if ~H = x2zy− y2xz;• in cylindrical coordinates at PCyl(1.5, 90◦ , 0.5), if ~H = 2

ρ (cos 0.2φ) ρ; and

• in spherical coordinates at PSpherical(2, 30◦ , 20◦), if ~H = 1sin θ θ.

Ans: −16x + 9y + 16z A/m2; 0.0549z A/m2; φ A/m2 [H1, D8.7, P246]



Gradient

1 Given the potential field, V = 2x2y− 5z, find the electric field intensity (~E = −∇V) at a givenpoint PCart (x, y, z).Ans: −4xyx− 2x2y + 5z V/m [H1, E4.3, P104]

2 Given the potential field, V = 100z2+1

ρ cos φ V, find the electric field intensity (~E = −∇V) at a

given point PCyl (3, 60◦ , 2).

Ans: −10ρ + 17.32φ + 24z V/m [H1, D4.8, P106]



Divergence Theorem

1 Evaluate both sides of the divergence theorem for the field , ~D = 2xyx + x2yand therectangular parallelepiped formed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3.Ans: 12 [H1, E3.5, P77]

2 Given the field, ~D = 6ρ sin φ2 ρ + 1.5ρ cos φ

2 φ, evaluate both sides of the divergence theoremfor the region bounded by ρ = 2, φ = 0, φ = π, z = 0, and z = 5.Ans: 225; 225 [H1, D3.9, P78]

˚V

(∇ ·~F

)dv =

‹S~F · ~ds



Stokes’ Theorem

1 Evaluate both sides of the Stokes’ theorem for the field , ~H = 6r sin φr + 18r sin θ cos φφ, andthe patch around the region, r = 4, 0 ≤ θ ≤ 0.1π, and 0 ≤ φ ≤ 0.3π.Ans: 22.2 [H1, E8.3, P248]

2 Evaluate both sides of the Stokes’ theorem for the field , ~H = 6xyx− 3y2y, and the patharound the region, 2 ≤ x ≤ 5, −1 ≤ y ≤ 1, and z = 0. Let the positive direction of ~ds be z.Ans: -126 [H1, D8.6, P251]

¨S

(∇×~F

)· ~ds =

˛C~F · ~dl


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