1. Chapter 2 Vector Calculus
Elementary
Vector Product
Differentiation of Vectors
Integration of Vectors
Del Operator or Nabla (Symbol )
Polar Coordinates

2. Chapter 2 Continued Line Integral Volume Integral Surface Integral
Green’s Theorem
Divergence Theorem (Gauss’ Theorem)
Stokes’ Theorem

3. 2.1 Elementary Vector Analysis
Definition 2.1 (Scalar and vector)
Scalar is a quantity that has magnitude but not direction.
For instance mass, volume, distance
Vector is a directed quantity, one with both magnitude and direction.
For instance acceleration, velocity, force

4. We represent a vector as an arrow from the origin O to a point A.
The length of the arrow is the magnitude of the vector written as or A A or O O

5. 2.1.1 Basic Vector System Perpendicular to each other
In the positive directions
of the axes
have magnitude (length) 1
Unit vectors , ,

6. Define a basic vector system and form a right-handed set, i.e

7. 2.1.2 Magnitude of vectors Let P = (x, y, z). Vector is defined by
with magnitude (length)

8. 2.1.3 Calculation of Vectors
1. Vector Equation
Two vectors are equal if and only if the corresponding components are equals

9. 2. Addition and Subtraction of Vectors
3. Multiplication of Vectors by Scalars

10. Example 2.1

11. 2.2 Vector Products 1) Scalar Product (Dot product)
2) Vector Product (Cross product)

12. 3) Application of Multiplication of Vectors
Given 2 vectors and , projection onto is defined by
compb a
b) The area of triangle

13. A = c) The area of parallelogram d) The volume of tetrahedrone
e) The volume of parallelepiped
A =

14. Example 2.3

15. 2.4 Vector Differential Calculus
Let A be a vector depending on parameter u,
The derivative of A(u) is obtained by differentiating each component separately,

16. The nth derivative of vector is given by
The magnitude of is

17. Example 2.4

18. Example 2.5 The position of a moving particle at time t is given
by x = 4t + 3, y = t2 + 3t, z = t3 + 5t2. Obtain
The velocity and acceleration of the particle.
The magnitude of both velocity and acceleration at t = 1.

19. Solution The parameter is t, and the position vector is
The velocity is given by
The acceleration is

20. At t = 1, the velocity of the particle is
and the magnitude of the velocity is

21. At t = 1, the acceleration of the particle is
and the magnitude of the acceleration is

22. 2.4.1 Differentiation of Two Vectors
If both and are vectors, then

23. 2.4.2 Partial Derivatives of a Vector
If vector depends on more than one parameter, i.e

24. Partial derivative of with respect to is given by
e.t.c.

25. Example 2.6

26. Exercise 2.1

27. 2.5 Vector Integral Calculus
The concept of vector integral is the same as the integral of real-valued functions except that the result of vector integral is a vector.

28. Example 2.7

29. Exercise 2.2

30. 2.6 Del Operator Or Nabla (Symbol )
Operator  is called vector differential operator, defined as

31. 2.6.1 Grad (Gradient of Scalar Functions)
If  x,y,z is a scalar function of three variables and f is differentiable, the gradient of f is defined as

32. Example 2.8

34. Exercise 2.3

35. Solution

36. Grad Properties
If A and B are two scalars, then

37. 2.6.2 Directional Derivative

38. Example 2.9

39. Solution
Directional derivative of  in the direction of

42. Unit Normal Vector
Equation  (x, y, z) = constant is a surface equation. Since  (x, y, z) = constant, the derivative of  is zero; i.e.

43. This shows that when  (x, y, z) = constant,
Vector grad  =   is called normal vector to the surface  (x, y, z) = constant y ds
grad  z x

44. Unit normal vector is denoted by
Example 2.10
Calculate the unit normal vector at (-1,1,1) for 2yz + xz + xy = 0.

45. Solution
Given 2yz + xz + xy = 0. Thus

46. Divergence of a Vector

47. Example 2.11

48. Exercise 2.4

49. Remarks

50. Curl of a Vector

51. Example 2.12

52. Solution

53. Exercise 2.5

54. Answer
Remark

55. 2.7 Polar Coordinates
Polar coordinate is used in calculus to calculate an area and volume of small elements in easy way.
Lets look at 3 situations where des Cartes Coordinate can be rewritten in the form of Polar coordinate.

56. 2.7.1 Polar Coordinate for Plane (r, θ)y ds d x 

57. 2.7.2 Polar Coordinate for Cylinder (, , z)ds dv z y   x

58. 2.7.3 Polar Coordinate for Sphere (r, q, f)z r  y  x

59. Example 2.13 (Volume Integral)z  y 
4  3

60. Solution
Since it is about a cylinder, it is easier if we use cylindrical polar coordinates, where

61. 2.8 Line Integral
Ordinary integral  f (x) dx, we integrate along the x-axis. But for line integral, the integration is along a curve.
 f (s) ds =  f (x, y, z) ds A O B

62. 2.8.1 Scalar Field, V Integral
If there exists a scalar field V along a curve C, then the line integral of V along C is defined by

63. Example 2.14

64. Solution

66. Exercise 2.6

67. 2.8.2 Vector Field, Integral Let a vector field and
The scalar product is written as

69. Example 2.15

70. Solution

73. Exercise 2.7

74. * Double Integral *

78. 2.9 Volume Integral 2.9.1 Scalar Field, F Integral
If V is a closed region and F is a scalar field in region V, volume integral F of V is

79. Example 2.20
Scalar function F = 2 x defeated in one cubic that has been built by planes x = 0, x = 1, y = 0, y = 3, z = 0 and z = 2. Evaluate volume integral F of the cubic. z x y 3 O 2 1

80. Solution

81. Vector Field, Integral
If V is a closed region and , vector field in region V, Volume integral of V is

82. Example 2.21
Evaluate , where V is a region bounded by x = 0, y = 0, z = 0 and 2x + y + z = 2, and also given

83. Solution
If x = y = 0, plane 2x + y + z = 2 intersects z-axis at z = 2.
(0,0,2)
If x = z = 0, plane 2x + y + z = 2 intersects y-axis at y = 2.
(0,2,0)
If y = z = 0, plane 2x + y + z = 2 intersects x-axis at x = 1.
(1,0,0)

84. z y x We can generate this integral in 3 steps :2 O 1
2x + y + z = 2
y = 2 (1  x)
We can generate this integral in 3 steps :
Line Integral from x = 0 to x = 1.
Surface Integral from line y = 0 to line y = 2(1-x).
Volume Integral from surface z = 0 to surface x + y + z = 2 that is z = 2 (1-x) - y

85. Therefore,

86. Example 2.22
Evaluate where and V is region bounded by z = 0, z = 4 and x2 + y2 = 9 x z  y 
4  3

87. Using polar coordinate of cylinder,; ; ;
where

88. Therefore,

89. Exercise 2.8

90. 2.10 Surface Integral 2.10.1 Scalar Field, V Integral
If scalar field V exists on surface S, surface integral V of S is defined by
where

91. Example 2.23
Scalar field V = x y z defeated on the surface S : x2 + y2 = 4 between z = 0 and z = 3 in the first octant.
Evaluate
Solution
Given S : x2 + y2 = 4 , so grad S is

92. Also,
Therefore,
Then,

93. Surface S : x2 + y2 = 4 is bounded by z = 0 and z = 3 that is a cylinder with z-axis as a cylinder axes and radius,
So, we will use polar coordinate of cylinder to find the surface integral. x z  y 2 3 O

94. Polar Coordinate for Cylinder
where
(1st octant) and

95. Using polar coordinate of cylinder,
From

96. Therefore,

97. Exercise 2.9

98. Vector Field, Integral
If vector field defeated on surface S, surface integral of S is defined as

99. Example 2.24

100. Solution x z y 3 O

103. Using polar coordinate of sphere,

105. Exercise 2.9

106. 2.11 Green’s Theorem
If c is a closed curve in counter-clockwise on plane-xy, and given two functions P(x, y) and
Q(x, y),
where S is the area of c.

107. Example 2.25  y 2 x C3 C2 C1 O
x2 + y2 = 22
Solution

115. 2.12 Divergence Theorem (Gauss’ Theorem)
If S is a closed surface including region V in vector field

116. Example 2.26

117. Solution x z y 2 4 O S3 S4 S2 S1 S5

128. 2.13 Stokes’ Theorem
If is a vector field on an open surface S and boundary of surface S is a closed curve c, therefore

129. Example 2.27
Surface S is the combination of

130. Solution z y x 3 4 O S3 C2 S2 C1 S1

131. We can also mark the pieces of curve C as
C1 : Perimeter of a half circle with radius 3.
C2 : Straight line from (-3,0,0) to (3,0,0).
Let say, we choose to evaluate first.
Given

132. So,

133. By integrating each part of the surface,

134. Then ,
and

135. By using polar coordinate of cylinder ( because
is a part of the cylinder),

136. Therefore,
Also,

138. (ii) For surface , normal vector unit to the
surface is
By using polar coordinate of plane ,

140. For surface S3 : y = 0, normal vector unit
to the surface is
dS = dxdz
The integration limits :
So,