1. 1 Chapter 2 Vector Calculus 1.Elementary 2.Vector Product 3.Differentiation of Vectors 4.Integration of Vectors 5.Del Operator or Nabla (Symbol  ) 6.Polar Coordinates

2. 2 Chapter 2 Continued 7.Line Integral 8.Volume Integral 9.Surface Integral 10.Green’s Theorem 11.Divergence Theorem (Gauss’ Theorem) 12.Stokes’ Theorem

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

4. 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. O A or O A

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

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

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

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

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

10. 10 Example 2.1

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

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

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

14. 14 Example 2.3

15. 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. 16 The n th derivative of vector is given by The magnitude of is

17. 17 Example 2.4

18. 18 Example 2.5 The position of a moving particle at time t is given by x  4t + 3, y  t 2 + 3t, z  t 3 + 5t 2. Obtain The velocity and acceleration of the particle. The magnitude of both velocity and acceleration at t  1.

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

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

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

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

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

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

25. 25 Example 2.6

26. 26 Exercise 2.1

27. 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. 28 Example 2.7

29. 29 Exercise 2.2

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

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

32. 32 Example 2.8

33. 33

34. 34 Exercise 2.3

35. 35 Solution

36. 36 2.6.1.1 Grad Properties If A and B are two scalars, then

37. 37 2.6.2 Directional Derivative

38. 38 Example 2.9

39. 39 Solution Directional derivative of  in the direction of

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42. 42 2.6.3 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. 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. 44 Unit normal vector is denoted by Example 2.10 Calculate the unit normal vector at (-1,1,1) for 2 yz  xz  xy  0.

45. 45 Solution Given 2 yz  xz  xy  0. Thus

46. 46 2.6.4 Divergence of a Vector

47. 47 Example 2.11

48. 48 Exercise 2.4

49. 49 Remarks

50. 50 2.6.5 Curl of a Vector

51. 51 Example 2.12

52. 52 Solution

53. 53 Exercise 2.5

54. 54 Answer Remark

55. 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. 56 2.7.1 Polar Coordinate for Plane (r, θ) x ds y  d d

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

58. 58 2.7.3 Polar Coordinate for Sphere (r,  y x r z  

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

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

61. 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. 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. 63 Example 2.14

64. 64 Solution

65. 65

66. 66 Exercise 2.6

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

68. 68

69. 69 Example 2.15

70. 70 Solution

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72. 72

73. 73 Exercise 2.7

74. 74 * Double Integral *

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78. 78 2.9Volume 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. 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. 80 Solution

81. 81 2.9.2 Vector Field, Integral If V is a closed region and, vector field in region V, Volume integral of V is

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

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

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

85. 85 Therefore,

86. 86 Example 2.22 Evaluate where and V is region bounded by z = 0, z = 4 and x 2 + y 2 = 9 x z  y  4  3 3

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

88. 88 Therefore,

89. 89 Exercise 2.8

90. 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. 91 Example 2.23 Scalar field V  x y z defeated on the surface S : x 2  y 2  4 between z  0 and z  3 in the first octant. Evaluate Solution Given S : x 2  y 2  4, so grad S is

92. 92 Also, Therefore, Then,

93. 93 Surface S : x 2  y 2  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 2 3 O

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

95. 95 Using polar coordinate of cylinder, From

96. 96 Therefore,

97. 97 Exercise 2.9

98. 98 2.10.2 Vector Field, Integral If vector field defeated on surface S, surface integral of S is defined as

99. 99 Example 2.24

100. 100 Solution x z y 3 3 3 O

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103. 103 Using polar coordinate of sphere,

104. 104

105. 105 Exercise 2.9

106. 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. 107 Example 2.25  y 2 x 2 C3C3 C2C2 C1C1 O x 2 + y 2 = 2 2 Solution

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

116. 116 Example 2.26

117. 117 Solution x z y 2 2 4 O S3S3 S4S4 S2S2 S1S1 S5S5

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

129. 129 Example 2.27 Surface S is the combination of

130. 130 Solution z y x 3 4 O S3S3 C2C2 S2S2 C1C1 S1S1 3

131. 131 We can also mark the pieces of curve C as C 1 :Perimeter of a half circle with radius 3. C 2 :Straight line from (-3,0,0) to (3,0,0). Let say, we choose to evaluate first. Given

132. 132 So,

133. 133 By integrating each part of the surface,

134. 134 and Then,

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

136. 136 Therefore, Also,

137. 137

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

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140. 140 (iii) For surface S 3 : y = 0, normal vector unit to the surface is dS = dxdz The integration limits : So,

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