1. Chapter 3 1. Line Integral Volume Integral Surface Integral
Green’s Theorem
Divergence Theorem (Gauss’ Theorem)
Stokes’ Theorem

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

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

4. 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

5. 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

6. Example

7. Solution

9. Exercise 2.6

10. 2.8.2 Vector Field, Integral Let a vector field and
The scalar product is written as

12. Example 2.15

13. Solution

16. Exercise 2.7

17. * Double Integral *

21. 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

22. 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

23. Solution

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

25. 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

26. 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)

27. 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

28. Therefore,

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

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

31. Therefore,

32. Exercise 2.8

33. 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

34. 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

35. Also,
Therefore,
Then,

36. 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

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

38. Using polar coordinate of cylinder,
From

39. Therefore,

40. Exercise 2.9

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

42. Example 2.24

43. Solution x z y 3 O

46. Using polar coordinate of sphere,

48. Exercise 2.9

49. 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.

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

58. 2.12 Divergence Theorem (Gauss’ Theorem)
If S is a closed surface including region V in vector field

59. Example 2.26

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

71. 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

72. Example 2.27
Surface S is the combination of

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

74. 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

75. So,

76. By integrating each part of the surface,

77. Then ,
and

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

79. Therefore,
Also,

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

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