Chapter 3

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

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

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

4. A B O 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

5. Scalar Field, V Integral If there exists a scalar field V along a curve C, then the line integral of Valong 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. z 2 y O 3 1 x 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.

23. Solution

24. 2.9.2 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 2 2x + y + z = 2 O y 2 y = 2 (1 x) 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 2x + y + z= 2 that is z= 2 (1-x) -y

28. Therefore,

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

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 Vof 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. z 3 O y 2  2 x

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

38. Using polar coordinate of cylinder, From

39. Therefore,

40. Exercise 2.9

41. 2.10.2 Vector Field, Integral If vector field defeated on surface S, surface integral of S is defined as

42. Example 2.24

43. z 3 O y 3 3 x Solution

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