1. 1 Line Integrals In this section we are now going to introduce a new kind of integral. However, before we do that it is important to note that you will need to be able to write down a set of parametric equations for a given curve. Here are some of the more basic curves that we will need to know how to do as well as limits on the parameter if they are required. Curves Parametric Equations

2. 2

3. 3 Definition of Line Integral If is defined in a region containing a smooth curve of finite length, then the line integral of along is given by Example: (Larson, Ex-2,Page-1006) Evaluate where C is the line segment from (0,0,0) to (1,2,1). Solution: The parametric equations of the given curve are

4. 4 Hence Thus, the line integral is

5. 5 Suppose is a path composed of smooth curves If is continuous on, then it can be shown that Example: (Larson, Ex-3,Page-1007) Evaluate the line integral where C is a path composed of the line segment given by from to and the curve given by from to. Solution: Here

6. 6 For : Parametric equations Thus

7. 7 For : Parametric equations Thus So that

8. 8 Substitute these values in equation (i), we obtain Ans.

9. 9 Example: (Larson, Ex-4,Page-1008) Evaluate where C is the curve represented by Solution: Do.

10. 10 Line Integrals of Vector Fields One of the most important physical application of line integrals is that of finding the work done on an object moving in a force field. Definition of Line Integral of a Vector Field Let be a continuous vector field defined on a smooth curve given by.The line Integral of on is given by Work Done= (Force)(Distance)

11. 11 Example: (Larson, Ex-6,Page-1010) Find the work done by the force field on a particle as it moves along the helix given by from the point to. Solution: Because Thus, the given force field can be written as By taking first derivative of (i), we get

12. 12 We know, work done Ans.

13. 13 Example: (Larson, Ex-7,Page-1011) Let and evaluate the line integral for each of the following parabolic curves. Solution: Do.

14. 14 Line Integrals in Differential Form If is a vector field of the form, and is a curve given by, then For two variables the differential form of line integrals can be written as

15. 15 Example: (Larson, Ex-8,Page-1012) Evaluate, where is a circle of radius given by Solution: Since, we obtain

16. 16 Thus, the line integral is

17. 17 Ans.

18. 18 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 15.2 Page no: 1013 Problems: 7,9;22,25;43,44;56,57;

19. 19 Green’s Theorem Green’s Theorem, named after the English mathematician George Green (1793-1841). This theorem states that the value of a double integral over a simply connected plane region is determined by the value of a line integral around the boundary of. R

20. 20 Definition of Green's Theorem Let be a simply connected region with a piecewise smooth boundary, oriented counterclockwise. If and have continuous partial derivatives in an open region containing, then

21. 21 Example: (Larson, Ex-1,Page-1023) Use Green’s theorem to evaluate the line integral, where is the path from to along the graph of and from to along the graph of Solution: Because, it follows that Applying Green’s theorem, we have

22. 22 Ans.

23. 23 Example: (Larson, Ex-2,Page-1029) While subject to the force, a particle travels once around the circle of radius. Use Green’s theorem to find the work done by. Solution: Because, it follows that Applying Green’s theorem, we have

24. 24 we have, Substituting these in (i), the work done is

25. 25 Ans.

26. 26 Example: (Larson, Ex-4,Page-1030) Evaluate where is the path enclosing the annular region is given by. Solution: Since,it follows that Applying Green’s theorem, we have

27. 27 Ans.

28. 28 Line Integral for Area If is a plane region bounded by a piecewise smooth simple curve, oriented counterclockwise, then the area of is given by Note:

29. 29 Example: (Larson, Ex-5,Page-1031) Use a line integral to find the area of the ellipse Solution: Parametric equations to the ellipse are Therefore, the area is

30. 30 Ans.

31. 31 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 15.4 Page no: 1033 Problems: 11. 13. 17; 21. 24; 25;

32. 32 Surface integral, In calculus, the integral of a function of several variables calculated over a surface. More generally, an Integral calculated over a plane or curved surface results in a surface integral representing a volume. Surface Integrals Let be a surface with equation and its projection on the. If are continuous on and is continu- ous on, then the surface integral of over is R Evaluating a Surface Integral:

33. 33 If is the graph of and is its projection on the xz-plane, then If is the graph of and is its projection on the yz-plane, then

34. 34 Example: (Larson, Ex-1,Page-1047) Evaluate the surface integral, where is the first-octant of the plane Solution: Begin with writing is

35. 35 We can write Therefore, the surface integral is

36. 36 Ans.

37. 37 Parametric Surface and Surface Integrals For a surface given by the vector-valued function defined over a region in the uv-plane is called parametric surface. Examples: Surfaces Parametric Equations

38. 38 We can show that the surface integral of over is given by

39. 39 Example: Evaluate the surface integral, where is the surface whose side is the cylinder, whose bottom is the disc in the xy-plane and whose top is the plane Example: Here surface integral

40. 40 Now The Cylinder In parametric form, the surface is given by

41. 41

42. 42

43. 43 Now Plane on top of the Cylinder

44. 44 And Bottom of the Cylinder Here

45. 45

46. 46 Now substitute these values in (i), we obtain Ans.

47. 47 Flux Integrals We can think of flux as the amount of something crossing a surface. This “something” can be fluid, wind, electric field etc. Definition of Flux Integral: Let, where have continuous first partial derivatives on the surface oriented by a unit normal vector. The flux integral of across is given by. If is density of the fluid, then the flux integral is given by S

48. 48 Evaluating a Flux Integral: Let be an oriented surface given by and let be its projection on the xy-plane. If the surface oriented upward. and If the surface oriented downward.

49. 49 Example: (Larson, Ex-5,Page-1053) Let be that portion of the paraboloid, lying above the xy-plane, oriented by an upward normal vector. A fluid of constant density is flowing through the surface according to the velocity field. Find the rate of mass flow through. S

50. 50 Solution: Given The rate of mass flow through the surface is

51. 51 Ans.

52. 52 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 15.6 Page no: 1055 Problems: 2;5;13,14;21,23;

53. 53 Divergence Theorem In vector calculus, the divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem, is a result that relates the flow (flux) of a vector field through a surface to the behavior of the vector field inside the surface. Let be a solid region bounded by a closed surface oriented by a unit normal vector directed outward from. If is a vector field whose component functions have continuous partial derivatives in. Then The Divergence Theorem

54. 54 Example: (Larson, Ex-1,Page-1060) Let be the solid region bounded by the coordinate planes and the plane, and let. Find, where is the surface of We have, Solution: Given plane and vector field

55. 55

56. 56 Ans.

57. 57 Example: (Larson, Ex-3,Page-1062) Let be the solid region bounded by the cylinder, the plane, and the xy-plane. Find, where is the surface of and Solution: Given cylinder and plane is and vector field We have,

58. 58

59. 59 Ans.

60. 60 Example: (Larson, Ex-4,Page-1064) Let be the region bounded by the sphere. Find the outward flux of the vector field through the sphere. Solution: Given sphere is and vector field We have,

61. 61 By the divergence theorem, we have

62. 62 Ans.

63. 63 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 15.7 Page no: 1064 Problems: 3,5,7,9,11;

64. 64 Stokes’s Theorem A second order higher-dimension analog of Green’s Theorem is called Stokes’s Theorem, after the English mathematical physicist George Gabriel Stokes (1819-1903). Stokes’s Theorem gives the relationship between a surface integral over an oriented surface and a line integral along a closed curve forming the boundary of. Stokes’s Theorem: Let be an oriented surface with unit normal vector, bounded by a piecewise smooth, simple curve. If is a vector field whose component functions have continuous partial derivatives on an open region containing and, then

65. 65 Example: (Larson, Ex-1,Page-1067) Let be the oriented triangle lying in the plane. Evaluate, where Solution: Given plane,

66. 66 Using Stokes’s Theorem, for an upward normal vector to obtain

67. 67 Ans. -9.

68. 68 Example: (Motion of a Swirling Liquid) (Larson, Ex-3,Page-1070) A liquid is swilling around in a cylindrical container of radius 2 such that its motion is described by the velocity field, as shown in the figure. Find, where is the upper surface of the cylindrical tank. z y x Solution: The Curl of is given by

69. 69

70. 70 Letting, we have Ans.

71. 71 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 15.8 Page no: 1071 Problems: 13,14,16;21,22;

72. 72 End (of Syllabus)

73. 73 Discussed Topics Line Integrals Green’s Theorem Surface Integrals Flux Integrals Divergence Theorems Stokes’s Theorem