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Vector Calculus

CHAPTER 9.10~9.17. Vector Calculus. Contents. 9.10 Double Integrals 9.11 Double Integrals in Polar Coordinates 9.12 Green’s Theorem 9.13 Surface Integrals 9.14 Stokes’ Theorem 9.15 Triple Integrals 9.16 Divergence Theorem 9.17 Change of Variables in Multiple Integrals.

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Vector Calculus

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  1. CHAPTER 9.10~9.17 Vector Calculus

  2. Contents • 9.10 Double Integrals • 9.11 Double Integrals in Polar Coordinates • 9.12 Green’s Theorem • 9.13 Surface Integrals • 9.14 Stokes’ Theorem • 9.15 Triple Integrals • 9.16 Divergence Theorem • 9.17 Change of Variables in Multiple Integrals

  3. 9.10 Double Integrals • Definition 9.10: Let f be a function of two variables defined on a closed region R. Then the double integral of f over R is given by (1) Integrability: If the limit in (1) exists, we say that f is integrable over R, and R is the region of integration. Area: When f(x,y)=1 on R. Volume: When f(x,y) 0 on R.

  4. Properties of Double Integrals • Theorem 9.11: Let f and g be functions of two variables that are integrable over a region R. Then (i) , where k is any constant (ii) (iii) where and

  5. Regions of Type I and II • Region of Type ISee the region in Fig 9.71(a) R: a x  b, g1(x)  y  g2(x) • Region of Type IISee the region in Fig 9.71(b) R: c  y  d, h1(y)  x  h2(y)

  6. Fig 9.71

  7. Iterated Integral • For Type I: (4) • For Type II: (5)

  8. THEOREM 9.12 Let f be continuous on a region R. (i) For Type I: (6)(ii) For Type II: (7) Evaluation of Double Integrals

  9. Note: • Volume =wherez = f(x, y) is the surface.

  10. Example 1 Evaluate over the region bounded by y = 1, y = 2, y = x, y = −x + 5.See Fig 9.73. SolutionThe region is Type II

  11. Fig 9.73

  12. Example 2 Evaluate over the region in the first quadrant bounded by y = x2, x = 0, y = 4. SolutionFrom Fig 9.75(a) , it is of Type IHowever, this integral can not be computed.

  13. Fig 9.75(a) Fig 9.75(b)

  14. Example 2 (2) Trying Fig 9.75(b), it is of Type II

  15. 9.11 Double Integrals in Polar Coordinates • Double IntegralRefer to the figure.The double integral is

  16. Refer to the figure.The double integral is

  17. Change of Variables • Sometimes we would like to change the rectangular coordinates to polar coordinates for simplifying the question. If and then (3)Recall: x2 + y2 = r2 and

  18. Example 2 Evaluate SolutionFrom the graph is shown in Fig 9.84.Using x2 + y2 = r2, then 1/(5 + x2 + y2 ) = 1/(5 + r2)

  19. Fig 9.84

  20. Example 2 (2) Thus the integral becomes

  21. Example 3 Find the volume of the solid that is under and above the region bounded byx2 + y2 – y = 0. See Fig 9.85. SolutionFig 9.85

  22. Example 3 (2) We find that and the equations becomeand r = sin . Now

  23. Example 3 (3)

  24. Area • If f(r, ) = 1,then the area is

  25. 9.12 Green’s Theorem • Along Simple Closed Curves For different orientations of simple closed curves, please refer to Fig 9.88.Fig 9.88(a) Fig 9.88(b) Fig 9.88(c)

  26. Notations for Integrals Along Simple Closed Curves • We usually write them as the following formswhere and represents in the positive and negative directions, respectively.

  27. THEOREM 9.13 • Partial ProofFor a region R is simultaneously of Type I and Type II, IF P, Q, P/y, Q/x are continuous on R, which is bounded by a simply closed curve C, then Green’s Theorem in the Plane

  28. Fig 9.89(a) Fig 9.89(b)

  29. Partial Proof Using Fig 9.89(a), we have

  30. Partial Proof Similarly, from Fig 9.89(b),From (2) + (3), we get (1).

  31. Note: • If the curves are more complicated such as Fig 9.90, we can still decompose R into a finite number of subregions which we can deal with. • Fig 9.90

  32. Example 2 Evaluate where C is the circle (x – 1)2 + (y – 5)2 = 4 shown in Fig 9.92.

  33. Example 2 (2) SolutionWe have P(x, y) = x5 + 3yandthenHence Since the area of this circle is 4, we have

  34. Example 3 • Find the work done by F = (– 16y + sin x2)i + (4ey + 3x2)j along C shown in Fig 9.93.

  35. Example 3 (2) SolutionWe have Hence from Green’s theorem In view of R, it is better handled in polar coordinates, since R:

  36. Example 3 (3)

  37. Example 4 • The curve is shown in Fig 9.94. Green’s Theorem is not applicable to the integralsince P, Q, P/x, Q/y are not continuous at the region.

  38. Fig 9.94

  39. Region with Holes • Green’s theorem cal also apply to a region with holes. In Fig 9.95(a), we show C consisting of two curves C1and C2. Now We introduce cross cuts as shown is Fig 9.95(b), R is divided into R1 and R2. By Green’s theorem: (4)

  40. Fig 9.95(a) Fig 9.95(b) • The last result follows from that fact that the line integrals on the crosscuts cancel each other.

  41. Example 5 Evaluate where C = C1C2 is shown in Fig 9.96. SolutionBecause

  42. Example 5 (2) are continuous on the region bounded by C, then

  43. Fig 9.96

  44. Conditions to Simplify the Curves • As shown in Fig 9.97, C1 and C2are two nonintersecting piecewise smooth simple closed curves that have the same orientation. Suppose that P and Q have continuous first partial derivatives such that P/y = Q/x in the region R bounded between C1 and C2, then we have

  45. Fig 9.97

  46. Example 6 Evaluate the line integral in Example 4. SolutionWe find P = – y / (x2 + y2)and Q = x / (x2 + y2)have continuous first partial derivatives in the region bounded by C and C’. See Fig 9.98.

  47. Fig 9.98

  48. Example 6 (2) Moreover,we have

  49. Example 6 (3) Using x = cos t, y = sin t, 0  t  2 , then Note: The above result is true for every piecewise smooth simple closed curve C with the origin in its interior.

  50. 9.13 Surface Integrals DEFINITION 9.11 Let f be a function with continuous first derivatives fx, fyon a closed region. Then the area of the surface z=f(x,y) over R is given by (2) Surface Area

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