23 Double Integral over Polar Coordinate.pptx

Double Integral over Polar Coordinate

1. Double Integrals Over Polar Equations

2. Double Integrals Over Polar Equations * Double Integrals Over Regions Defined in Polar Coordinates * The Cylindrical Coordinates

3. Given a domain D defined by r 1 ( ) and r 2 () where r 1 < r 2 and A <  < B as shown in the figure, D = {(r,  )| r 1 < r < r 2 , A <  < B}. Double Integrals Over Polar Equations Y Z A B X D r 2 () r 1 ()

4. Double Integrals Over Polar Equations Y Z A B X Partition the xy-plane with Δ r and Δ . D r 2 () r 1 () Given a domain D defined by r 1 ( ) and r 2 () where r 1 < r 2 and A <  < B as shown in the figure, D = {(r,  )| r 1 < r < r 2 , A <  < B}.

5. … Double Integrals Over Polar Equations Partition the xy-plane with Δ r and Δ . Given a domain D defined by r 1 ( ) and r 2 () where r 1 < r 2 and A <  < B as shown in the figure, D = {(r,  )| r 1 < r < r 2 , A <  < B}. Y Z A B X D r 2 () r 1 ()

6. … Δ r Double Integrals Over Polar Equations Partition the xy-plane with Δ r and Δ . Given a domain D defined by r 1 ( ) and r 2 () where r 1 < r 2 and A <  < B as shown in the figure, D = {(r,  )| r 1 < r < r 2 , A <  < B}. Y Z A B X D r 2 () r 1 ()

7. … Δ  Δ r Double Integrals Over Polar Equations Partition the xy-plane with Δ r and Δ . Given a domain D defined by r 1 ( ) and r 2 () where r 1 < r 2 and A <  < B as shown in the figure, D = {(r,  )| r 1 < r < r 2 , A <  < B}. Y Z A B X D r 2 () r 1 ()

8. arbitrary Δ r Δ – tile as shown and l et (r * ,  * ) be an arbitrary point in the tile and Δ A be area of the tile, then Δ A ≈ r * Δ r Δ . … Δ  Δ r ( r* , *) r * Double Integrals Over Polar Equations Partition the xy-plane with Δ r and Δ . Select an Given a domain D defined by r 1 ( ) and r 2 () where r 1 < r 2 and A <  < B as shown in the figure, D = {(r,  )| r 1 < r < r 2 , A <  < B}. Y Z A B X D r 2 () r 1 ()

9. arbitrary Δ r Δ – tile as shown and l et (r * ,  * ) be an arbitrary point in the tile and Δ A be area of the tile, then Δ A ≈ r * Δ r Δ . (HW. Why? In fact Δ A = r * Δ r Δ  if r * is the center.) … Δ  Δ r ( r* , *) r * Double Integrals Over Polar Equations Partition the xy-plane with Δ r and Δ . Select an Given a domain D defined by r 1 ( ) and r 2 () where r 1 < r 2 and A <  < B as shown in the figure, D = {(r,  )| r 1 < r < r 2 , A <  < B}. Y Z A B X D r 2 () r 1 ()

10. Double Integrals Over Polar Equations Let z = f(r, ) ≥ 0 be a function over D. Y Z A B X z = f(r, ) D r 2 () r 1 ()

11. The volume of the column over one tile having z as the cover is Δ V ≈ f(r * ,  * ) r * Δ r Δ . Double Integrals Over Polar Equations Y A B X Let z = f(r, ) ≥ 0 be a function over D. D Z z = f(r, ) r 2 () r 1 ()

12. The volume of the column over one tile having z as the cover is Δ V ≈ f(r * ,  * ) r * Δ r Δ . lim Σ ( Δ V) Δ r, Δ  0 Double Integrals Over Polar Equations Y A B X Let z = f(r, ) ≥ 0 be a function over D. D Z Hence the volume V over D is the sum of all z = f(r, ) r 2 () r 1 ()

13. The volume of the column over one tile having z as the cover is Δ V ≈ f(r * ,  * ) r * Δ r Δ . lim Σ ( Δ V) = lim Σ f(r * ,  * )r * Δ r Δ  = ∫∫ f(r, )dA. Δ r, Δ  0 Δ r, Δ  0 Δ A Double Integrals Over Polar Equations Y A B X Let z = f(r, ) ≥ 0 be a function over D. D D Z Hence the volume V over D is the sum of all z = f(r, ) r 2 () r 1 ()

14. The volume of the column over one tile having z as the cover is Δ V ≈ f(r * ,  * ) r * Δ r Δ . lim Σ ( Δ V) = lim Σ f(r * ,  * )r * Δ r Δ  = ∫∫ f(r, )dA. Δ r, Δ  0 Δ r, Δ  0 ∫ ∫ f(r, ) r drd, which is V, r=r 1 ( ) r 2 ( ) =A B Δ A Double Integrals Over Polar Equations Y A B X Let z = f(r, ) ≥ 0 be a function over D. D D Z Hence the volume V over D is the sum of all z = f(r, ) in term of drd, we have r 2 () r 1 () ∫∫ f(r, )dA = D

15. The volume of the column over one tile having z as the cover is Δ V ≈ f(r * ,  * ) r * Δ r Δ . lim Σ ( Δ V) = lim Σ f(r * ,  * )r * Δ r Δ  = ∫∫ f(r, )dA. Δ r, Δ  0 Δ r, Δ  0 ∫ ∫ f(r, ) r drd, which is V, r=r 1 ( ) r 2 ( ) =A B Δ A Double Integrals Over Polar Equations Y A B X Let z = f(r, ) ≥ 0 be a function over D. D D Z Hence the volume V over D is the sum of all z = f(r, ) in term of drd, we have = volume of the solid over D = {A <  < B; r 1 < r < r 2 }. r 2 () r 1 () ∫∫ f(r, )dA = D

16. Cylindrical Coordinates The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

17. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. Cylindrical Coordinates The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

18. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. 3 120 o x y z Cylindrical Coordinates The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

19. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. 3 120 o 4 x y z (3, 120 o , 4 ) Cylindrical Coordinates The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

20. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. 3 120 o 4 x y z (3, 120 o , 4 ) x = 3cos( 120 o ) = –3/2 y = 3sin(120 o ) = 3 Hence the point is (–3/2, 3, 4) Cylindrical Coordinates The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

21. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. 3 120 o 4 x y z (3, 120 o , 4 ) x = 3cos( 120 o ) = –3/2 y = 3sin(120 o ) = 3 Hence the point is (–3/2, 3, 4) b. Convert (3, –3, 1) into to cylindrical coordinate. Cylindrical Coordinates The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

22. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. 3 120 o 4 x y z (3, 120 o , 4 ) x = 3cos( 120 o ) = –3/2 y = 3sin(120 o ) = 3 Hence the point is (–3/2, 3, 4) b. Convert (3, –3, 1) into to cylindrical coordinate. Cylindrical Coordinates The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

23. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. 3 120 o 4 x y z (3, 120 o , 4 ) x = 3cos( 120 o ) = –3/2 y = 3sin(120 o ) = 3 Hence the point is (–3/2, 3, 4) b. Convert (3, –3, 1) into to cylindrical coordinate.  = 315 o , r = 9 + 9 = 18 Hence the point is (18, 315 o , 1) the cylindrical coordinate. Cylindrical Coordinates The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

24. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. 3 120 o 4 x y z (3, 120 o , 4 ) x = 3cos( 120 o ) = –3/2 y = 3sin(120 o ) = 3 Hence the point is (–3/2, 3, 4) b. Convert (3, –3, 1) into to cylindrical coordinate. ( 18, 315 o , )  = 315 o , r = 9 + 9 = 18 Hence the point is (18, 315 o , 1) the cylindrical coordinate. x y Cylindrical Coordinates z The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

25. Example A. a. Plot the point (3,120 o , 4) in cylindrical coordinate. Convert it to rectangular coordinate. 3 120 o 4 x y z (3, 120 o , 4 ) x = 3cos( 120 o ) = –3/2 y = 3sin(120 o ) = 3 Hence the point is (–3/2, 3, 4) b. Convert (3, –3, 1) into to cylindrical coordinate. ( 18, 315 o , )  = 315 o , r = 9 + 9 = 18 Hence the point is (18, 315 o , 1) the cylindrical coordinate. x y Cylindrical Coordinates z (18, 315 o , 1 ) = (3, –3, 1 ) 1 The cylindrical coordinate system is the combination of using polar coordinates for points in the xy–plane with z as the 3 rd coordinate.

26. Cylindrical Coordinates The constant equations r = k describes cylinders of radius k, thus the name "cylindrical coordinate".

27. The constant equations r = k describes cylinders of radius k, thus the name "cylindrical coordinate". Example B. a. Sketch r = 2 2 Cylindrical Coordinates x y z

28. The constant equations r = k describes cylinders of radius k, thus the name "cylindrical coordinate". x y Example B. a. Sketch r = 2 2 The constant equations  = k describes the vertical plane through the origin, at the angle k with x-axis. b. Sketch θ = 3 π /4 3 π /4 Cylindrical Coordinates z x y z

29. The constant equations r = k describes cylinders of radius k, thus the name "cylindrical coordinate". x y Example B. a. Sketch r = 2 2 The constant equations  = k describes the vertical plane through the origin, at the angle k with x-axis. b. Sketch θ = 3 π /4 3 π /4 Cylindrical Coordinates z x y z

30. Cylindrical Coordinates Graphs may be given in the cylindrical coordinate using polar coordinates for points in the domain in the xy–plane and z = f(r,  ). Here are some examples of cylindrical graphs.

31. Cylindrical Coordinates Graphs may be given in the cylindrical coordinate using polar coordinates for points in the domain in the xy–plane and z = f(r,  ). Here are some examples of cylindrical graphs. Example C. a. z = sin(r) , D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π } → 0 ≤ z ≤ 1

32. Cylindrical Coordinates Graphs may be given in the cylindrical coordinate using polar coordinates for points in the domain in the xy–plane and z = f(r,  ). Here are some examples of cylindrical graphs. Example C. a. z = sin(r) , D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π } → 0 ≤ z ≤ 1 y z = sin(r) x ( π , 0, 1)

33. Cylindrical Coordinates Graphs may be given in the cylindrical coordinate using polar coordinates for points in the domain in the xy–plane and z = f(r,  ). Here are some examples of cylindrical graphs. Example C. a. z = sin(r) , D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π } → 0 ≤ z ≤ 1 y z = sin(r) b. z =  sin(r), D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π } → 0 ≤ z ≤ 2 π x ( π , 0, 1)

34. Cylindrical Coordinates Example C. a. z = sin(r) , D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π } → 0 ≤ z ≤ 1 x y z = sin(r) x y z = sin(r) b. z =  sin(r), D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π } → 0 ≤ z ≤ 2 π ( π /2, 2 π , 2 π ) p ( π , 0, 1) Graphs may be given in the cylindrical coordinate using polar coordinates for points in the domain in the xy–plane and z = f(r,  ). Here are some examples of cylindrical graphs.

35. Double Integrals Over Polar Equations D D Let’s set up the volume calculation for both solids. I II

36. Double Integrals Over Polar Equations The base D D D r = π Let’s set up the volume calculation for both solids. We note that D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π }. I II

37. ∫ ∫ f(r, ) r drd r=r 1 ( ) r 2 ( ) =A B Double Integrals Over Polar Equations Vol(I) = The base D D D r = π Let’s set up the volume calculation for both solids. We note that D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π }. Using the formula I II V = Vol(II) = , their volumes are

38. ∫ ∫ f(r, ) r drd r=r 1 ( ) r 2 ( ) =A B Double Integrals Over Polar Equations Vol(I) = The base D D D r = π Let’s set up the volume calculation for both solids. We note that D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π }. Using the formula I II V = ∫ ∫ sin(r) r drd, r=0 π =0 2 π Vol(II) = , their volumes are

39. ∫ ∫ f(r, ) r drd r=r 1 ( ) r 2 ( ) =A B Double Integrals Over Polar Equations Vol(I) = The base D D D r = π Let’s set up the volume calculation for both solids. We note that D = {0 ≤ r ≤ π , 0 ≤  ≤ 2 π }. Using the formula I II V = ∫ ∫ sin(r) r drd, r=0 π =0 2 π Vol(II) = ∫ ∫  sin(r) r drd. (HW: Finish the problems.) r=0 π =0 2 π , their volumes are

40. Example D. Let z = f(r, ) = cos () over the domain D = {(r, )| 0 < r < sin( ), 0 ≤  ≤ π /2}. Find the volume of the solid defined by z over D. Double Integrals Over Polar Equations x

41. Example D. Let z = f(r, ) = cos () over the domain D = {(r, )| 0 < r < sin( ), 0 ≤  ≤ π /2}. Find the volume of the solid defined by z over D. Double Integrals Over Polar Equations 1 D r=sin( ) The domain D = {0 < r < sin( ), 0 ≤  ≤ π /2} is the right half of a circle as shown. x

42. Example D. Let z = f(r, ) = cos () over the domain D = {(r, )| 0 < r < sin( ), 0 ≤  ≤ π /2}. Find the volume of the solid defined by z over D. Double Integrals Over Polar Equations 1 D r=sin( ) The domain D = {0 < r < sin( ), 0 ≤  ≤ π /2} is the right half of a circle as shown. The surface z = cos () makes a 90 o twist and the solid with base D below z are shown here. x y x z = f(r, ) = cos() D (r, 0, 1)

43. Example D. Let z = f(r, ) = cos () over the domain D = {(r, )| 0 < r < sin( ), 0 ≤  ≤ π /2}. Find the volume of the solid defined by z over D. Double Integrals Over Polar Equations 1 D r=sin( ) The domain D = {0 < r < sin( ), 0 ≤  ≤ π /2} is the right half of a circle as shown. The surface z = cos () makes a 90 o twist and the solid with base D below z are shown here. x y x x z = f(r, ) = cos() D D y (r, 0, 1) (r, 0, 1)

44. = ∫ ∫ cos( )r drd r=0 =0 Convert the integral to iterated integral, we get ∫∫ cos( )dA D r=sin( ) π /2 = ∫ cos( )r 2 /2 | d r=0 =0 π /2 sin( ) = ½ ∫ cos( )sin 2 ()d =0 π /2 Change variable, set u = sin( ). = sin 3 ()/6 | =1/6 =0 π /2 Double Integrals Over Polar Equations y x x z = f(r, ) = cos() D D y (r, 0, 1) (r, 0, 1) 1 D r=sin( ) x

45. Example E. Evaluate by converting it into polar integral Double Integrals Over Polar Equations ∫ ∫ (x 2 + y 2 ) 3/2 dydx y= - 4 – x 2 -2 2 y= 4 – x 2

46. Example E. Evaluate by converting it into polar integral The domain D is: r=2 Double Integrals Over Polar Equations ∫ ∫ (x 2 + y 2 ) 3/2 dydx y= - 4 – x 2 -2 2 y= 4 – x 2

47. Example E. Evaluate by converting it into polar integral The domain D is: r=2 Its defined by the polar equation r = 2 & r=0 Double Integrals Over Polar Equations ∫ ∫ (x 2 + y 2 ) 3/2 dydx y= - 4 – x 2 -2 2 y= 4 – x 2

48. Example E. Evaluate by converting it into polar integral The domain D is: (x 2 + y 2 ) 3/2 = (r 2 ) 3/2 = r 3 in polar form. r=2 Its defined by the polar equation r = 2 & r=0 Double Integrals Over Polar Equations ∫ ∫ (x 2 + y 2 ) 3/2 dydx y= - 4 – x 2 -2 2 y= 4 – x 2

49. Example E. Evaluate by converting it into polar integral The domain D is: (x 2 + y 2 ) 3/2 = (r 2 ) 3/2 = r 3 in polar form. Hence the integral is r=2 Its defined by the polar equation r = 2 & r=0 ∫ ∫ r 3 * rdrd r= 0 2 π r= 2 Double Integrals Over Polar Equations ∫ ∫ (x 2 + y 2 ) 3/2 dydx y= - 4 – x 2 -2 2 y= 4 – x 2

50. Example E. Evaluate by converting it into polar integral The domain D is: (x 2 + y 2 ) 3/2 = (r 2 ) 3/2 = r 3 in polar form. Hence the integral is r=2 Its defined by the polar equation r = 2 & r=0 ∫ ∫ r 3 * rdrd r= 0 2 π r= 2 ∫ r 5 /5 | d r= 0 2 π r= 2 = Double Integrals Over Polar Equations ∫ ∫ (x 2 + y 2 ) 3/2 dydx y= - 4 – x 2 -2 2 y= 4 – x 2

51. Example E. Evaluate by converting it into polar integral The domain D is: (x 2 + y 2 ) 3/2 = (r 2 ) 3/2 = r 3 in polar form. Hence the integral is r=2 Its defined by the polar equation r = 2 & r=0 ∫ ∫ r 3 * rdrd r= 0 2 π r= 2 ∫ r 5 /5 | d r= 0 2 π r= 2 = ∫ 32/5 d 2 π = Double Integrals Over Polar Equations ∫ ∫ (x 2 + y 2 ) 3/2 dydx y= - 4 – x 2 -2 2 y= 4 – x 2

52. Example E. Evaluate by converting it into polar integral The domain D is: (x 2 + y 2 ) 3/2 = (r 2 ) 3/2 = r 3 in polar form. Hence the integral is ∫ ∫ (x 2 + y 2 ) 3/2 dydx y= - 4 – x 2 -2 2 y= 4 – x 2 r=2 Its defined by the polar equation r = 2 & r=0 ∫ ∫ r 3 * rdrd r= 0 2 π r= 2 ∫ r 5 /5 | d r= 0 2 π r= 2 = ∫ 32/5 d 2 π = = 64 π /5 Double Integrals Over Polar Equations

53. For more integration examples of changing from dxdy form to the polar r drd r –form may be found at the following link: Double Integrals Over Polar Equations http://ltcconline.net/greenl/courses/202/multipleIntegration/doublePolarIntegration.htm