## anonymous 5 years ago Consider the given curves to do the following. y=5sqrt(x) y=0 x=1 Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about x = -3.

1. amistre64

wha!?!

2. amistre64

:) just kiddin' this ones easy too

3. anonymous

yup :( i went on this online tutor thing from the hw website i use, they got me set up but i think i have the boundaries wrong

4. anonymous

lol xD the integral they told me to use is

5. anonymous

$2\pi \int\limits5x^(3/2) + 15x^(1/2)$

6. anonymous

sorry it should be 5x^(3/2) and 15x^(1/2)

7. anonymous

use {} when using exponent

8. anonymous

kay i will

9. anonymous

i thought the boundaries for the integral would be 1 and -3, but when i plug in -3 to the integral itll give me trouble lol

10. amistre64

11. amistre64

this is a rough idea of what you got and since we are spinning around an x; we integrate with respect to y.....lets move that x line; and I got the rotation axis in the wrong spot

12. anonymous

so if we integrate with respect to y, that means we have to get x = in terms of y ?

13. amistre64

like this

14. amistre64

the shells have a radius along the x; so lets stick to that.... i had to draw the pic for clarity in me head :)

15. amistre64

lets +3 to all the 'x' values; to move this thing; NOT f(x) + 3; but f(x+3)

16. anonymous

17. amistre64

y=5sqrt(x+3) y=0 = x axis x=1+3; you see that instruction to spin about x=-3? get that to zero dear lol

18. anonymous

LOL oh. haha thank youu

19. amistre64

our inteval is then ; [0+3,1+3], or simply; [3,4] right?

20. anonymous

yes

21. amistre64

Our farthest Raduis =4 and our inner radius = 3 so we integrate from r to R roght?

22. anonymous

so like lets say the instruction told us to spin about x = 1, would we subtract 1 from the x values?

23. anonymous

and correct

24. amistre64

and our height only changes as 5sqrt(x+3) changes now; so lets do this :)

25. amistre64

make that 5sqrt(x-3) ; you know why?

26. anonymous

mmm, im not too sure why

27. amistre64

we want 5sqrt(0) to equal 5sqrt(3+___) so remember that we moved it but we dont want to change its value; just its position

28. amistre64

the value at 3 away from center of rotation nedds to be the same regardless of where we put this thing right?

29. anonymous

correct

30. amistre64

so; lets integrate this thing like this: 2pi {S} x*5sqrt(x-3) dx ; [3,4]

31. anonymous

you see the 5sqrt(x-3)

32. amistre64
33. anonymous

can we put it as 5(x-3)^1/2

34. amistre64

i see it :) we can pull out that 5 right? nothing is stopping that from not being a constant

35. anonymous

ooh that makes it easier than what i was about to do lol

36. amistre64

we can put that lonely x inside the sqrt

37. amistre64

x*sqrt(x-3) = sqrt(x^2(x-3)) (x^3 -3x^2)^(1/2) right?

38. anonymous

can i ask one thing though. can we use that integral of 2pi x f(x) dx for any cylindrical shell problem? because when i went to the hw tutor from the hw website, they told me to use the 2pi r h or something like that

39. amistre64

r = x or y depending on if you spin it up or sideways h = f(x) regardless

40. amistre64

well; maybe h can = f(y) too lol

41. anonymous

ohh so they're basically the same thing, just different way of putting it

42. anonymous

x*sqrt(x-3) = sqrt(x^2(x-3)) (x^3 -3x^2)^(1/2) right? and yes lol xD

43. amistre64

yes; its like calling your dad by his given name; hell respond to either right? the person is the same, the name can change

44. anonymous

haha nice comparison xD

45. amistre64

we could integrate this numercally instead; with like the simpsons rule or trap rule if trying to find a morph that works proves too difficult

46. anonymous

i havent learned those rules yet :(

47. amistre64

you have, you just havent learned how to apply them yet lol its like this shell methid; you already knew how to find the area of a sheet of paper; but dint realize how that fact could aid you

48. anonymous

lol oh xD im sorry, im still a little confused on why we added 3. i understand to get x=0 but im still confused on why we did it.

49. amistre64

becasue in analysis it is always easier to bring things to a standard working table that is right in front of you. If we dont move it, it is stuck up high on a shelf and that is no way to examine it right? so, we bring it to the table by moving it off the shelf

50. amistre64

we change its position ; but retain its value

51. amistre64

this is the basic of the trapazoid rule

52. anonymous

you mean everything is the same, the value is the same, its just on a different point on the graph?

53. amistre64

exactly :)

54. amistre64

so the answer we receive from it will be the same; if you move your breakfast form the stove to your plate to the table; what has changed?

55. anonymous

the place of where its at lol xD i gotcha, thank you for clarifying that for me. i was so confused before lol :p

56. amistre64

lets see if we can int this function :) x*sqrt(x-3) then :) any tricks up your sleeve thatll be useful for us?

57. amistre64

ohohoh...maybe int by parts lol

58. amistre64

u = x and v = sqrt(x-3) ??

59. anonymous

are we using substitution right now?

60. amistre64

its a kind of subsitution based on the product rule for derivatives

61. amistre64

remember the product rule? Dx(uv) = uv' + vu' right?

62. anonymous

yesss i remember that :)

63. anonymous

so it would be equal to x/2 (x-3)^-1/2 + sqrt of x-3 ?

64. amistre64

{S} (uv)' = {S} u v' + {S} v u' right? uv = {S} u dv + {S} v du

65. amistre64

uv - {S} v du = {S} u dv

66. amistre64

we want to integrate " u = x sqrt(x-3)" and it has 2 parts; so we use this trick to find it lol

67. anonymous

how did you get from {S} (uv)' = {S} u v' + {S} v u' right? to uv = {S} u dv + {S} v du

68. amistre64

69. amistre64

the {S} of a derivate (y') = y; {S} y' = y; right?

70. anonymous

correct

71. amistre64

v' = dv; just different notations

72. amistre64

lets focus on that part of my "u" column... we start with: (x-3)^(1/2) ; and that ints up to: 2/3 (x-3)^(3/2) ; in tthat again... pull out the 2/3 and do: (x-3)^(3/2) goes to what?

73. amistre64

(2/5) (x-3)^(5/2) right?

74. anonymous

so we take the antiderivative twice for it?

75. amistre64

$\int\limits_{}(x.\sqrt{x-3}).dx = [\frac{2x}{3}.\sqrt{(x-3)^3}] - [\frac{4}{15}.\sqrt{(x-3)^{5}}]$

76. amistre64

right?

77. amistre64

we integrate it till we can stop pretty much

78. amistre64

if we pick a value for u that derives to 0 then it doesnt matter what v suits up to cause when it gets to u=0 its done

79. anonymous

okay i see what you mean now

80. amistre64

u =x u' = 1 u'' = 0 right?

81. anonymous

correct

82. amistre64

v = sqrt(...) {S} v = {S} sqrt(...) {S} [{S} v] = {S} [{S} sqrt(...)]

83. amistre64

+u * {S} v - u' * {S} [{S} v] + 0 * ...doesnt matter lol

84. amistre64

this is the method known as integration by parts; and it can get tricky but the key is to pick smart values for u and v to begin with and itll work itself out in the end :)

85. amistre64

int by parts is just using the product rule as a basis; and then using it again, and again, till you get to 0 lol

86. anonymous

seems so hard :( lol but i think i got a jist of what you mean. thank you for all your help and for staying and explaining all of it to me. :))

87. amistre64

youre welcome :) the more i explain it the better i understand it meself ;)

88. amistre64

its says your typing.... its like waiting for the waffle iron