## anonymous 5 years ago Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the y-axis. y = x^2 y=0 x=1

1. amistre64

HA!! shells, i knew they were comin

2. amistre64

since we is going around the y axis; its simpler right now

3. anonymous

yup :( i never used this method before ><

4. amistre64

the radius of each shell moves from x=0 to x=1 right?

5. anonymous

yup

6. amistre64

and the height of each shell is determined by: f(x)

7. anonymous

wait, i know why its x=1, but why is it x=0

8. amistre64

so the area of each shell, when flattened out is height* width...

9. amistre64

umm... y axis means x=0 they are the same thing

10. anonymous

ohh okay gotcha

11. amistre64

the height = f(x) the width = 2pi [x] the area of any given shell = 2pi x[f(x)] so integrate that from [a,b] = [0,1] in this case

12. amistre64

2pi [S] x(x^2) dx ; [0,1] 2pi x^4 ------- = (pi x^4)/4 = pi/4 4

13. anonymous

how did you get rid of the 4th power in the numerator?

14. amistre64

15. amistre64

you dont... why would you?

16. anonymous

but you put (pi x^4)/4 = pi/4 where did the 4th power go?

17. amistre64

F(0) = 0 F(1) = pi/4 F(1) - F(0) = pi/4

18. amistre64

over 2 lol

19. amistre64

i forgot i changed 2/4 to 1/2.... doh!

20. amistre64

$\frac{2\pi x^4}{4} = \frac{\pi x^4}{2}$

21. anonymous

much clearer, lol thank you :D

22. amistre64

when im wrong; say im wrong...not ask me how im right lol. I just assume im right ;)

23. anonymous

lol will do xD

24. amistre64

you understand the mechanics of the shell method tho?

25. anonymous

all the methods like the washer, disk, etc all those methods seem so similiar to eachother and i get them all confused. because in the end youre always taking the antiderivative and plugging in the boundaries right?

26. anonymous

like i understand theres certain formulas you have to follow for each method, which i have to memerize lol all of them seem really similiar

27. amistre64

28. anonymous

so for the shell method, basically youre taking half of it, and spreading it out which is why it turns to be a rectangle shape?

29. amistre64

when we flatten out the "shell" we get a flat rectangular piece that the area is eay to determine right?

30. anonymous

correct

31. amistre64

the width = 2pi x...can you tell me why?

32. amistre64

we aint using half the shell, but the whole thing

33. anonymous

i think it has to do with the radius, is that why the width is 2 pi x?

34. anonymous

ohh is it the circumference of the cylnder?

35. amistre64

it does, but i want to make sure you understand this :) its very basic and easy...exactly lol

36. anonymous

lol i took a careful look at the cylinder and realized haha xD

37. amistre64

you did good :)

38. amistre64

the shell method makes things easier for some problems

39. anonymous

thanks ;) so basically for the shell method we're plugging in f(x) into the formula and taking the antiderivative of it? then plug in the boundaries and solve?

40. amistre64

lets think that thruough; we want to add up all the areas...which is what intagrating does; each area = 2pi x[f(x)]

41. amistre64

{S} 2pi x[f(x)] dx is what we do right?

42. amistre64

if there is a constant we can pull it aside right?

43. anonymous

mhmm

44. anonymous

the 2pi

45. amistre64

2 pi goes out and what are we left with inside?

46. anonymous

x*f(x)

47. amistre64

excatly; so we integrate 'x*f(x)'

48. amistre64

do you se how that differes from your first statement of ; we int f(x)?

49. anonymous

yesss, thanks for clearing that up for me :) this method seems a bit easier than the others lol

50. amistre64

look at the disc method

51. anonymous

oh, i have one question. when you solve for an integral, is it possible for it to be negative? or does it always have to be positive

52. amistre64

the disc method take all the areas of a given circle; with a radius of f(x) and adds them up

53. amistre64

volume is always a positive number

54. amistre64

what is the formula for the the area of a circle ?

55. anonymous

uh oh. darn i should of took the absolute value. i was so sure i did it right too. lol but thank you :)

56. anonymous

and area of circle is pi r^2

57. amistre64

so when we add up all the areas of the cicles thru integration we do: {S} pi [f(x)]^2 dx right? adding up the areas of circles

58. anonymous

59. amistre64

look at the drawing i did in this one

60. amistre64

and tell me what the radius of each given circle is?

61. anonymous

oh, f(x) xD haha

62. amistre64

that is all these volume of rotation problems amopunt to :)

63. anonymous

i remember there was this one method i think where you had to divide by two, is there a method that involves that? or am i making stuff up lol :p

64. amistre64

i dont think thats a volume of rotation one :) in some area propblems we are asked to find the area under a curve that hopes the y axis... and if the symmetry is the same from left to right side of the y axis; we get a skewed result

65. amistre64

lets try something simple and see; suppose we want to find the area of a square that is sitting halfway between the sides

66. amistre64

lets say 4 high and 4 wide; the area should be 16 right?

67. anonymous

correct

68. amistre64

if we integrate f(x) = y = 4; from -2 to 2 we know we should get 16 right?

69. anonymous

how come its from -2 and 2?

70. amistre64

{S} 4 dx -> 4x 4(2) - 4(-2) = 8 -- 8 = 16, so that one doesnt apply to the /2 thing you discussed lol

71. amistre64

the distance from -2 to 2 = 4 right?

72. anonymous

yupp

73. amistre64

and a square with sides = 4 would straddle the y axis would sit from -2 to 2 on the x axis right?

74. anonymous

correct. i gotcha now :) lol

75. anonymous

if the prob asked to rotate around the x-axis, its generally the same idea of solving it like we did with the y axis right?

76. amistre64

i pic is worth a thousand words :)

77. anonymous

it really is. thank you for that :D

78. amistre64

yes, but we just need to make sure we are rotating it properly; make sure the numbers come out correctly

79. amistre64

once we can draw a picture, the rest is just intuitive

80. anonymous

gotcha. i shall try and attempt the rest of these shell method questions. thank you so so much for all your help. :D you make it possible for me to understand calculus lol xD thank you times infinity !

81. amistre64

youre welcome :)