## anonymous 4 years ago The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the y-axis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36. These are two separate rotation axis and answers.

1. amistre64

which method?

2. amistre64

|dw:1327531365366:dw|

3. amistre64

we need to know the point of intersection here

4. anonymous

Hey sorry, keyboard died. The point of intersection is (4,16).

5. amistre64

really? i see it as (6,36) meself (y=6) = (y=sqrt(x)) 6 = sqrt(x) at x=36

6. anonymous

|dw:1327531523131:dw|

7. amistre64

8. anonymous

It's a rotation of y=6, but the shape is bounded by y=4, so the intersection of y=4 and y=sqrt(x) would be (16,4).

9. anonymous

|dw:1327531686984:dw|

10. anonymous

So I'm thinking of using the cylindrical shell method, 2Pi(radius)(height)

11. amistre64

ok, since typos happen; lets review what youve posted: The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the y-axis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36

12. anonymous

Yes, that is correct.

13. amistre64

|dw:1327531783399:dw|

14. amistre64

then this is our graph of that region

15. amistre64

shell is fine, but we need to establish that this is our representation

16. anonymous

so you this is a graph for rotation around x=36?

17. amistre64

yes, in fact its the same graph for rotate y=6 AND x=36

18. anonymous

Oh oh

19. amistre64

the inverse of y=sqrt(x) that we can use for this is x=y^2 that will give us the shell height for y=6 rotation

20. anonymous

so when rotating around a Y-axis, we should put the curve as Y too?

21. amistre64

first we should simply draw out the graphs; after we have the graphs down, then we worry about where to rotate at

22. amistre64

the graph itself has nothing to do with what we rotate around to begin with

23. anonymous

true

24. anonymous

so the graph I had illustrated was inccorect

25. anonymous

visualizing the graphs has been the toughest part for me!

26. amistre64

|dw:1327532250390:dw|this is our shell representing rotate about y=6

27. amistre64

the graph you drew was something else altogther :)

28. anonymous

haha, I'm sorry.

29. anonymous

I HAVE NO CLUE WHY I PUT Y = 4!!

30. amistre64

'sok :) so$Area_{shell}=\int_{0}^{6}2pi\ y\ f(y)\ dy$ $Area_{shell}=2pi\int_{0}^{6} y\ y^2\ dy$ $Area_{shell}=2pi\int_{0}^{6} y^3\ dy$ we agree so far?

31. anonymous

|dw:1327532422099:dw|

32. amistre64

thats better lol

33. amistre64

i think i want to "move" our graph stuff to a better location casue I think im forgetting something; moving the graph changes no value, just positions

34. anonymous

Since we are rotating around y=6, wouldn't we do pi(6-y)^2

35. amistre64

|dw:1327532555854:dw|

36. amistre64

ugh, yes, your right; been awhile :) x=(y-6)^2

37. anonymous

so we could do the cross-section method

38. amistre64

$A=2pi\int_{0}^{6}y(y-6)^2dy$ $A=2pi\int_{0}^{6}y(y^2-12y+36)dy$ $A=2pi\int_{0}^{6}(y^3-12y^2+36y)dy$ looks better to me for the shell method.

39. anonymous

Ahhh, it does..

40. amistre64

disc would be: $A=pi\int_{0}^{36}(x^{1/2}-6)dx$

41. anonymous

|dw:1327532793666:dw|

42. anonymous

what i posted above is for the cross-section, or it works for it..

43. amistre64

$A=pi\int_{0}^{36}(x^{1/2}-6)^2dx$ forgot to r^2 the area :)

44. amistre64

the disc gives me: 216 pi the shell gives me: 216 pi

45. anonymous

That is right!

46. amistre64

yay!!

47. anonymous

ok so I'm still a little confused how we got there..

48. amistre64

lets turn this on its side so it looks like an upsidedown cone; the math doesnt care how it looks :)

49. anonymous

2pi(r)(h) 2pi$2\pi \int\limits_{0}^{6} y(6-y)$

50. amistre64

|dw:1327533114400:dw|

51. amistre64

good, r = y and h=f(y), which is the inverse of f(x) which we moved down by 6

52. amistre64

y=sqrt(x)-6 y+6 = sqrt(x) (y+6)^2 = x = f(y) I used the other side of the line so I got (y-6)^2 which makes no difference here

53. amistre64

well, it does according to wolfram but .... fer some reason i did it right to begin with lol

54. anonymous

hmm.. so what would be the equation we are trying to integrate and with what limits? 0 to 36?

55. amistre64

Volume, not Area; named it wrong .... it bites getting old$V=2\pi\int_{0}^{-6}y(y+6)^2dy=216\pi$

56. amistre64

the 0 to 36 is for the disc method for this thing

57. amistre64

i need to downlaod google chrome; this thing is just way too slow in IE

58. anonymous

haha it runs great in chrome

59. anonymous

so why is the (y+6) being squared?

60. anonymous

Thanks so much for your help, you are great!

61. amistre64

college computers here clear out and reboot so any downloads are erased

62. amistre64

w have to tell the integration to look at the the proper rendition of out equation for height

63. anonymous

ohhhhh I see!! I was looking at the limit from x=y instead of the axis of revolution.

64. amistre64

the direction we are looking at for the shell measuers up from the y axis, or over ; but we have to rewrite the height line in terms of y for it to make sence in out integration using shells

65. amistre64

since w is the inverse of a normal xy graph we inverse our y=sqrt(x) to match it

66. amistre64

since our view is the inverse ....

67. anonymous

do you mean height instead of width?

68. anonymous

I have the correct answer on my paper now, sweet!

69. anonymous

Could you help me do a couple more? :)

70. amistre64

i can, i got the chrome downlaoded now; doesnt help the typos, but its a bit quicker lol

71. anonymous

cool, should I start an entirely new question so I can give you another medal?

72. amistre64

not that the medals matter, but starting a new one might help my computer to keep up the pace :)

73. amistre64

we didnt get thru rotate about the x=36 yet tho, so lets do that one on another post

74. anonymous

Alright I'll do that!