## anonymous 5 years ago find the volume of the region bounded by the graphs y = sqrt(x), y = 0, and x = 6, revolved about the line x = 6

1. amistre64

:) again :)

2. anonymous

hello! This problem is similar to the one we did together yesterday, but for some reason I am still doing this wrong

3. anonymous

I know that I am supposed to do the disk method, and that I need to solve for each equation in terms of y

4. amistre64

rewrite y= sqrt(x) in terms of x = f(y) y = sqrt(x) ; ^2 both side y^2 = x this we can solve for easily right?

5. anonymous

so I have v = pi (the integral from 0 to 6) (y+9)^2 - (9)^2 dx

6. amistre64

graph this and see that it is the same

7. anonymous

yes

8. amistre64

x = 6 is a pain, we want to move it to the x=0 line to examine it; how do we move x = 6 to x=0? x= 6 ; -6 from both sides x-6 = 0 do the same for the x=y^2 to keep it in the same position

9. anonymous

minus 6 from both sides in the equation I wrote?

10. amistre64

x = y^2 ; -6 both sides x-6 = y^2 -6

11. anonymous

whoops---I am revolving around x = 9, sorry about that

12. amistre64

brb..

13. anonymous

okay

14. amistre64

back...

15. anonymous

hi

16. amistre64

since it 9 we just -9 fro itall

17. anonymous

so v = pi (intergral from 0 to 6) (y + 9)^2 - 9^2 ?

18. amistre64

move it all to the left by 9 to examine it at x=0; does that make sense? Its over on a shelf and we want to examine it at the table where we can see it clearly; the table is the origin of the graph where x=0 and y=0, so we move it

19. anonymous

So now I am looking at a graph that is shifted backwards by 9 on my calculator

20. amistre64

y = sqrt(x) ; change to an equal form to accomodate for the y axis y^2 = x ; now move it over to the x=0 by subtracting 9 y^2 -9 = x-9 ; the x-9 can be rewritten as just x

21. anonymous

sqrt(x +9)

22. anonymous

okay so it is y^2 -9 not +9

23. amistre64

sqrt(x+9) is good; now reform it y = sqrt(x+9) y^2 = x+9 y^2 -9 = x

24. amistre64

whats our bounds now: for y=0 to how far?

25. anonymous

x = 6....?

26. anonymous

I've never seen anyone do transformations of a graph this way..

27. amistre64

y = sqrt(9) y = 3

28. anonymous

we have to redo them when we solve for a different variable...keep forgetting

29. amistre64

integrate the discs from 0 to 3

30. anonymous

okay

31. amistre64

pi {S} y^2-9 dy ; [0,3] = volume of revolution

32. amistre64

forgot to square the function lol

33. amistre64

pi {S} [y^2-9]^2 dy ; [0,3]

34. anonymous

so you don't have to subtract another 9?

35. amistre64

why would you subtract two 9s?

36. amistre64

we only move the graph over to x=0 from x=9 right?

37. anonymous

one from the first eqn and one from eqn y = 0?

38. amistre64

if i tellyou to move 9 steps to the left; what do you do? move 18?

39. anonymous

ohhhh

40. amistre64

y=0 is already bounding the volume; no need to adjust that part lol

41. anonymous

42. amistre64

dunno; lets check..

43. anonymous

no, its' not

44. amistre64

pi {S} (y^2-9)^2 dy pi {S} (y^4 -18y^2 +81) dy pi(y^5/5 -18y^3/3 +81y) at y=3

45. anonymous

46. anonymous

and got 648/5, but the homework grader marked it wrong

47. amistre64

i get 129.6pi

48. anonymous

same answer, and the wrong one. What did we do wrong?

49. anonymous

Hang on. You're rotating the graph of y= sqrt(x) from x=6 to x=9 about the line x=9 ? Is that right?

50. amistre64

243 27 243 pi( ---- - --- + ---) right? 5 3 1

51. amistre64

x = 0 to 9

52. anonymous

right, and we are rotating the region about the line x = 9

53. amistre64

ilove; retype the problem in its entirety so that we have a clear picture

54. anonymous

Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines. y = sqrt(x) y = 0 x = 6 about the line x = 9

55. amistre64

ohhh... so its a donut shape; i assumed you meant y=0; x=9 about the x=9

56. anonymous

57. anonymous

I don't think it's a doughnut

58. amistre64

its a torus alright :)

59. anonymous

More like a mountain.

60. amistre64

we have to subtract the middle of what we found to get the volume then

61. anonymous

oh wait. I'm looking at wrong graph

62. anonymous

Small hill. Not a doughnut. There is no hole in the middle.

63. amistre64

x=6 to x=9 is empty tho

64. anonymous

there is no hole it the middle? isn't the shape a bunt cake?

65. anonymous

no

66. anonymous

graph y = sqrt(x) from x = 6 to x=9 y = 0 to y = 3

67. anonymous

but the region is cut off at x = 6, and we have to get it to revolve about line x = 9

68. anonymous

that's where it starts

69. anonymous

not where it ends. it doesn't say it's bounded by x=0

70. anonymous

hmmm...then how to we caluclate where it begins and ends?

71. anonymous
72. amistre64

73. anonymous

If the problem said bounded by x=0, x=6 and y= 0 and y = sqrt(x) I would agree with that picture. It doesn't say x=0.

74. anonymous

so using what you just said, the boundries would be?

75. anonymous

x = 6, x = 9, y = 0, y = sqrt(x) and rotate around x = 9

76. amistre64

then we would double our results right?

77. anonymous

no, you have the wrong part of the graph I think.

78. amistre64

with this boundary; y = [0,sqrt(6)]

79. anonymous

You want to integrate over the white part in your picture rather than the grey part.

80. anonymous

what wouuld the equation look like?

81. anonymous

$v = \pi \int\limits_{0}^{3}...$

82. anonymous

You're using disks or shells?

83. anonymous

We can only use the disk method, right?

84. anonymous

where did you get the boundaries 0 to 3 from , again?

85. anonymous

do you have the value for the answer so we can be sure we're doing the right region?

86. anonymous

nope----online h/w problem

87. anonymous

dumb.

88. anonymous

I'm pretty sure I'm reading it right though. For this I would use cylindrical shells

89. amistre64

y = sqrt(x); curve of sqrt(x) y = 0 ; the x axis is y=0 x = 6 ; the left boundary x=9; axis to spin around

90. amistre64

x=6 the right bound

91. anonymous

x = 6 is the left bound in my interpretation of the problem, and the axis of rotation would be the right bound.

92. anonymous

93. anonymous

I thought the entire region was left bound, and the axis is right bound.

94. anonymous

like left-right or top-bottom

95. amistre64

the region to cut from my original; if im right is: is the cylindar of r = 3, and height = sqrt(6) 216 sqrt(6) pi ------------ 5

96. anonymous

okay, this is the right answer. Can you go over it one more time?

97. amistre64
98. amistre64

99. anonymous

So it was the doughnut?

100. anonymous

both are correct, but how do you do it correctly?

101. anonymous

They should have been more specific about the left boundary... grr..

102. amistre64

lol..... the only way we are both correct is if the volume of each version are identical :)

103. anonymous

I have not subtracted by 9sqrt(6) pi at all in what I have been doing though

104. anonymous

okay, so i should have$\pi \int\limits_{0}^{3}(....?....)$

105. amistre64

106. amistre64

your interval of [0,3] is wrong, since we only need the the sqrt(6)

107. anonymous

how do I know when to remove a cylinder?

108. amistre64

when the spin axis is not the same is the boundary axis; there is a gap that is formed

109. anonymous

So you've confirmed the answer is what amistre has?

110. amistre64

that gap is not a part of the volume when revolved so it must be cut out; like makeing a solid donut then removing the middle

111. anonymous

it's what both of you said. Now I am trying to determine what is the best way to approach this problem

112. anonymous

So I remove the middle when the region is bounded off by a number x = # that is smaller that the x = # that it is rotating about?

113. anonymous

it's not what I said if it's what he said. I have a very different answer.

114. amistre64

115. amistre64

when the spin is different from the bound, you get an offset that needs to be accounted for....

116. anonymous

.....

117. anonymous

polpak, with your graph, how did you attempt the problem?

118. anonymous

did you eliminate the middle part in the volume equation?

119. amistre64

polpak eliminated the outer part

120. anonymous

can I get this step by step? I am totally confused. I need to see an equation

121. anonymous

well we're still not clear on what problem you're solving ;p

122. amistre64

for mine: we did the stuff above and then I subtracted the middle in an equation like this: $\pi.\int\limits_{0}^{\sqrt{6}} (y^4 -18y^2+81) - (-3)^2.dy$

123. anonymous

Okay,that i what I was looking for. I had every thing except the -3. where did that come from?

124. amistre64

that is the y = -3 line that result from the boundary between x=6 and x=9

125. amistre64

teh 3 and 6 in my circle graph should be switched :) and then you can see the "3"

126. anonymous

OHHHHHHH

127. anonymous

okay, so now I finally get it. Thanks everyone! :D Thanks for staying on so long amistre64

128. amistre64

hey, if you dont figure it out, neither do I lol

129. anonymous

see ya later!

130. anonymous