## anonymous 5 years ago Find volume of revolution for region bounded by y = x^2, y = x , in first quadrant, revolve it about the line x = 6 using shell method

1. anonymous

i think i got it , integral 2pi ( x - x^2) (6-x) dx from 0 to 1 ?

2. anonymous

the part i dont get is the 6 - x

3. anonymous

4. anonymous

like with the washer method , integral pi f(x)^2 - pi (g(x)) ^2 , where f(x) > g(x)

5. anonymous

where exactly does 6-x come from

6. anonymous

volume will be $2\pi \int\limits_{0}^{1} (5+x)(x-x^2).dx -\pi 5^2 .1$

7. anonymous

i believe thats wrong

8. anonymous

integral 2pi ( x - x^2) (6-x) dx from 0 to 1

9. anonymous

i'm giving you the graph...

10. anonymous

thats not correct, what do you get as an answer

11. watchmath

I agree with cantor set. The 6-x coming from the distance from a slice to the axis of symmetry.

12. anonymous

first find the volume made by revolving only y=x, then subtract the volume made by y=x^2. and that also due to the cylinder of radius 5.

13. anonymous

, watchmath, what about this question Determine the volume of the solid obtained by rotating the region bounded by x^(1/3), x=8, y=0, about the x-axis.

14. anonymous

$\pi \int\limits_{x=0}^{8}y^2 dx$ where y=x^1/3 $\pi \int\limits_{0}^{8}x ^{2/3}dx =3\pi 8^{5/3}/2=48\pi$

15. anonymous

thats wrong too

16. anonymous

17. anonymous

the question is here http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

18. anonymous

sorry i did a mistake in doing the integration...$\pi \int\limits_{0}^{8}x ^{2/3}dx=3\pi (x^{5/3}|_{0}^{8})/5=3\pi 8^{5/3}/5$$=3\pi 2^5 /5 =96\pi /5$

19. anonymous

oh nevermind

20. anonymous

ok small error

21. anonymous

but i should have said, use shell method :)

22. anonymous

i make it 48pi

23. anonymous

shell method stumps me on this problem

24. watchmath

Shell method $$\int_0^2 2\pi y (8-y^3)\, dy$$

25. anonymous

$2\pi \int\limits_{x=0}^{8}y (8-x)dy$

26. anonymous

yeah, how come its 8-y^3

27. anonymous

use x=y^3

28. anonymous

29. anonymous

sorry - i made the same mistake as dipank

30. anonymous

how come normally we dont do 8 - y^3 , like shell method for region bounded by y = x^2 , x axis, y axis

31. anonymous

you need one more boundary...

32. watchmath

Because the axis of rotation now is the x-axis (before the rotation axis was parallel to the y-axis)

33. anonymous

is there a similiar problem to this revolving about y axis

34. watchmath

Assuming all the information are the same but now we rotate about the y-axis then the volume is given by $$2\pi\int_0^8 x\cdot x^{1/3}\, dx$$

35. anonymous

shell method for region y= x^2 bounded by y=0, x=2, revovling about y=-2

36. watchmath

$$\int_0^4 2\pi (y+2)(2-\sqrt{y})\,dy$$

37. anonymous

thx

38. anonymous

whats your trick for shell methods, i have trouble setting them up

39. anonymous

whats the shell method for region bounded by y = e^(-x^2) , x = 2, y = 0. revolved about x axis . only need to set up

40. anonymous

$2\pi \int\limits_{0}^{2}x e^{-x^2/2}dx$

41. watchmath

Well there is not trick. First make a slice that is parallel to the axis of symmetry. In your last problem we slice along the y-axis. Then imagine that you rotate that slice. You will have a cyllinder. Try to find the radius of the cyllinder. If the slice is of distance y from the x-axis, then it is of distance y+2 from the axis of rotation (y=-2). So the circumference of the base of the cyllinder is 2pi(y+2). Now the height of the cyllinder is bounded to the right by x=8 and to the left by x= sqrt(y). So the height is (8-sqrt(y)).

42. anonymous

i think this will be the ans...

43. watchmath

Give it a try Cantorset for you very last problem.

44. anonymous

wait, which problem are you explaining , you said the height of the cylinder is bounded by x = 8, for which problem

45. anonymous

So the circumference of the base of the cyllinder is 2pi(y+2). Now the height of the cyllinder is bounded to the right by x=8 and to the left by x= sqrt(y). So the height is (8-sqrt(y)). for which problem was this?

46. watchmath

Oh sorry, the height should be 2- sqrt(y)

47. anonymous

dont you mean, width of the cylinder?

48. anonymous

oh height, if you look at it sideways

49. watchmath

Try this one. Find the volume of solid resulted from rotating the region bounded by the circle (x-2)^2+y^2=1 around the y-axis (the solid is a torus).

50. anonymous

ok one sec, before i do that one. what if i have a region bounded y = x^2 , y=4, x = 2, about x axis

51. anonymous

then it would be integral 2pi * y *sqrt y dy ?

52. anonymous

from 0 to 4

53. anonymous

for the second problem the height is same for all the shells and it is 2 and the radius is equal to y so volume will $2\pi \int\limits_{y=0}^{1} y *2dy$be

54. anonymous

please tell me is the integral correct for the problem y=e^(-x^2/2)

55. anonymous

no

56. anonymous

for which one? first solve for x

57. watchmath

Need to go ... I'll be back in 20 minute(ish)

58. anonymous

ok

59. anonymous

graph the curve y = e^(-x^2)

60. anonymous

and we are revolving about x axis

61. anonymous

for the volume bounded by the curve y=e^(-x^2/2), x=2,y=0, around x-axis..

62. anonymous

63. anonymous

oh wait, i meant y = e^(-x^2)

64. anonymous

i have done a rough graph can you check it...

65. anonymous

not the right graph, should look like a bell shape

66. anonymous

ok i will try it.......

67. anonymous
68. anonymous

thanx

69. anonymous

its easier to check with disc method, so lets try that

70. anonymous

integral pi *e^(-x^2) ^2 from 0 to 2

71. anonymous

918800

72. watchmath

If you want to use the cyllindrical shell to the e^(-x^2/2) problem, then remember we need to make a slice parallel to the x-axis. This slice is bounded to the right by x=2 and to the left by $$x=\sqrt{-\ln y}$$.

73. watchmath

sorry I mean to the right by $$x=\sqrt{-2\ln y}$$

74. watchmath

to the left ... :)

75. anonymous

for the sake of graphing and checking i changed the problem

76. anonymous

y = 4 e^(-x^2/4) (which stretches it horizontally and vertically)

77. anonymous

bounded by x = 2 , x = 0 , y =0

78. anonymous

sorry i keep changing problem , im checking this with calculator

79. anonymous

using the disc method, just to make sure we are doing the shell method right (to check later)

80. watchmath

and we rotate about which axis?

81. anonymous

disc method is int pi ( 4 e^(-x^2/4) ) ^2 , from 0 to 2 , about x axis

82. anonymous

i get 60.13 as my volume

83. anonymous

ok shell method ... x= sqrt (- 4 ln (y/4) )

84. watchmath

It is not nice using the shell method here. You need to split into two integrals.

85. anonymous

so there are two pieces to the shell method int 2pi* y*sqrt (- 4 ln (y/4) ) from 2 to 4 +

86. anonymous

+ int 2pi * y* 2 from 0 to 2

87. watchmath

0 to 4e^(-1) and from 4e^(-1) to 4

88. anonymous

oh

89. anonymous

int 2pi * y* 2 from 0 to 4e^-1 + int 2pi* y*sqrt (- 4 ln (y/4) ) from 4e^-1 to 4

90. anonymous

answer does not match disc method

91. anonymous

ok it does,

92. anonymous

wow, what a problem !!!

93. anonymous

int 2pi * y* 2 from 0 to 4e^-1 + int 2pi* y*sqrt (- 4 ln (y/4) ) from 4e^-1 to 4 = int pi (4*e^(-x^2/4)) ^2 from 0 to 2

94. anonymous

i guess i have problems with upper and lower limits, i confuse them , which ones we are doing

95. anonymous

im just doing these problems to be thorough,