## anonymous 5 years ago consider the region bounded by the graphs of f(x)=x^1/2, y=0, and x=2. find the volume of the soild formed by rotating the region about the x-axis

1. amistre64

ok.... y = sqrt(x) good

2. amistre64

between [0,2]? right

3. anonymous

yes

4. amistre64

we find an area of rotation by taking the radius and adding it up as we go thru the interval (S) 2pi(f(x)) dx or: 2pi (S) f(x) dx ... right?

5. anonymous

i have no idea....

6. amistre64

lol .... ok... I can work with that :)

7. anonymous

is the graph bounded by x $\in$ [0, 2]?

8. amistre64

if we take one section of this graph, we have a radius = to the value of f(x) at the point x right?

9. anonymous

$x \in [0, 2]$ ?

10. anonymous

yea

11. amistre64

rdoshi, yes

12. anonymous

alrighty...

13. amistre64

if we take this radius and integrate 2pi(r) we will get the area of this "slice" lets use an example we are familiar with ok?

14. amistre64

lets say the radius = 5; the area of a circle would be equal to pi(r^2) right?

15. anonymous

yess

16. anonymous

I understand that, i'm just confused on how to do it with this problem...because I just AM DUMB

17. amistre64

if we take the Circumference of a circle and "integrate" it we get this: (S) 2pi(r) dr 2pi (S) r dr 2pi (r^2/2) = pi(r^2) which gives us the area of our circle.... 25pi

18. anonymous

$\int\limits_{0}^{2}\pi \sqrt{x}$ That should do it?

19. amistre64

what we do by integrating along an interval is taking each "slice" finding its area and then adding up all the slices to get the whole volume

20. anonymous

sorry sqrt(x) is squared for area

21. anonymous

so its just integral (pi * x) from 0 to 2?

22. amistre64

our radius is equal to y = x^(1/2) does that make sense to you?

23. anonymous

you find the equation for the cross sectional area of the solenoid that is formed by rotating the graph which is A(x) = pi (sqrt x)^2 then you just find the integral of that area along x for the volume

24. anonymous

but to find the integral do you find the anti derviative?

25. amistre64

yes... an integral and an antiderivative are the same thing...

26. amistre64

beth: whats the antiderivative of x^(1/2)

27. anonymous

nd then i just plug in 2 and 0 and solve right

28. amistre64

actually, 0 will equal 0 so you just have to solve for x=2

29. anonymous

you dont take integral of sqrt(x)

30. anonymous

because to find area you have to square the radius

31. anonymous

sqrt(x) ^ 2 is just x... so you find the integral of pi(x) from 0 to 2

32. anonymous

but i am trying to find volume not area...or does that even matter?

33. amistre64

volume is just 2d area.... add up all the volumes of the slices to get the 3d area.... same thing

34. anonymous

alright here's how it works.... integrating area will give you volume. in order to find area A = pi radius^2... radius in this case is sqrt(x)

35. amistre64

2pi (S) x^(1/2) dx [0,2] 2pi(x^3/2)(2/3) when x = 2 is all you need to find

36. amistre64

(4/3)pi 2^(3/2) = ??

37. anonymous

pi (S) (x^1/2)^2 dx from 0 to 2

38. anonymous

What rdoshi said. Your should only need integrals for this question. by using the formula below it will find you the volume. when rotating around x: $V = \pi \int\limits_{b}^{a} y^{2} dx$ [therefore this means f(x) is squared] when rotating around the y: $V = \pi \int\limits_{b}^{a} x^{2} dy$ [therefore this means you make x the subject of the function and then square it]

39. anonymous

so the final answer should be 2pi?

40. amistre64

we already went thru what we need to do to get the "area"... so follow along :)

41. amistre64

4 pi 2^(3/2) ------------- 3

42. anonymous

you guys are telling me two different things...i dont know which one is right haaha

43. anonymous

its 2pi

44. anonymous

for area you just exclude pi from the formula while taking into account that if the function goes into a negative area you use absolute values (seeing you can't have negative an area).

45. anonymous

yea but i am trying to find volume, but wait that doesnt matter right?

46. amistre64

the area of each slice is the antiderivative of 2pi(x^(1/2))

47. anonymous

i thought the area of each slice is the antiderivative of pi((x^1/2)^2)

48. amistre64

the antiderivative of 2pi(r) = 2pi(r^2)/2

49. anonymous

2piR is circumfrence not area

50. anonymous

pi R^2 is area

51. amistre64

why would it be the antiderivative of pi(x^(1/2)^2)?? that makes no sense

52. anonymous

R in this case is y = sqrt(x)

53. amistre64

rdoshi... you antiderive the circumference of a circle to get the formula for the area of a circle.... its just that simple

54. anonymous

anti deravative of 2pi R is pi R^2

55. amistre64

yes... and the R value is y = x^(1/2) BEFORE you get the antiderivative...

56. anonymous

that doesn't change anything

57. amistre64

it does in my mind :)

58. anonymous

you can substitute before or after... method of substitution

59. amistre64

I perfer before... makes my life easier :)

60. anonymous

nonetheless you end up to solve for pi * integral of x from [0,2]

61. anonymous

amistre64 does the equation look like this $\int\limits_{2}^{0} \pi x^3/2 divided by 3/$

62. anonymous

that should give you (x^2)/2 evaluated from 0 to 2 thus (2^2)/2 - (0^2)/2

63. anonymous

can someone just type out the work from start to finish so i undertand it bplease....:)

64. amistre64

the volume of a cone is 1/3 basearea times h.... because the "3" in 1/3 comes from the antiderivative of the equation for the line.... never tried it the other way around tho

65. amistre64

I know that the area of each slice is (S) 2pi(x^(1/2)) dx ... so ill use that:

66. amistre64

2pi (2/3) (x^(3/2)) is what I will use to find the volume of rotation...

67. anonymous

area of each slice A = pi * (Radius = sqrt x)^2

68. anonymous

thus A = pi*x

69. amistre64

4pi (2)^(2/3) 0 ----------- - --- = area under the curve 3 3

70. amistre64

4pi sqrt(8) --------- = area 3 fliiped my exponent around there by accident...

71. amistre64

8pi sqrt(2) -------- should be the area 3

72. amistre64

about 11.85 if I did it right, but maybe im wrong... ;)

73. anonymous

are you normally right...god i hope so

74. anonymous

for a solenoid with maximum radius 2?

75. amistre64

lol ..... me too :)

76. anonymous

ok well can you now teach me how to find the volume of the solid form by rotating the region sbout the y axis

77. anonymous

dude its pretty much the same

78. amistre64

put your equation in terms of "y" and solve it the same way

79. anonymous

when you rotate a graph, and want to find volume, you take the cross section of the 3D solid formed and find the equation for that section. Then you integrate that equation for cross section.

80. anonymous

In your given case, your y = sqrt (x) You are rotating it around x-axis

81. anonymous

it will give you a solenoid... kinda like a sideways bowl

82. anonymous

but now it need to be rotated about the y....

83. anonymous

i'm saying what amistre64 said is not quite right

84. anonymous

when you have a cross sectional area of a solenoid, it is always a circle

85. anonymous

and Area of that circle is A = pi * y(x)

86. anonymous

when you do integral of A from given bounds for x, you get the volume

87. anonymous

y(x)^2 sorry... i forgot the power in my A equation

88. anonymous

if you are rotating about Y axis, all you need to do is to put the given equation in x(y) form

89. anonymous

and evaluate for your range... i.e y belongs to [a, b]

90. amistre64

ill admit I could be wrong... so I will go "re-look" at my understanding of this :)

91. anonymous

ok that makes sense but what if i fine the volme of the solid ford b rotating the region about the line y= -2

92. amistre64

yeah, I was mistaken.... apparently I was confusing a couple a methods... for instance, take the line y=5 on the interval of [0,10]. this should produce a volume of 250pi. But if I do it the way I had in my head I get: 2pi (S) 5 dx 2pi(5x) -> 2pi5(10) = 100pi.... So I was wrong.....