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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

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  1. amistre64
    • 5 years ago
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    ok.... y = sqrt(x) good

  2. amistre64
    • 5 years ago
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    between [0,2]? right

  3. anonymous
    • 5 years ago
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    yes

  4. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    i have no idea....

  6. amistre64
    • 5 years ago
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    lol .... ok... I can work with that :)

  7. anonymous
    • 5 years ago
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    is the graph bounded by x \[\in\] [0, 2]?

  8. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    \[x \in [0, 2]\] ?

  10. anonymous
    • 5 years ago
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    yea

  11. amistre64
    • 5 years ago
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    rdoshi, yes

  12. anonymous
    • 5 years ago
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    alrighty...

  13. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    lets say the radius = 5; the area of a circle would be equal to pi(r^2) right?

  15. anonymous
    • 5 years ago
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    yess

  16. anonymous
    • 5 years ago
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    I understand that, i'm just confused on how to do it with this problem...because I just AM DUMB

  17. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    \[\int\limits_{0}^{2}\pi \sqrt{x}\] That should do it?

  19. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    sorry sqrt(x) is squared for area

  21. anonymous
    • 5 years ago
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    so its just integral (pi * x) from 0 to 2?

  22. amistre64
    • 5 years ago
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    our radius is equal to y = x^(1/2) does that make sense to you?

  23. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    but to find the integral do you find the anti derviative?

  25. amistre64
    • 5 years ago
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    yes... an integral and an antiderivative are the same thing...

  26. amistre64
    • 5 years ago
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    beth: whats the antiderivative of x^(1/2)

  27. anonymous
    • 5 years ago
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    nd then i just plug in 2 and 0 and solve right

  28. amistre64
    • 5 years ago
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    actually, 0 will equal 0 so you just have to solve for x=2

  29. anonymous
    • 5 years ago
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    you dont take integral of sqrt(x)

  30. anonymous
    • 5 years ago
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    because to find area you have to square the radius

  31. anonymous
    • 5 years ago
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    sqrt(x) ^ 2 is just x... so you find the integral of pi(x) from 0 to 2

  32. anonymous
    • 5 years ago
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    but i am trying to find volume not area...or does that even matter?

  33. amistre64
    • 5 years ago
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    volume is just 2d area.... add up all the volumes of the slices to get the 3d area.... same thing

  34. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    (4/3)pi 2^(3/2) = ??

  37. anonymous
    • 5 years ago
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    pi (S) (x^1/2)^2 dx from 0 to 2

  38. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    so the final answer should be 2pi?

  40. amistre64
    • 5 years ago
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    we already went thru what we need to do to get the "area"... so follow along :)

  41. amistre64
    • 5 years ago
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    4 pi 2^(3/2) ------------- 3

  42. anonymous
    • 5 years ago
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    you guys are telling me two different things...i dont know which one is right haaha

  43. anonymous
    • 5 years ago
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    its 2pi

  44. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    yea but i am trying to find volume, but wait that doesnt matter right?

  46. amistre64
    • 5 years ago
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    the area of each slice is the antiderivative of 2pi(x^(1/2))

  47. anonymous
    • 5 years ago
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    i thought the area of each slice is the antiderivative of pi((x^1/2)^2)

  48. amistre64
    • 5 years ago
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    the antiderivative of 2pi(r) = 2pi(r^2)/2

  49. anonymous
    • 5 years ago
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    2piR is circumfrence not area

  50. anonymous
    • 5 years ago
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    pi R^2 is area

  51. amistre64
    • 5 years ago
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    why would it be the antiderivative of pi(x^(1/2)^2)?? that makes no sense

  52. anonymous
    • 5 years ago
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    R in this case is y = sqrt(x)

  53. amistre64
    • 5 years ago
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    rdoshi... you antiderive the circumference of a circle to get the formula for the area of a circle.... its just that simple

  54. anonymous
    • 5 years ago
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    anti deravative of 2pi R is pi R^2

  55. amistre64
    • 5 years ago
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    yes... and the R value is y = x^(1/2) BEFORE you get the antiderivative...

  56. anonymous
    • 5 years ago
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    that doesn't change anything

  57. amistre64
    • 5 years ago
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    it does in my mind :)

  58. anonymous
    • 5 years ago
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    you can substitute before or after... method of substitution

  59. amistre64
    • 5 years ago
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    I perfer before... makes my life easier :)

  60. anonymous
    • 5 years ago
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    nonetheless you end up to solve for pi * integral of x from [0,2]

  61. anonymous
    • 5 years ago
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    amistre64 does the equation look like this \[ \int\limits_{2}^{0} \pi x^3/2 divided by 3/\]

  62. anonymous
    • 5 years ago
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    that should give you (x^2)/2 evaluated from 0 to 2 thus (2^2)/2 - (0^2)/2

  63. anonymous
    • 5 years ago
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    can someone just type out the work from start to finish so i undertand it bplease....:)

  64. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    I know that the area of each slice is (S) 2pi(x^(1/2)) dx ... so ill use that:

  66. amistre64
    • 5 years ago
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    2pi (2/3) (x^(3/2)) is what I will use to find the volume of rotation...

  67. anonymous
    • 5 years ago
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    area of each slice A = pi * (Radius = sqrt x)^2

  68. anonymous
    • 5 years ago
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    thus A = pi*x

  69. amistre64
    • 5 years ago
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    4pi (2)^(2/3) 0 ----------- - --- = area under the curve 3 3

  70. amistre64
    • 5 years ago
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    4pi sqrt(8) --------- = area 3 fliiped my exponent around there by accident...

  71. amistre64
    • 5 years ago
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    8pi sqrt(2) -------- should be the area 3

  72. amistre64
    • 5 years ago
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    about 11.85 if I did it right, but maybe im wrong... ;)

  73. anonymous
    • 5 years ago
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    are you normally right...god i hope so

  74. anonymous
    • 5 years ago
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    for a solenoid with maximum radius 2?

  75. amistre64
    • 5 years ago
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    lol ..... me too :)

  76. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    dude its pretty much the same

  78. amistre64
    • 5 years ago
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    put your equation in terms of "y" and solve it the same way

  79. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    In your given case, your y = sqrt (x) You are rotating it around x-axis

  81. anonymous
    • 5 years ago
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    it will give you a solenoid... kinda like a sideways bowl

  82. anonymous
    • 5 years ago
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    but now it need to be rotated about the y....

  83. anonymous
    • 5 years ago
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    i'm saying what amistre64 said is not quite right

  84. anonymous
    • 5 years ago
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    when you have a cross sectional area of a solenoid, it is always a circle

  85. anonymous
    • 5 years ago
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    and Area of that circle is A = pi * y(x)

  86. anonymous
    • 5 years ago
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    when you do integral of A from given bounds for x, you get the volume

  87. anonymous
    • 5 years ago
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    y(x)^2 sorry... i forgot the power in my A equation

  88. anonymous
    • 5 years ago
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    if you are rotating about Y axis, all you need to do is to put the given equation in x(y) form

  89. anonymous
    • 5 years ago
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    and evaluate for your range... i.e y belongs to [a, b]

  90. amistre64
    • 5 years ago
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    ill admit I could be wrong... so I will go "re-look" at my understanding of this :)

  91. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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.....

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