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anonymous

  • 5 years ago

can i ask a quick question. im helping a student who thinks that you can form a solid of revolution , bounded by y=x^2, y = x+2, and revolve it about the y axis. this is not possible, since part of the region is in quadrant 2

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  1. anonymous
    • 5 years ago
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    You are correct.

  2. amistre64
    • 5 years ago
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    its possible; you just get a different shape

  3. anonymous
    • 5 years ago
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    no but you are revolving about itself twice

  4. amistre64
    • 5 years ago
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    then subtract he part that gets doubled...

  5. anonymous
    • 5 years ago
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    the limits of itnegration are -1 to 2 and we are revolving about y axis (using shell method )

  6. anonymous
    • 5 years ago
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    hmmmm, that doesnt make sense to me

  7. amistre64
    • 5 years ago
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    if I have a region in the form of a rectangle that overlaps the y axis and its domain is from -1 to 5 with a height of 3; when I rotate it about the yaxis; I just have to account for the part that get counted twice...

  8. anonymous
    • 5 years ago
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    ive never seen a book say, subtract the region that gets doubled ,

  9. anonymous
    • 5 years ago
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    i would say the question is ill defined

  10. amistre64
    • 5 years ago
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    you dont need a book to tell you that the volume of a solid isnt counted twice and to adjust for the computational errors :) do you?

  11. anonymous
    • 5 years ago
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    because you have to ignore the negative part, or quadrant 3 and 4 of the curve

  12. amistre64
    • 5 years ago
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    if that region is consumed by the rest, then yes, ignore it lol

  13. amistre64
    • 5 years ago
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    it part of that region is left exposed; count the part that is exposed ;)

  14. anonymous
    • 5 years ago
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    then one should just adjust the limits, start at x = 0 for lower limit

  15. amistre64
    • 5 years ago
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    that all depends onhow the curves and their boundaries interact

  16. anonymous
    • 5 years ago
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    but you said to ignore the region counted twice. we can do this by starting at x = 0 i have a different question, revolve region bounded by y = sqrt x, y = 0, x = 4, x = 6 , about the line x = 6 (using shell method i think)

  17. amistre64
    • 5 years ago
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    in this case; you can safely ignore the left part

  18. anonymous
    • 5 years ago
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    ok , so then its ok to start at x = 0 ?

  19. anonymous
    • 5 years ago
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    when would i not, i think im going to state a general theorem soon, but can you help me with that other shell problem

  20. amistre64
    • 5 years ago
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    it is for the first question becasue the rotation covers the left parts completely and nulls them

  21. anonymous
    • 5 years ago
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    right, thats a good word. it annhilates :)

  22. anonymous
    • 5 years ago
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    nulls them

  23. amistre64
    • 5 years ago
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    when the rotation eats into itself, you adjust for the part that gets doubled...

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  24. anonymous
    • 5 years ago
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    whoa

  25. anonymous
    • 5 years ago
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    how did you graph that?

  26. amistre64
    • 5 years ago
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    i used paint in windows lol

  27. anonymous
    • 5 years ago
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    nice :)

  28. amistre64
    • 5 years ago
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    i created some curves that would account for a condition where you spin and have some parts redundant

  29. anonymous
    • 5 years ago
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    can you state the problem though, what are the regions, etc

  30. amistre64
    • 5 years ago
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    if you really wanna find the volume; dunk it in a bathtub filled with water and measure the volume of overflow lol

  31. anonymous
    • 5 years ago
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    haha, by displacement,

  32. anonymous
    • 5 years ago
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    can you state the problem though, what are the regions, etc

  33. amistre64
    • 5 years ago
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    i wouldnt be able to state that graph with any accuracy becasue I didnt use any defined functions to create it; I used artisitic liscense lol

  34. amistre64
    • 5 years ago
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    but then again, a function and its equation are not the same thing are they...

  35. anonymous
    • 5 years ago
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    i mean can you set up a problem where your situation happens , like in this graph

  36. amistre64
    • 5 years ago
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    sure, its some cubic coupled with a linear an bounded at their common points; then spin around the y axis..

  37. amistre64
    • 5 years ago
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    the image as is just shows the accounting for spin by applying the mirror image apon the original positions

  38. amistre64
    • 5 years ago
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    all the steps to accont for the red portion tho; might take longer than what I got for tonight lol

  39. anonymous
    • 5 years ago
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    ok , i think you will end up in a contradiction though , i suspect it

  40. amistre64
    • 5 years ago
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    nah; its goblet shaped in the end; so a goblet is just as feesible as any other shape created by rotation.. right?

  41. anonymous
    • 5 years ago
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    what method of integration are you doing, disc?

  42. anonymous
    • 5 years ago
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    also the shell method will produce a negative answer for your quadrant 2 curve

  43. amistre64
    • 5 years ago
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    you dont have to confine yourself to one method; use the tools available to you to determine the easiest way to arrive at the solution; break it up, and glue it back when your done right?

  44. amistre64
    • 5 years ago
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    negative answers are not a problem when accounting for volume; since volume doesnt care about direction...

  45. anonymous
    • 5 years ago
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    volume is not defined for negative

  46. amistre64
    • 5 years ago
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    Do you know how to account for the 'contradiction' of multiplying an inequaility by a negative?

  47. anonymous
    • 5 years ago
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    you have a contradictory curve

  48. anonymous
    • 5 years ago
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    the quadrant 1 and 4 overlaps will intersect quadrant 1 and 3 right?

  49. amistre64
    • 5 years ago
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    the curve is perfectly representative of any other; its a line and a cubic;

  50. anonymous
    • 5 years ago
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    no your curves are redundant

  51. amistre64
    • 5 years ago
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    the volume produced by creating this shape on a potters wheel does not produce a contradiction does it?

  52. anonymous
    • 5 years ago
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    the portion in Q 1 and 4 will revolve right over Q 2 and 3

  53. anonymous
    • 5 years ago
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    your goblet is just the portion in Q 1 and 4 revolving

  54. amistre64
    • 5 years ago
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    the mirror image is not a part of the original functions; they are simply there to show what happens during rotation

  55. amistre64
    • 5 years ago
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    the curve itself is bounded from q1 into q2 and ends in q3; then spun around the y axis

  56. amistre64
    • 5 years ago
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  57. anonymous
    • 5 years ago
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    oh can you graph this without shading , that got me confused

  58. amistre64
    • 5 years ago
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    thats the original concept for the curves to be used in the rotation and the grey are is the area that creates the volume of rotation

  59. toxicsugar22
    • 5 years ago
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    hey contrasat can u help me

  60. amistre64
    • 5 years ago
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    i gotta get to sleep; so yall have fun :)

  61. anonymous
    • 5 years ago
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    amistre, ok , there are two porblems with this, if youre revolving about y axis, it should be in Q1 or Q4 onlye

  62. anonymous
    • 5 years ago
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    i think thats a theorem, im going to make it a theorem :)

  63. amistre64
    • 5 years ago
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    it is inwhat it is in; you can adjust it up and down for sure; but thats a given

  64. anonymous
    • 5 years ago
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    i guess im not as flexible as you are

  65. amistre64
    • 5 years ago
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    if mathis only good for a few solutions; it isnt good for any solutions right ;)

  66. anonymous
    • 5 years ago
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    well math is idealized, so you have set up your problem to match it i guess

  67. anonymous
    • 5 years ago
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    well, the problem here is that volume is always positive, and does not annul itself

  68. amistre64
    • 5 years ago
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    i have a cubic and a line that can create a bounded region that canbe solved by integration thru rotation about the y axis... math can do it :) just have to adjust the graphs for common sense and solve

  69. amistre64
    • 5 years ago
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    put the curves on the table in a way that they can be analysed; and then the rest is easy

  70. anonymous
    • 5 years ago
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    amistre, two people disagree with you

  71. anonymous
    • 5 years ago
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    so far

  72. anonymous
    • 5 years ago
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    lets ask watchmath

  73. toxicsugar22
    • 5 years ago
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    can u help me contrsat

  74. amistre64
    • 5 years ago
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    does the area of a circle disappear when its at the origin?

  75. anonymous
    • 5 years ago
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    yes it does

  76. amistre64
    • 5 years ago
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    lol.... area is an absolute value

  77. anonymous
    • 5 years ago
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    wait, waht do you mean?

  78. amistre64
    • 5 years ago
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    i mean, does the area of a circle vanish if its center is at the origin?

  79. anonymous
    • 5 years ago
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    whats the radius?

  80. amistre64
    • 5 years ago
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    radius doesnt matter; make it 10 if you want, or 500000

  81. anonymous
    • 5 years ago
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    what do you mean vanish? do you mean if you revolve the region about the x axis?

  82. anonymous
    • 5 years ago
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    its not possible to do that, using disc method

  83. amistre64
    • 5 years ago
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    its possible; you jsut gotta use common sense and see that area or volume is an absolute value

  84. amistre64
    • 5 years ago
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    does the area above the xaxis null the area below it? or do quadrants cancel each other out? or what?

  85. watchmath
    • 5 years ago
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    so what's going on here? :)

  86. amistre64
    • 5 years ago
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    lol..... well...

  87. amistre64
    • 5 years ago
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    cantor says I cant determine the volume of my curves i think

  88. watchmath
    • 5 years ago
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    I think so far I agree with you Amistre. The important think is that do we get a solid object is some of the region is on the left and some on the right of the axis of rotation? yes we do? But how to compute the volume of that object maybe not as simple as applying the standard method of computing volume.

  89. watchmath
    • 5 years ago
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    Also mathematically speaking we want to allow that kind of situation because we can have a more complex solid object.

  90. amistre64
    • 5 years ago
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    exactly :)

  91. amistre64
    • 5 years ago
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    just as when we have an off centered rotation; we adjust the function to get them so that we can examine them from a reasonable stage

  92. amistre64
    • 5 years ago
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    if it gets to complicated; we can break it up, examine parts and come to a conclusion by summing up our findings

  93. watchmath
    • 5 years ago
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    One way to find the volume is to reflect the one of the region about the axis of the symmetry so that now everything is on one side only. In the first cantorset example when we reflect the left part to the right, we don't get any new region on the right. So to compute volume we just integrate from x=0 to x=2.

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