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anonymous

  • 5 years ago

volume of a torus...?

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  1. anonymous
    • 5 years ago
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    cross section of the circle has a radius of .75 and radius from center of torus to center of cross section circle is 1.5

  2. anonymous
    • 5 years ago
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    i got the equations but i cant figure out the positive negative problem with the square root

  3. anonymous
    • 5 years ago
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    arman, you can solve this using cylindrical shells. \[\delta V = 2\pi (radius) (height)(elemental.thickness)\]\[\delta V = 2\pi x \left( \sqrt{r^2-(x-3/2)^2}-\left( -\sqrt{r^2-(x-3/2)^2} \right) \right) \delta x\]\[\delta V = 2\pi x \left( 2\sqrt{r^2-(x-3/2)^2}\right) \delta x\]where r = 3/4 (0.75)

  4. anonymous
    • 5 years ago
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    Your limits of integration would be x=3/4 to x=9/4.

  5. anonymous
    • 5 years ago
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    dang yo i just reposted this up top

  6. anonymous
    • 5 years ago
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    i think im supposed to use washers though

  7. anonymous
    • 5 years ago
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    oh

  8. anonymous
    • 5 years ago
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    Well, that can be done too.

  9. anonymous
    • 5 years ago
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    at the end he has us compare our answer via washer to the answer via cylindes

  10. anonymous
    • 5 years ago
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    but how?? if you can just tell me what integrals to use and why i can figure it out but i keep having my areas cancell out of i have a root of a negative

  11. anonymous
    • 5 years ago
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    ok, just let me do it on paper.

  12. anonymous
    • 5 years ago
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    Okay, ready?

  13. anonymous
    • 5 years ago
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    I'm going to let a=3/2 and r=3/4.

  14. anonymous
    • 5 years ago
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    yesss, thank you so much

  15. anonymous
    • 5 years ago
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    \[\delta V = \pi \left[ \left( a+\sqrt{r^2-y^2} \right)^2-\left( a-\sqrt{r^2-y^2} \right)^2 \right]\delta y\]

  16. anonymous
    • 5 years ago
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    Did you get something like that at first?

  17. anonymous
    • 5 years ago
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    oh wow... i know my problem now

  18. anonymous
    • 5 years ago
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    nope nvrmnd i still dont.. yeah thats what i got

  19. anonymous
    • 5 years ago
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    Okay, well all you have to do is expand that and integrate. \[\delta V=\pi \left( 4a \sqrt{r^2-y^2} \right)\delta y\]

  20. anonymous
    • 5 years ago
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    i integrated then expanded...

  21. anonymous
    • 5 years ago
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    \[V=4a \pi \int\limits_{-r}^{r}\sqrt{r^2-y^2}dy\]

  22. anonymous
    • 5 years ago
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    x and y are interchangeable (technically speaking) in this case?

  23. anonymous
    • 5 years ago
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    No, you're integrating over the interval on the y-axis.

  24. anonymous
    • 5 years ago
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    right, but like as long as i sub for that variable, it doesn't matter if you write x or y physical interpretation will rotate obviously)?

  25. anonymous
    • 5 years ago
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    well...if you keep to the understanding of what x is in this situation and you stick to the same limits of integration. Otherwise, if you want to integrate over the x-axis, you're going to have to make a substitution for y in terms of x, and then find the differential of y with respect to x. It's more work.

  26. anonymous
    • 5 years ago
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    I think I know where you were having problems too.

  27. anonymous
    • 5 years ago
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    when you sub you get a negative numberunder the radical

  28. anonymous
    • 5 years ago
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    or 0 in this case

  29. anonymous
    • 5 years ago
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    If you integrate that expression for y (indefinite integral), you'll derive: \[\int\limits{}_{}\sqrt{r^2-y^2}dy=\frac{1}{2}\left[ y \sqrt{r^2-y^2}+r^2 \tan^{-1}\left( \frac{y}{\sqrt{r^2-y^2}} \right) \right]+c\]

  30. anonymous
    • 5 years ago
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    plus constant (didn't show up).

  31. anonymous
    • 5 years ago
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    arctan is wayyy too complex for this O.o

  32. anonymous
    • 5 years ago
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    and the directions request that it be rotated around the x axis >.<

  33. anonymous
    • 5 years ago
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    Now...when you approach the limit r from within your interval, you are going to approach zero in the radical from the positive side...yes, the denominator will approach zero, by arctan approach positive pi/2. When y approaches -r, again, the denom. will approach zero, but arctan will approach NEGATIVE pi/2. When you subtract (i.e. limits), you'll be adding 2 lots of pi/2...there'll be no cancelling.

  34. anonymous
    • 5 years ago
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    You need to draw me a picture of how you're supposed to orient it. The volume you get will be the same.

  35. anonymous
    • 5 years ago
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    You can just switch axes.

  36. anonymous
    • 5 years ago
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    The volume I get is \[2\pi^2ar^2\]

  37. anonymous
    • 5 years ago
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    which is the volume of a torus.

  38. anonymous
    • 5 years ago
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    Is this for class or fun?!

  39. anonymous
    • 5 years ago
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    it can be both you know lol

  40. anonymous
    • 5 years ago
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    i dont know anymore than the following: measure your torus: radius of circular cross section of the "tube" = a = _____ (i put .75in) radius rom center of torus to center of tube= b =______ (i put 1.5in) write the integral that calculates the volume of the torus by revolving the circle around the x/axis. show steps. show what A(x) is.

  41. anonymous
    • 5 years ago
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    what you got with the 2pi^2ar^2 is correct, hat much i know

  42. anonymous
    • 5 years ago
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    Okay...that's no problem...just switch the x's for y's.

  43. anonymous
    • 5 years ago
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    I rotated around the y-axis, but if you consider the geometry, by switching labels, 'suddenly' you're revolving the torus around the x-axis.

  44. anonymous
    • 5 years ago
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    i understand that... i dont understand why my equations are wrong ><

  45. anonymous
    • 5 years ago
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    Well, your setup is right, since we both agreed, so it must be the algebra in between.

  46. anonymous
    • 5 years ago
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    i should be using: v= int(pi(root(r2-x2)+1.5) - int(pi(-root(r2-x2)+1.5))

  47. anonymous
    • 5 years ago
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    That's the same as what we did above. I used y. You can use x. Whereas I put my circle on the x-axis to rotate, you're putting it on the y-axis. All you need to do is switch y for x in the formulas above. You're not changing the mathematics because, in this context, x and y and dummy variables.

  48. anonymous
    • 5 years ago
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    You know, I will set the problem up with a circle on the y-axis. You can punch your teacher for me.

  49. anonymous
    • 5 years ago
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    lol i completely agree with everything youve said i just dont know why you got a different answer, ill try it another dozen times and repost if i still fail miserably

  50. anonymous
    • 5 years ago
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    This might surprise you, but I got the same formula like I did for y, except now it has x's...

  51. anonymous
    • 5 years ago
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    Okay, you have your circle on the y-axis and it's centered at (0,a).

  52. anonymous
    • 5 years ago
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    The circle's equation is \[x^2+(y-a)^2=r^2\]

  53. anonymous
    • 5 years ago
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    The washer method says that the area of the washer is that of a punctured disc; that is, \[\pi y_2^2- \pi y_1^2\]

  54. anonymous
    • 5 years ago
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    haha funny almost like i didnt know you can switch x and y -.- armaan not stupid ><

  55. anonymous
    • 5 years ago
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    Solving the equation of the circle for those y-values gives\[y_2=a+\sqrt{r^2-x^2}\]\[y_1=a-\sqrt{r^2-x^2}\]

  56. anonymous
    • 5 years ago
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    The area of the punctured disc is\[\pi(y_2^2-y_1^2)\]\[=\pi [(a+\sqrt{r^2-x^2})^2-(a-\sqrt{r^2-x^2})^2]\]

  57. anonymous
    • 5 years ago
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    right

  58. anonymous
    • 5 years ago
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    happy?

  59. anonymous
    • 5 years ago
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    this makes sense

  60. anonymous
    • 5 years ago
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    The volume will just be the area multiplied by the 'width', delta x:\[\delta V = \pi[(a+\sqrt{r^2-x^2})^2-(a-\sqrt{r^2-x^2})^2] \delta x\]

  61. anonymous
    • 5 years ago
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    From here on in it is algebra and integration. The problem's been set up.

  62. anonymous
    • 5 years ago
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    The bit in the brackets:\[a^2+2\sqrt{r^2-x^2}+r^2-x^2-(a^2-2\sqrt{r^2-x^2}+r^2-x^2)\]

  63. anonymous
    • 5 years ago
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    sorry for causing the trouble on this :/

  64. anonymous
    • 5 years ago
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    yeah, after you punch your teacher, punch yourself

  65. anonymous
    • 5 years ago
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    :p

  66. anonymous
    • 5 years ago
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    brackets continued...

  67. anonymous
    • 5 years ago
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    haha well said

  68. anonymous
    • 5 years ago
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    \[4a \sqrt{r^2-x^2}\]

  69. anonymous
    • 5 years ago
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    When I expanded above, I forgot the 'a' next to the sqrt expressions.

  70. anonymous
    • 5 years ago
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    \[a^2+2a \sqrt{r^2-x^2}+r^2-x^2-(a^2-2a \sqrt{r^2-x^2}+r^2-x^2)\]

  71. anonymous
    • 5 years ago
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    should be this ^

  72. anonymous
    • 5 years ago
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    Okay, so the volume is now,\[V=4 \pi a \int\limits_{-r}^{r}\sqrt{r^2-x^2}dx\]

  73. anonymous
    • 5 years ago
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    But you can recognize the integral as the area of a semi-circle of radius r, which would be\[\pi \frac{r^2}{2}\]

  74. anonymous
    • 5 years ago
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    which means your volume is\[V=4\pi a \times \pi \frac{r^2}{2}=2 \pi^2 a r^2\]

  75. anonymous
    • 5 years ago
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    Okay?

  76. anonymous
    • 5 years ago
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    ok i see, but then why does it not work if you just subtract the original integrals without combining or simplifying

  77. anonymous
    • 5 years ago
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    You mean without doing the algebraic expansion way up there?

  78. anonymous
    • 5 years ago
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    It should.

  79. anonymous
    • 5 years ago
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    But looking at it should make you go, "No way, I'm expanding" because you have radicals and squares.

  80. anonymous
    • 5 years ago
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    iight thanks, id say i feel retarded, but given what im trying to learn, i feel pretty normal

  81. anonymous
    • 5 years ago
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    yeah, you're doing fine. just need to calm down.

  82. anonymous
    • 5 years ago
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    There aren't many people (proportionately speaking) who can understand this stuff, so you're doing exceptionally well.

  83. anonymous
    • 5 years ago
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    id say thats a bit of exaggeration haha i seriously want to sit in your shoes though, given what you know

  84. anonymous
    • 5 years ago
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    in time...

  85. anonymous
    • 5 years ago
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    how much time did it take you?

  86. anonymous
    • 5 years ago
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    depends...maths is so large, you're learning every day.

  87. anonymous
    • 5 years ago
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    do you have a phD?

  88. anonymous
    • 5 years ago
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    Not yet.

  89. anonymous
    • 5 years ago
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    so thats what your research is for?

  90. anonymous
    • 5 years ago
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    pretty much + work!

  91. anonymous
    • 5 years ago
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    = money = food + accommodation + clothes...

  92. anonymous
    • 5 years ago
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    do you have a math based job?

  93. anonymous
    • 5 years ago
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    Yes, I'm teaching it.

  94. anonymous
    • 5 years ago
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    i see, you spend your time at the college either researching or lecturing

  95. anonymous
    • 5 years ago
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    what is your phD on?

  96. anonymous
    • 5 years ago
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    I was doing some research into biophysics and something called Takahashi's problem. It's a mathematical physics problem.

  97. anonymous
    • 5 years ago
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    do you intend to solve or something of that sort the problem?

  98. anonymous
    • 5 years ago
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    sorry?

  99. anonymous
    • 5 years ago
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    yes

  100. anonymous
    • 5 years ago
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    but we'll see.

  101. anonymous
    • 5 years ago
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    I'll be round for a while if you need anything else.

  102. anonymous
    • 5 years ago
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    i need to go soon DX its 235 here

  103. anonymous
    • 5 years ago
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    iight i have no idea how you got 4a[etcetc]

  104. anonymous
    • 5 years ago
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    i can understand how, but how did you know it was 4a without actually expanding it?

  105. anonymous
    • 5 years ago
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    I did expand it. The 4a comes from the fact you have \[2a \sqrt{r^2-x^2}+2a \sqrt{r^2-x^2}=4a \sqrt{r^2-x^2}\]

  106. anonymous
    • 5 years ago
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    oh ok, i thought there was some hint or trick to it

  107. anonymous
    • 5 years ago
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    No...you should probably go to bed.

  108. anonymous
    • 5 years ago
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    after i finish my doughnut -.-

  109. anonymous
    • 5 years ago
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    iight im done, so ima head to bed, thanks for the help, as always, gnight, ttyl

  110. anonymous
    • 5 years ago
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    and idk about the punching the teacher... hes pretty cool

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