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Homework Help: Solid of revolution about other lines

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Hey we have started solids of revolutions using disk, washer, and shell methods. But I came across a problem i cannot figure out. "The region of the graph of y=x^2 and the x-axis, for 0<x<2, rotated about the line y=4.



    2. Relevant equations

    Area of a circle = ∏r^2

    3. The attempt at a solution

    Ive drawn the picture but don't even know how to begin with this one. Thank you in advance for the help.
     
  2. jcsd
  3. Feb 5, 2012 #2

    tiny-tim

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    hi schapman22! :smile:

    they're washers, aren't they? …

    vertical washers, all with outer radius 4, and thickness dx :wink:
     
  4. Feb 5, 2012 #3

    Deveno

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    or:

    consider that you get the same solid of revolution (in terms of volume) if you rotate the region between y = -4 and y = x2 - 4 about the x-axis.

    if you do this both ways, and get the same answer, chances are you're correct.
     
  5. Feb 5, 2012 #4
    Thank you TinyTim, I see what your saying but I am still having trouble coming up with the formula for the washer. To my understanding the formula for the washer method is:
    V = PI*rout^2*h - PI+rin^2*h
    I am still confused on what to use for rout and rin.
     
  6. Feb 5, 2012 #5
    i could be totally wrong, cause i am just learned this in class not to long ago,
    but would the outer radius be 4-0 and the inner would be 4-x^2?
     
  7. Feb 5, 2012 #6

    tiny-tim

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    hi schapman22! :smile:

    (try using the X2 and X2 buttons just above the Reply box, and the "Quick symbols" to the right :wink:)
    they are always measured from the axis

    the axis here is y = 4, so you go from there to the two curves specified in the question, the x-axis for rout, and y = x2 for rin :wink:
     
  8. Feb 5, 2012 #7
    Thank you guys! That helped a lot.
     
  9. Feb 5, 2012 #8
    basically what I did was V=∏(42)-∏(x2)2
    so dV/dx = ∏(42-x4)
    ∫dV=∏∫02(16-x4)dx
    V=[16x-(x5/5]02
    so V=25.6∏≈80.4248

    Does that look right or am I still off?
     
  10. Feb 5, 2012 #9

    tiny-tim

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    you rin is wrong :redface:
     
  11. Feb 5, 2012 #10
    Didn't you say in your last post to use y=x2 for rin?
     
  12. Feb 5, 2012 #11

    tiny-tim

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    nooo …
    (but isn't that obvious anyway from the diagram you've drawn for yourself?)
     
  13. Feb 5, 2012 #12
    Oh ok so would that make it 4-x2?
     
  14. Feb 6, 2012 #13

    tiny-tim

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    (just got up :zzz: …)

    yup! :biggrin:
     
  15. Feb 6, 2012 #14
    ugh sorry its taking me so long to get this one, but I did that and now i got a negative answer for my volume. because after imntegrating I got ∏[(x5/5) - (8x3/3)] from 0 to 2. Which is -46.9145
     
  16. Feb 6, 2012 #15

    tiny-tim

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    no, you should have got minus that…

    ∫ π[(4)2 - (4 - x2)2] dx :wink:
     
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