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Solids of revolution problem

  1. Apr 23, 2013 #1
    1. The problem statement, all variables and given/known data
    o6xrn4.jpg

    The region R shown in Fig. 1 is bounded by the line y=8(x-2), the axes, and the line y=h. Find by integration the volume formed when R is rotated through 360° about the y axis(see Fig. 2)

    A whisky glass has the shape indicated in Fig 2 where the units are centimetres. A whisky taster
    holds the glass upright and pours in whisky to a depth of 2cm. He then adds water to a further depth of 2cm. Show that if he had poured in the water first and then the whisky, each to a depth of 2cm the glass would have contained approximately 25% more whisky than by the first method.





    The attempt at a solution

    For the volume generated I got 55∏/6 cubic units.

    It's the 2nd part of the question that is puzzling me.
    Well I know realize that the cup is not uniform so.. there would be more whiskey in the 2nd method. But how do I show this mathematically?

    HELP MUCH APPRECIATED!!
     
  2. jcsd
  3. Apr 23, 2013 #2

    Dick

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    I'm confused as to how the volume you got for the first part doesn't depend on h. What did you use for h? For the second part you just need to compute the volumes of the two halves.
     
    Last edited: Apr 23, 2013
  4. Apr 23, 2013 #3
    to get the volume I did this

    ∏∫x2δy (from 0 to 2)

    = ∏[ y3/48 + y2/4 + 4y]
     
  5. Apr 23, 2013 #4
    Also how would I calculate the volumes of the halves?

    Would I call the top of the liquids y=2 and y=4 for each liquid?

    since 4cm of liquid was added? hmm and integrate x^2 = (y^2/8 + 2)

    for the certain limits?
     
  6. Apr 23, 2013 #5
    Check your volume. Firstly, you are integrating with respect to y, but you are using x=0 and x=2 as bounds. You want to convert these into y-values (Hint: The fact that your answer did not include an h when it obviously should affect your answer tells you that something is wrong). Also check your algebra. If y=8(x-2), then x= y/8+2 and x^2=y^2/64+y/2+4. You integrated the last two terms correctly, but ∫y^2/64 dy=y^3/(64*3) ≠ y^3/48
     
  7. Apr 23, 2013 #6
    Wow sorry, I get (h^3 + 48h^2 + 768h)/192 Pi cubic units
     
  8. Apr 23, 2013 #7

    Dick

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    That's right. If you evaluate that at h=2 you get the volume of the first half. h=4 gives you the sum of the first half and the second half. So subtract to get the volume of the second half.
     
  9. Apr 23, 2013 #8
    OHHHHHHH I see now because h wasn't in my volume , it made it SEEM SO CONFUSING !!!! When this question is somewhat easy... thank you guys!! I feel soo dumb for using the x's as bounds why it was with respect to y :( .
     
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